# Abelian $p$-Groups with
Minimal Characteristic Inertia

Peter V. Danchev  Institute of Mathematics and Informatics, Bulgarian Academy
of Sciences, 1113 Sofia, Bulgaria<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Patrick W. Keef  Department of Mathematics, Whitman College, Walla Walla,
WA 99362, USA<EMAIL_ADDRESS>
###### Abstract.

For Abelian $p$-groups, Goldsmith, Salce, et al., introduced the notion of
minimal full inertia. In parallel to this, we define the concept of minimal
characteristic inertia and explore those $p$-primary Abelian groups having
minimal characteristic inertia. We establish the surprising result that, for
each Abelian $p$-group $A$, the square $A\oplus A$ has the minimal
characteristic inertia if, and only if, it has the minimal full inertia. We
also obtain some other relationships between these two properties.
Specifically, we exhibit groups which do not have neither of the properties,
as well as we show via a concrete complicated construction that, for any prime
$p$, there is a $p$-group possessing the minimal characteristic inertia which
does not possess the minimal full inertia.

###### Key words and phrases:

Abelian $p$-groups, Fully inert and characteristically inert subgroups,
Minimal full inertia, Minimal characteristic inertia, Fully transitive and
transitive groups

###### 2010 Mathematics Subject Classification:

20K10

##  1\. Introduction and Conventions

All groups considered will be Abelian $p$-primary for some arbitrary but a
fixed prime $p$. Our notation and terminology will generally agree with the
classical books [6], [7], [11] and [13]. If $X$ is a set, $|X|$ will denote
its cardinality, and if $x$ is an element of a group, $|x|$ will denote its
$p$-height.

As part of a more general investigation into the concept of algebraic entropy
(see, for example, [8]), Goldsmith, Salce, et al., introduced the important
concept of minimal full inertia (see [10] and [9]). In particular, they proved
some significant results like these: the direct sums of cyclic groups have
minimal full inertia (see [9] Corollary 3.3) as well as that the class of
groups which have minimal full inertia is not closed under taking finite
direct sums (see [9] , Example 3.6). They also construct numerous groups that
either do or do not have minimal full inertia (see [9], Proposition 3.7,
Example 3.8 and Proposition 4.1, respectively).

Furthermore, the fact that direct sums of cyclic groups have minimal full
inertia was greatly generalized in [14] where a class of groups properly
containing the totally projective groups (just called “countably totally
projective”) was also shown to have this property.

These ideas utilize ideas that go back at least as far as [14]. The collection
of endomorphisms of a group are used to define when a subgroup is fully
invariant. If the collection of endomorphisms is replaced by the set of
automorphisms of the group, the parallel notion is that of a characteristic
subgroup. The central theme of this paper is to investigate how this change of
perspectives affects the property of when a group has minimal inertia.

We turn to formally defining the above terms.

######  Definition 1.1.

Two subgroups of a given group are called commensurable, provided their
intersection has finite index in each of them.

Throughout the text, if $X$ is a subgroup of a group $G$ and $\phi$ is an
endomorphism of $G$, then let $\hat{\phi}(X)=(\phi(X)+X)/X$. So, $X$ is
characteristic or fully invariant in $G$ if, and only if, $\hat{\phi}(X)=0$
for every automorphism or endomorphism, respectively, of $G$. It is obvious
and well-known that fully invariant subgroups are always characteristic, while
the converse implication fails. It is then of some interest to consider those
groups whose characteristic subgroups are fully invariant and, even more
generally, when all characteristic subgroups are commensurable with fully
invariant subgroups. Thus the objective of our work is to examine exactly such
groups. To that goal, the following notion appeared in [2].

######  Definition 1.2.

A subgroup $X$ of $G$ is characteristically inert (or fully inert) if
$\hat{\phi}(X)$ is finite for every automorphism (respectively, endomorphism)
of $G$.

It is easy to check that, if the subgroup $X$ is commensurable with a
characteristic or fully invariant subgroup of the whole group $G$, then it is
too characteristically or fully inert, respectively.

Continuing our parallel between characteristic versus fully invariant and
characteristically versus fully inert, we are led to the following idea.

######  Definition 1.3.

The group $G$ has minimal characteristic inertia (or minimal full inertia) if
every characteristically inert subgroup (respectively, fully inert subgroup)
is commensurable with some characteristic (respectively, fully invariant)
subgroup of $G$.

Our further work is organized thus: In the next section, we formulate and
prove our main assertions on groups with minimal characteristic inertia and
some closely related properties (see, e.g., Theorems 2.7 and 2.13). In the
subsequent section, we construct a series of examples illustrating these ideas
(see, e.g., Examples 3.1, 3.2 and 3.6). We finish our exposition by stating
two unresolved problems of some interest and importance which, hopefully, will
stimulate a further investigation of the subject (see Problems 1 and 2).

##  2\. Statements and Proofs

We begin with a simple but useful observation.

######  Lemma 2.1.

Suppose $X$ is a subgroup of $G$ and $\phi_{1},\dots,\phi_{k}$ are
endomorphisms of $G$ such that $\hat{\phi}_{i}(X)$ is finite for
$i=1,\dots,k$. If $\gamma=\phi_{1}+\cdots\phi_{k}$, then $\hat{\gamma}(X)$ is
finite and its cardinality satisfies the inequality

$|\hat{\gamma}(X)|\leq|\hat{\phi}_{1}(X)|\cdots|\hat{\phi}_{k}(X)|.$

###### Proof.

Simple addition gives a homomorphism

$\sigma:Z:=\hat{\phi}_{1}(X)\oplus\cdots\oplus\hat{\phi}_{k}(X)\to G/X.$

The condition $\gamma=\phi_{1}+\cdots\phi_{k}$ easily implies that
$\hat{\gamma}(X)\subseteq\sigma(Z)$. Therefore,

$|\hat{\phi}_{1}(X)|\cdots|\hat{\phi}_{k}(X)|=|Z|\geq|\sigma(Z)|\geq|\hat{\gamma}(X)|,$

as required. ∎

######  Proposition 2.2.

Suppose $G$ has the property that every endomorphism of $G$ is the sum of
automorphisms, and $X$ is a subgroup of $G$.

(a) $X$ is fully invariant if, and only if, it is characteristic.

(b) $X$ is commensurable with a fully invariant subgroup if, and only if, it
is commensurable with a characteristic subgroup.

(c) $X$ is fully inert if, and only if, it is characteristically inert.

###### Proof.

The forward direction in all these equivalences follows immediately from the
definitions.

If $\gamma$ is any endomorphism of $G$, then
$\gamma=\phi_{1}+\cdots+\phi_{k}$, where each $\phi$ is an automorphism. So,
assuming $X$ is characteristic, each $\hat{\phi}_{i}(X)=0$, which by Lemma
2.1, implies that $\hat{\gamma}(X)=0$. Letting $\gamma$ range over all
endomorphisms of $G$ gives the converse in (a). And clearly, the converse in
(b) follows from the converse in (a).

Similarly, if $X$ is characteristically inert, $\hat{\phi}_{i}(X)$ is finite
for each $i=1,\dots,k$. So, again by Lemma 2.1, $\hat{\gamma}(X)$ will always
be finite, as required. ∎

By letting $X$ range overall subgroups of $G$, we obtain the following result.

######  Corollary 2.3.

If $G$ has the property that every endomorphism of $G$ is the sum of
automorphisms, then $G$ has minimal full inertia if and only if it has minimal
characteristic inertia. And in this case, for any subgroup $X$ of $G$, the two
conditions in Proposition 2.2(b) are logically equivalent to the two
conditions in Proposition 2.2(c).

We recall the following result, due to Paul Hill:

######  Theorem 2.4 ([12], Theorem 4.2).

A totally projective $p$ group, $p\neq 2$, has the property that any
endomorphism is the sum of two automorphisms.

In [14] it was shown that any totally projective group has minimal full
inertia. Putting this together with Hill’s result leads to the next
observation.

######  Corollary 2.5.

If $G$ is a totally projective group $p$-group with $p\neq 2$, then $G$ has
minimal characteristic inertia and, for a subgroup $X$ of $G$, the four
statements in Proposition 2.2 (b) and (c) are equivalent.

The next result is often attributed to Kaplansky (see, e.g., [13] as well as
the proof of [2, Theorem 4.3]), however for the reader’s convenience and
completeness of the exposition we shall provide a simple proof.

######  Lemma 2.6.

Suppose $A$ is a group and $G=A\oplus A$. Then any endomorphism of $G$ is the
sum of at most four automorphisms.

###### Proof.

Suppose $R$ is the endomorphism ring of $A$ and $1_{A}\in R$ is the identity.
Writing homomorphisms on the right, any endomorphism $G$ can be thought of as
multiplication on the right by a 2 by 2 matrix with entries in $R$. For any
such endomorphism we have a decomposition:

$\left[\begin{matrix}\alpha&\beta\cr\delta&\epsilon\end{matrix}\right]=\left[\begin{matrix}1_{A}&\beta\cr
0&1_{A}\end{matrix}\right]+\left[\begin{matrix}-1_{A}&0\cr\delta&-1_{A}\end{matrix}\right]+\left[\begin{matrix}\alpha&1_{A}\cr
1_{A}&0\end{matrix}\right]+\left[\begin{matrix}0&-1_{A}\cr-1_{A}&\epsilon\end{matrix}\right].$

Clearly, the last four matrices represent automorphisms of $G$, as expected. ∎

In particular, this means that Corollary 2.3 applies to any such “squared”
group.

We now arrive at the following curious assertion, which actually somewhat
extends [9, Lemma 3.4] and is in parallel to a well-known result from [5].

######  Theorem 2.7.

If $A$ is a group and $G=A\oplus A$, then the following three points are
equivalent:

(a) $A$ has minimal full inertia.

(b) $G$ has minimal full inertia;

(c) $G$ has minimal characteristic inertia.

###### Proof.

The equivalence of (b) and (c) is an immediate consequence of Corollary 2.3
and Lemma 2.6. To show they are equivalent to (a) we first fix some notation.
Let $\pi_{1}:G\to A$ be the projection onto the first summand and
$\rho_{1}:A\to G$ be the obvious inclusion into the first summand; define
$\pi_{2}$ and $\rho_{2}$ similarly. We also let
$\kappa_{i}=\rho_{i}\circ\pi_{i}:G\to G$ ($i=1,2$) be the corresponding
idempotent endomorphisms and $\sigma:G\to G$ be the automorphism given by
$\sigma((a_{1},a_{2}))=(a_{2},a_{1})$.

We first assume (b) holds. To prove (a), suppose $Y$ is a fully inert subgroup
of $A$. We claim that $X:=Y\oplus Y$ is fully inert in $G$. If $\gamma$ is an
endomorphism of $G$, and for $i,j=1,2$ we let
$\gamma_{i,j}=\pi_{i}\circ\gamma\circ\rho_{j}:A\to A$, it is clear that

$\gamma(X)\subseteq[(\gamma_{1,1})(Y))+(\gamma_{1,2})(Y))]\oplus[(\gamma_{2,1})(Y))+(\gamma_{2,2})(Y))]:=L$

For $i,j\in\\{1,2\\}$, since $Y$ is fully inert in $A$, it follows that
$\hat{\gamma}_{i,j}(Y)$ will be finite. Adding in the two summands clearly
determines a surjection

$(\hat{\gamma}_{1,1}(Y)\oplus\hat{\gamma}_{1,2}(Y))\oplus(\hat{\gamma}_{2,1}(Y)\oplus\hat{\gamma}_{2,2}(Y))\twoheadrightarrow[L+X]/X.$

Therefore, $[L+X]/X$ must be finite, so that $\hat{\gamma}(X)\subseteq[L+X]/X$
is also finite.

Letting $\gamma$ range over all endomorphisms of $G$, we conclude that $X$ is
fully inert in $G$. So, by hypothesis, $X\sim W$, where $W$ is fully invariant
subgroup in $G$. Since $\kappa_{i}(W)\subseteq W$ ($i=1,2$) and
$\sigma(W)\subseteq W$, we can conclude that $W=V\oplus V$, where
$V=\pi_{i}(W)$ is fully invariant in $A$. It readily follows that $Y\sim V$,
giving the result.

For the converse, suppose $A$ has minimal full inertia. Let $X$ be any fully
inert subgroup of $G$. Note that if

$Y_{1}=\kappa_{1}(X)+\sigma(\kappa_{2}(X))\subseteq A\oplus 0,\ \ {\rm and}\ \
Y_{2}=\kappa_{2}(X)+\sigma(\kappa_{1}(X))\subseteq 0\oplus A,$

then $\sigma$ restricts to an isomorphism between $Y_{1}$ and $Y_{2}$; let
$Y=\pi_{1}(Y_{1})=\pi_{2}(Y_{2})\subseteq A$. Since
$X\subseteq\kappa_{1}(X)+\kappa_{2}(X)$, we can conclude that $X\subseteq
Y_{1}+Y_{2}=Y\oplus Y$. And since $X$ is fully inert, for $i=1,2$, both groups

$R_{i}:=\hat{\kappa}_{i}(X)=(\kappa_{i}(X)+X)/X\ \ {\rm and}\ \
S_{i}:=\hat{\sigma\circ\kappa}_{i}(X)=(\sigma(\kappa_{i}(X))+X)/X$

are finite. Since addition in each summand gives a surjective homomorphism

$(R_{1}\oplus S_{2})\oplus(R_{2}\oplus
S_{1})\twoheadrightarrow(Y_{1}+Y_{2})/X=(Y\oplus Y)/X,$

we can conclude that $(Y\oplus Y)/X$ is finite; i.e., $X\sim Y\oplus Y$. Now,
since $X$ is fully inert in $G$, $Y\oplus Y$ will be, as well. And from this,
we can easily conclude that $Y$ is fully inert in $A$. Therefore, there is a
fully invariant subgroup $V\subseteq A$ such that $Y\sim V$. It follows that
$W:=V\oplus V$ is fully invariant in $G$ and $X\sim Y\oplus Y\sim V\oplus
V=W$. Thus, $G$ has minimal full inertia, completing the proof. ∎

Note that the equivalence of conditions (a) and (b) is clearly related to
Lemma 3.4 of [9], where the groups were assumed to be fully transitive.

In regard to the last theorem and [9, Theorem 3.5], a question which directly
arises is what can be said for the group $G$ being an infinite direct sum of
copies of the group $A$?

We now consider when the one property implies the other. The following
observation shows that in one important case, the property of having minimal
characteristic inertia is stronger than having minimal full inertia.

######  Proposition 2.8.

Suppose $G$ has the property that every characteristic subgroup of $G$ is
fully invariant. If $G$ has minimal characteristic inertia, then it has
minimal full inertia.

###### Proof.

Suppose $X$ is a fully inert subgroup of $G$; we need to show that it is
commensurable with a fully invariant subgroup.

Certainly, $X$ is characteristically inert, as well. Therefore, there is a
characteristic subgroup $Y\subseteq G$ that is commensurable with $X$. By
hypothesis, we know that $Y$ will also be fully invariant in $G$, completing
the argument. ∎

The last result has an important special case.

######  Corollary 2.9.

Suppose $p$ is odd and $G$ is transitive. If $G$ has minimal characteristic
inertia, then it has minimal full inertia.

###### Proof.

It is well known that, when $p$ is odd and $G$ is transitive, a characteristic
subgroup must be fully invariant. Indeed, suppose $X$ is characteristic,
$\gamma$ is an endomorphism of $G$ and $x\in X$. The height sequences satisfy
the inequality $\|x\|\leq\|\gamma(x)\|$. By the argument of ([14], Theorem
26), this implies that $\phi(x)=y_{1}+y_{2}$, where
$\|x\|=\|y_{1}\|=\|y_{2}\|$. So, there are automorphisms $\phi_{i}$ ($i=1,2$)
such that $y_{i}=\phi_{i}(x)\in X$. Thus, $\gamma(x)=y_{1}+y_{2}\in X$, as
required. ∎

Since separable groups are always transitive, we have the following
consequence.

######  Corollary 2.10.

Suppose $p$ is odd and $G$ is separable. If $G$ has minimal characteristic
inertia, then it has minimal full inertia.

So, if we want to find a group $G$ that has minimal full inertia, but not
minimal characteristic inertia, we must have either $p=2$ or $p^{\omega}G\neq
0$. For instance, if $B=\oplus_{n\in\mathbb{N}}\mathbb{Z}_{2^{n}}$ is the
standard direct sum of cyclic $2$-groups, then it is known with the aid of [9]
that both $B$ and $\overline{B}$ has minimal full inertia. So, if either fails
to have the minimal characteristic inertia, then we would have our wanted
example.

We end this discussion with one last observation which expresses the question
entirely in terms of minimal characteristic inertia.

######  Corollary 2.11.

The group $A$ has minimal full inertia, but not minimal characteristic inertia
if, and only if, $G=A\oplus A$ has minimal characteristic inertia, but $A$
itself does not have minimal characteristic inertia.

Using techniques from [14], our next goal is to verify that every separable
group is a summand of a group with both minimal characteristic inertia and
minimal full inertia. The following easy result from that work will be useful.

######  Lemma 2.12 ([14], Lemma 2.3).

Suppose $L$ is a group and $B,C$ are subgroups of $L$.

(a) If $B\subseteq C$ and $k<\omega$, then $B\sim C$ if, and only if,
$B[p^{k}]\sim C[p^{k}]$ and $p^{k}B\sim p^{k}C$.

(b) If $n<\omega$ and $B[p^{n}]\subseteq C[p^{n}]$, then $B[p^{n}]\sim
C[p^{n}]$ if, and only if, $(p^{k}B)[p]\sim(p^{k}C)[p]$ for all $k<n$.

If $H$ is a group, then any endomorphism or automorphism on $H$ restricts to
an endomorphism or automorphisms, respectively, on $p^{\omega}H$. The
following construction applies when all of these restricted automorphisms are
simply multiplications, which implies that every subgroup of $p^{\omega}H$
will be characteristic in $H$.

######  Theorem 2.13.

Suppose $H$ is a group such that $p^{\omega+1}H=0$ and every automorphism of
$H$ when restricted to $p^{\omega}H$ is multiplication by some (non-zero)
element of $\mathbb{Z}_{p}$. Then there is a separable group $K$ such that the
group $G=H\oplus K$ has minimal characteristic inertia.

###### Proof.

If $H$ is bounded, then $H\oplus H$ is a direct sum of cyclic groups, and
hence this square has minimal full inertia (cf. [9], [14]). Therefore,
according to Theorem 2.7, it also has minimal characteristic inertia, and thus
we can just let $K=H$. So, without loss of generality, we may assume that $H$
is unbounded and, in particular, that it is infinite.

Let $\kappa>|H|^{\aleph_{0}}$ be some cardinal with
$\kappa^{\aleph_{0}}=\kappa$. Next, let $B_{1}$ be a direct sum of cyclic
groups all of whose Ulm invariants (at finite ordinals) are equal to $\kappa$;
so the torsion completion, $\overline{B}_{1}$, will also have cardinality
$\kappa$. Let $B_{2}$ be a direct sum of cyclic groups, all of whose (again,
finite) Ulm invariants equal $\aleph_{1}$. We let $K=\overline{B}_{1}\oplus
B_{2}$ and show that $G:=H\oplus\overline{B}_{1}\oplus B_{2}=H\oplus K$ has
the required properties.

Define an ordered set $\mathcal{O}$ as follows: $\mathcal{O}$ is the union of
$\omega=\\{0,1,2,\dots\\}$, and $\mathcal{S}$, the set of subgroups of
$M:=p^{\omega}H$. We identify the symbol $\infty$ with the zero subgroup. The
elements of $\omega\subseteq\mathcal{O}$ will (naturally) be called integers
and the others will (again, naturally) be called subgroups. We define
$\alpha<\beta$ as follows: if $\alpha$ and $\beta$ are integers and this is
true in the usual sense; if $\alpha$ is an integer and $\beta$ is a subgroup
and; if $\beta=\infty$. If $\alpha\in\mathcal{S}$, we agree that
$\alpha+1=\infty$. If $E\in\mathcal{S}$, we define $p^{E}G=E$; if $E$ is
$\infty=\\{0\\}$, this agrees with the usual definition. For each
$\beta\in\mathcal{O}$ let $S_{\beta}=(p^{\beta}G)[p]$. If
$\beta\in\mathcal{O}$ and $X$ is a subgroup of $G$, let $X(\beta)=X\cap
p^{\beta}G$ and $X_{\beta}=X/X(\beta)$; if $\beta$ is an integer or $\infty$,
these agree with the usual definitions. Note that $X(M)$ agrees with the more
usual $X(\omega)$ and $X_{M}$ with the more usual $X_{\omega}=X/X(\omega)$.

We will show that any characteristically inert subgroup is commensurable with
a characteristic subgroup of a particular form. If
$\alpha_{0}<\alpha_{1}<\alpha_{2}<\cdots$ are in $\mathcal{O}$ and
$\overline{\alpha}=(\alpha_{0},\alpha_{1},\dots)$, let $G(\overline{\alpha})$
be the set of $x\in G$ such that for all $n<\omega$, $p^{n}x\in
p^{\alpha_{n}}G$. So, for example, if $E\in\mathcal{O}$ is a subgroup, then
$G(1,3,E,\infty,\infty,\dots)$ is the collection of all $x$ such that $|x|\geq
1$, $|px|\geq 3$, $p^{2}x\in E$ and $p^{3}x=0$.

Since all $E\in\mathcal{S}$ are characteristic, it is relatively
straightforward to verify that any subgroup of the form $G(\overline{\alpha})$
will be characteristic, as well.

Our arguments will be based on the following technical observation which will
provide us with an abundance of automorphisms to use. It will play a role
similar to that of Lemmas 2.15 and 2.16 in [14].

######  Claim 2.14.

Suppose $X\subseteq G$ is a subgroup, $G=V\oplus W$ is a decomposition and
$\gamma:V\to W$ is a homomorphism. Let $\phi_{\gamma}$ be the automorphism of
$G$ given by $\phi_{\gamma}(v+w)=v+\gamma(v)+w$ for all $v\in V$, $w\in W$.

(a) If for all $j<\omega$, $v_{j}\in V\cap X$ with $w_{j}:=\gamma(v_{j})\in
W\subseteq G$ and the elements $w_{j}+X\in G/X$ are distinct, then
$\hat{\phi}_{\gamma}(X)$ is infinite, so that $X$ is not characteristically
inert.

(b) If $\\{n_{j}\\}_{j<\omega}$ is a strictly increasing sequence of integers
and for all $j<\omega$, $v_{j}\in V\cap X$ with
$w_{j}:=\gamma(p^{n_{j}}v_{j})\in W[p]\setminus p^{n_{j}}X$, then
$\hat{\phi}_{\gamma}(X)$ is infinite, so that $X$ is not characteristically
inert.

Clearly, $\phi_{\gamma}$ is an automorphism (its inverse is $\phi_{-\gamma}$.)

Regarding (a), consider the elements of $\hat{\phi}_{\gamma}(X)$ of the form

$\phi_{\gamma}(v_{j})+X=v_{j}+w_{j}+X=w_{j}+X.$

Since we are assuming these are distinct, the result follows.

For (b), suppose $j>k$ and $\phi_{\gamma}(v_{n_{j}})$ and
$\phi_{\gamma}(v_{n_{k}})$ represent the same element of
$\hat{\phi}_{\gamma}(X)$. So
$\phi_{\gamma}(v_{n_{j}})=v_{n_{j}}+\gamma(v_{n_{j}})$ and
$\phi_{\gamma}(v_{n_{k}})=v_{n_{k}}+\gamma(v_{n_{k}})$ are congruent modulo
$X$. That is, $\gamma(v_{n_{j}})\in W$ and $\gamma(v_{n_{k}})\in W$ are
congruent modulo $X$. Therefore,

$w_{j}=w_{j}-p^{n_{j}-n_{k}}w_{k}=p^{n_{j}}(\gamma(v_{n_{j}})-\gamma(v_{n_{k}}))\in
p^{n_{j}}X$

contrary to hypothesis.

Throughout, we will let $X$ be some characteristically inert subgroup of $G$;
our goal is to construct a sequence $\overline{\alpha}$ such that $X\sim
G(\overline{\alpha})$.

######  Claim 2.15.

One of three things happens:

(a) $X$ is finite. In this case, $X\sim 0$ and we let
$\alpha_{X}=\infty\in\mathcal{O}$.

(b) $X$ is infinite and $X_{\omega}$ is finite. In this case, $X\sim
X(\omega)\in\mathcal{S}\subseteq\mathcal{O}$ and we let
$\alpha_{X}=X(\omega)$.

(c) $X_{\omega}$ is infinite. In this case there is an integer
$\alpha_{X}\in\mathcal{O}$ such that $S_{\alpha_{X}}\sim X(\alpha_{X})[p]$ and
$X\sim X(\alpha_{X})$, i.e., $S_{\alpha_{X}}/X(\alpha_{X})[p]$ and
$X_{\alpha_{X}}$ are finite. (c.f., [14], Lemma 2.7)

Clearly, exactly one of the three conditions holds, and the conclusions in (a)
and (b) are routine. So assume $X_{\omega}$ is infinite. Note that
$|X(\omega)|\leq|H|<\kappa$. We first show that the cardinality of
$|X_{\omega}|=|X|=\kappa$. Otherwise, there is a decomposition
$\overline{B}_{1}=\overline{B}_{3}\oplus W$, where $\overline{B}_{3}$ and $W$
are copies of $\overline{B}_{1}$, and in the decomposition
$G=H\oplus\overline{B}_{3}\oplus W\oplus B_{2}$, we have $X\subseteq
V:=H\oplus\overline{B}_{3}\oplus 0\oplus B_{2}.$

If $\overline{V}_{\omega}$ is the torsion-completion of $V_{\omega}$, then
there is an isomorphism $\overline{V}_{\omega}\to W$. Let $\gamma$ be the
composition of the natural map $V\to V_{\omega}\to\overline{V}_{\omega}$ with
this isomorphism. If we choose any sequence of element
$\\{v_{j}\\}_{j<\omega}$ of $X$ that are pairwise not congruent modulo
$X(\omega)$, then the elements $w_{j}=\gamma(v_{j})$ are distinct elements of
$W$. So if $j<k$, then $w_{j}-w_{k}$ is a non-zero element of $W$, so in
particular, it is not in $X\subseteq V$. Since $X$ is characteristically
inert, this contradicts Claim 2.14(a), showing that $X_{\omega}$ must have
cardinality $\kappa$, as claimed.

There is a valuated decomposition $X[p]\cong X_{h}\oplus X(\omega)$. Since
$|X[p]|=\kappa$ and $|X(\omega)|<\kappa$, $X_{h}$ must have cardinality
$\kappa$. Now, $X_{h}$ is a separable valuated vector space, so it has a basic
subspace. If each of its (finite) Ulm invariants were finite, then $X_{h}$
(which is contained in a completion of this basic subspace) would also have
cardinality at most $2^{\aleph_{0}}<\kappa$ contrary to what we just
established. Choose the integer $\alpha_{X}$ minimal with the property that
the $\alpha_{X}$th Ulm invariant of $X_{p}$ is infinite. For simplicity, we
will just denote $\alpha_{X}$ by $\alpha$; we need to show
$S_{\alpha}/X(\alpha)[p]$ and $X/X(\alpha)$ are finite.

Suppose that $S_{\alpha}/X(\alpha)[p]$ is infinite. Since the $\alpha$th Ulm
invariant of $X_{h}$ is infinite, we can construct a decomposition $G=V\oplus
W$, where $V\cong\oplus_{j<\omega}\langle
v_{j}^{\prime}\rangle\cong\oplus_{j<\omega}\mathbb{Z}_{p^{\alpha+1}}$ such
that $V[p]\subseteq X$. Since $S_{\alpha}/X(\alpha)[p]$ is infinite, we can
also find elements $\\{w_{j}\\}_{j<\omega}$ of $W(\alpha)[p]$ that are
pairwise not congruent modulo $X(\alpha)[p]$, and hence not congruent modulo
$X$. If $v_{j}=p^{\alpha}v_{j}^{\prime}\in V[p]\subseteq X$, there is clearly
a homomorphism $\gamma:V\to W$ such that $\gamma(v_{j})=w_{j}$ for $j<\omega$.
So by Claim 2.14(a) we can conclude that $X$ is not characteristically inert,
contrary to hypothesis.

We now need to show that $X/X(\alpha)$ is finite, so assume it is not. Let
$\beta<\alpha$ have the property that $X(\beta)/X(\beta+1)$ is infinite; such
a value must clearly exist. Let $\\{v_{j}\\}_{j<\omega}\subseteq X(\beta)$ be
linearly independent modulo $X(\beta+1)$.

There is clearly a decomposition $G=V\oplus W$ where
$\\{v_{j}\\}_{j<\omega}\subseteq V$; and the $\beta$th Ulm invariant of $W$ is
infinite.

Note that $V_{\beta+1}=V/p^{\beta+1}V$ is $p^{\beta+1}$ bounded. This easily
implies that there is a decomposition $V_{\beta+1}=Y\oplus Y^{\prime}$ for
which $Y[p]=\langle v_{j}+p^{\beta+1}V:j\in\omega\rangle$. Since $\alpha$ is
the first infinite Ulm invariant of $X_{h}$, we can conclude that the image of
$X(\beta)[p]/X(\beta+1)[p]\to S_{\beta}/S_{\beta+1}$ is finite. On the other
hand, the image of $p^{\beta}W[p]/p^{\beta+1}W[p]\to S_{\beta}/S_{\beta+1}$ is
infinite. Therefore, there are elements $w_{j}\in(p^{\beta}W)[p]$ that are
pairwise not congruent modulo $X(\beta)[p]$, and hence pairwise not congruent
modulo $X$.

Clearly, $v_{j}+p^{\beta+1}V\mapsto w_{j}$ can be extended first to a
homomorphism $\lambda:Y\to W$ and then, by setting $\lambda(Y^{\prime})=0$, to
a homomorphism $\lambda:V_{\beta+1}\to W$. Let $\gamma$ be the composition
$V\to V_{\beta+1}\to W.$ As before, the elements
$\gamma(v_{j})=\lambda(v_{j}+p^{\beta+1}V)=w_{j}\in\lambda(Y)$ are not
congruent modulo $X$. Since $X$ is characteristically inert, this contradicts
Claim 2.14(a), completing the proof of Claim 2.15.

We now define an entire sequence $\overline{\alpha}$ for $X$ (c.f., [14],
Theorem 2.11). For each $n<\omega$, $p^{n}X$ will also be characteristically
inert in $G$ and we let $\alpha_{n}=\alpha_{p^{n}X}$ (in fact, this is just a
simplification of notation). We first verify that
$\alpha_{n}+1\leq\alpha_{n+1}$. If $\alpha_{n}=\infty$, then $p^{n}X$ is
finite, which implies that $p^{n+1}X$ is finite, so that
$\alpha_{n+1}=\infty$, as required. Similarly, if $\alpha_{n}$ is a subgroup,
then $(p^{n}X)_{\omega}$ is finite. Since
$(p^{n}X)(\omega)\subseteq(p^{n}X)[p]$ this implies that
$p^{n+1}X\cong(p^{n}X)/(p^{n}X)[p]$ is also finite, Therefore, again,
$\alpha_{n+1}=\infty$, as required.

Finally, suppose $\alpha_{n}$ is an integer. If $\alpha_{n+1}$ is a subgroup,
then the result is trivial. So assume $\alpha_{n}$, $\alpha_{n+1}$ are
integers. Since $p^{n}X/p^{n}X(\alpha_{n})$ is finite and multiplication by
$p$ gives a surjection onto $p^{n+1}X/p^{n+1}X(\alpha_{n}+1)$, the latter
quotient is also finite. And since $\alpha_{n+1}$ is the largest integer such
that $p^{n+1}X/p^{n+1}X(\alpha_{n+1})$ is finite, we must have
$\alpha_{n}+1\leq\alpha_{n+1}$.

######  Claim 2.16.

(cf. [14], Theorem 2.11(4)) For all $n<\omega$:

$X\sim X_{n}:=X\cap
G(\alpha_{0},\alpha_{1},\alpha_{2},\dots,\alpha_{n-1},\alpha_{n-1}+1,\alpha_{n-1}+2,\dots).$

If $n=1$, this just says $X\sim X(\alpha_{X})$, which we know is true. We now
show that for $n\geq 1$ we have $X_{n}\sim X_{n+1}$, which will complete the
argument. Consider the homomorphism $\nu$ given by the composition

$X_{n}\,\smash{\mathop{\longrightarrow}\limits^{\times
p^{n}}}\,p^{n}X_{n}\subseteq p^{n}X\to(p^{n}X)/(p^{n}X)(\alpha_{p^{n}X}).$

It readily follows that $X_{n+1}$ is the kernel of $\nu$. And since we know
this last group is finite, we have $X_{n}\sim X_{n+1}$, as desired.

Let $A=G(\overline{\alpha})$. Our goal is to show $X\sim A$.

######  Claim 2.17.

(cf. [14], Theorem 2.11(5)) If $n<\omega$, then $X[p^{n}]\sim A[p^{n}]$.

We know that $X_{n}[p^{n}]\subseteq A[p^{n}]$. For all $0\leq k<n$ we have
$p^{k}X_{n}[p]\sim p^{k}X[p]\sim S_{\alpha_{n}}=p^{k}A[p]=p^{k}(A[p^{n}])[p]$.
So by Lemma 2.12(b), $X_{n}[p^{n}]\sim A[p^{n}]$, which implies that
$X[p^{n}]\sim A[p^{n}]$.

Note that if $\alpha_{n}=\infty$ for some $n<\omega$, then it follows that
$X\sim X_{n+1}=X_{n+1}[p^{n}]\sim A[p^{n}]=A$ and the result follows.

So, from here on, we may assume $\alpha_{n}$ is an integer for all $n$.

######  Claim 2.18.

There is an $N<\omega$ such that for every $n\geq N$ we have $p^{n}X\subseteq
p^{\alpha_{n}}G$:

We suppose this fails and derive a contradiction. So, there must exist a
strictly ascending sequence $n_{0}<n_{1}<n_{2}<\cdots$ and elements
$v_{n_{j}}\in X$ such that $|p^{n_{j}}v_{n_{j}}|<\alpha_{n_{j}}$.

Find a decomposition $\overline{B}_{1}=\overline{B}_{3}\oplus W$ such that for
all $j<\omega$ we have $v_{n_{j}}\in V:=H\oplus\overline{B}_{3}\oplus 0\oplus
B_{2}$ and each Ulm factor of $W$ is infinite.

For $j<\omega$, let $v_{n_{j}}^{\prime}=v_{n_{j}}+p^{\omega}V\in V_{\omega}$;
so each $p^{n_{j}}v^{\prime}_{n_{j}}\neq 0$. After possibly restricting to a
subsequence, we may assume that for all $j<\omega$, if $n_{j}\leq k\in\omega$
and $p^{k}v^{\prime}_{n_{j}}\neq 0$, then
$|p^{k}v_{n_{j}}^{\prime}|<|p^{n_{j+1}}v_{n_{j+1}}^{\prime}|$. Using this
condition, it can be checked that there is a basic subgroup of $V_{\omega}$
the form $C=\oplus_{j<\omega}C_{j}$ such that for each $j<\omega$ we have
$p^{n_{j}}v_{n_{j}}^{\prime}\in C_{j}$.

For each $j<\omega$, if $\beta=|p^{n_{j}}v^{\prime}_{n_{j}}|<\alpha_{n_{j}}$,
then the image of

$p^{n_{j}}X(\beta)[p]/p^{n_{j}}X(\beta+1)[p]\to S_{\beta}/S_{\beta+1}$

is finite. On the other hand, the image of

$p^{\beta}W[p]/p^{\beta+1}W[p]\to S_{\beta}/S_{\beta+1}$

is infinite. So we can find a $w_{j}\in W[p]\setminus p^{n_{j}}X$ with
$|w_{j}|=\beta=|p^{n_{j}}v^{\prime}_{n_{j}}|$. The assignment
$p^{n_{j}}v^{\prime}_{n_{j}}\mapsto w_{j}$ clearly extends to a homomorphism
$C_{j}\to W$. Summing extends these all to a homomorphism $C\to W$. And since
$C$ is pure in $V_{\omega}$ and $W$ is pure-injective for torsion groups, this
extends to a homomorphism $V_{\omega}\to W$. We let $\gamma:V\to W$ be the
composition $V\to V_{\omega}\to W$, so that
$\gamma(p^{n_{j}}v_{n_{j}})=w_{n_{j}}$.

Since $X$ is characteristically inert, this contradict Claim 2.14(b) and
establishes the claim.

It follows from this that $X\sim X_{N}\subseteq A$. So, replacing $X$ by
$X_{N}$ there is no loss of generality in assuming that $X\subseteq A$.

######  Claim 2.19.

There is an $N<\omega$ such that for all $n\geq N$ we have
$(p^{n}X)[p]=S_{\alpha_{n}}=(p^{n}A)[p]$.

If the claim fails, then we can find a strictly increasing sequence of
elements of $\omega$, $n_{0}<n_{1}<n_{2}<\cdots$ and elements
$w_{n_{j}}\in(p^{n_{j}}A)[p]\setminus(p^{{n_{j}}}X)$. There is clearly a
decomposition $B_{2}=B_{4}\oplus V$ such that every Ulm factor of $V$ is
infinite and for each $j<\omega$ we have $w_{n_{j}}\in
H\oplus\overline{B}_{1}\oplus B_{4}\oplus 0:=W$.

Since by Claim 2.17, $X[p^{n_{j}+1}]\sim A[p^{n_{j}+1}]$, we can conclude that

$(X\cap V)[p^{n_{j}+1}]\sim V(\overline{\alpha})[p^{n_{j}+1}].$

Consequently,

$p^{n_{j}}((X\cap V)[p^{n_{j}+1}])\sim
p^{n_{j}}(V(\overline{\alpha})[p^{n_{n}+1}])=(p^{\alpha_{n_{j}}}V)[p].$

Therefore, we can find $v_{j}\in X\cap V$ such that $p^{n_{j}}v_{j}\in V[p]$
and $|p^{n_{j}}v_{j}|=\alpha_{n_{j}}$.

Again, $|p^{n_{j}}v_{j}|$ goes to infinity as $j$ does. So if we start with a
decomposition of $V$ into cyclic summands, then after possibly restricting to
a subsequence, we may assume that the supports of the $p^{n_{j}}v_{j}$ in this
decomposition are disjoint. Therefore, we can find a decomposition of $V$ into
$\oplus_{j<\omega}V_{j}$ such that $p^{n_{j}}v_{j}\in V_{j}$.

As before, the assignment $p^{n_{j}}v_{n_{j}}\mapsto w_{j}$ clearly extends to
a homomorphism $V_{j}\to W$. Summing extends these all to a homomorphism
$\gamma:V\to W$ with $\gamma(p^{n_{j}}v_{n_{j}})=w_{n_{j}}$. Since $X$ is
characteristically inert, this contradicts Claim 2.14(b), proving the claim.

One sees that Claim 2.19 implies that $p^{N}X$ is pure in $p^{N}A$, and since
$(p^{N}X)[p]=S_{\alpha_{N}}=(p^{N}A)[p]$ this means that $p^{N}X=p^{N}A$. And
since $X\subseteq A$, and by Claim 2.17 $X[p^{N}]\sim A[p^{N}]$, by Lemma
2.12(a) we can conclude that $X\sim A$, as required. ∎

If $H$ is separable, then it clearly satisfies the hypotheses of Theorem 2.13.
And in that proof, any subgroup that is characteristically inert (and hence
also any subgroup that is fully inert) is commensurable with the fully
invariant subgroup $G(\overline{\alpha})$. We, therefore, have the following
consequence.

######  Corollary 2.20.

Any separable group is a summand of a separable group with minimal
characteristic inertia as well as with minimal full inertia.

Observe that the last result holds for the prime $2$, even though a separable
2-group may have characteristic subgroups that fail to be fully invariant.

If $p$ is odd, we already showed in Corollary 2.10, listed above, that any
separable group that has minimal characteristic inertia also has minimal full
inertia. However, the subsequent Example 3.2 will demonstrate in the sequel
that this does not extend to non-separable groups.

##  3\. Examples and Problems

The assertions quoted above allow us to extract an example of a group that has
neither minimal characteristic inertia nor minimal full inertia.

######  Example 3.1.

For any prime $p$ there exists a $p$-group that has neither minimal full
inertia nor minimal characteristic inertia.

###### Proof.

Let $B=\oplus_{n<\omega}\mathbb{Z}_{p^{n+1}}$ be the standard direct sum of
cyclic groups and let $\overline{B}$ be its torsion completion. In [9] it was
noted that $B\oplus\overline{B}$ does not have minimal full inertia. In that
proof, it was shown that $X=0\oplus\overline{B}[p]$ is fully inert but not
commensurable with a fully invariant subgroup.

We claim that $B\oplus\overline{B}$ does not have minimal characteristic
inertia, as well. First, since $X$ is fully inert, it follows that it is
characteristically inert. So it will suffice to verify that $X$ is not
commensurable with a characteristic subgroup.

Let $k<\omega$ be arbitrary. Clearly, there is a decomposition

$B=B_{1}\oplus
B_{2}:=\left(\oplus_{n<k}\mathbb{Z}_{p^{n+1}}\right)\oplus\left(\oplus_{k\leq
n}\mathbb{Z}_{p^{n+1}}\right)$

Define $\phi_{k}$, an automorphism of $G$, using the decomposition

$G=(B_{1}\oplus\overline{B}_{2})\oplus(B_{1}\oplus\overline{B}_{2})$

as follows. If $x_{1},x_{1}^{\prime}\in B_{1}$ and
$x_{2},x_{2}^{\prime}\in\overline{B}_{2}$, let

$\phi_{k}((x_{1},x_{2},x_{1}^{\prime},x_{2}^{\prime}))=(x^{\prime}_{1},x_{2},x_{1},x_{2}^{\prime}).$

It is now easy to verify that $\hat{\phi}_{k}$ is isomorphic to the socle
$B_{1}[p]$, which has order $p^{k}$.

Letting $k$ vary, we can infer that $X$ is not uniformly characteristically
inert (where that term is defined as in the case of uniformly fully inert by
using automorphisms instead of endomorphisms – see, for instance, [9]).
Consequently, $X$ is not commensurable with a characteristic subgroup, as
stated. ∎

We observed in Proposition 2.8, that if the group $G$ has the property that
all of its characteristic subgroups are fully invariant, then if $G$ has
minimal characteristic inertia, then it must have minimal full inertia. So, if
$G$ is a group with minimal characteristic inertia, but does not have minimal
full inertia, then it must have characteristic subgroups that are not fully
invariant. The following shows that this can happen.

######  Example 3.2.

There exists a group with minimal characteristic inertia that fails to have
minimal full inertia. In fact, for any prime $p$ (even for $p=2$), there is a
group of length $\omega+1$ with these properties.

###### Proof.

Let $R$ be the polynomial ring $\mathbb{Z}_{p}[z]$. Next, let $M$ be the
$R$-module $\mathbb{Z}_{p}[z,z^{-1}]=\\{z^{k}r(z):k\in\mathbb{Z},r(z)\in R\\}$
(so $M$ is a submodule of the quotient ring of $R$). By a classical result of
Corner ([4]), there is a group $H$ such that $p^{\omega}H=M$ and the
endomorphism ring on $H$ restricts to $R$ on $M$ in such a way that the
automorphism group of $H$ restricts to the units of $R$, i.e., the non-zero
elements of $\mathbb{Z}_{p}$.

Let $G=H\oplus K$ be as in Theorem 2.13, so that $G$ has minimal
characteristic inertia.

We, therefore, need to show that $G$ does not have minimal full inertia. Let

$E=\langle 1,z^{-1},z^{-2},z^{-3},\dots\rangle\subseteq M=p^{\omega}G.$

Any endomorphism of $G$ restricted to $M$ is simply multiplication by some
polynomial

$r(x)=a_{0}+a_{1}z+\cdots+a_{k}z^{k}.$

If for $i=0,\dots,k$, $\phi_{i}$ is the endomorphism given by multiplication
by $z^{i}$, then it easily follows that $\hat{\phi}_{i}(E)$ is finite (in
fact, it is naturally isomorphic to the $\mathbb{Z}_{p}$ span of
$z,z^{2},\dots,z^{i}$, so that it has dimension $i$). It follows from Lemma
2.1 that $\hat{r}(E)$ is also finite. Thus, $E$ has full inertia. On the other
hand, since $\hat{\phi}_{k}(E)\cong\mathbb{Z}_{p}^{k}$ for each $k$, $E$ does
not have uniform full inertia as defined in [9], so it is not commensurable
with a fully invariant subgroup. Therefore, $G$ does not have minimal full
inertia, as asserted. ∎

By combining Corollary 2.20 and Example 3.2, we observe that the class of
groups with minimal characteristic inertia is not closed under taking direct
summands.

In Example 3.1 it was noted that $B\oplus\overline{B}$ has neither minimal
full inertia nor minimal characteristic inertia. This immediately implies that
neither class is closed under taking direct sums. On the other hand, in
Theorem 2.7 it was observed that the group $A$ has minimal full inertia if,
and only if, its “square”, $G=A\oplus A$, has this property. The last result
immediately implies that this does not hold for the property of having minimal
characteristic inertia and thus we arrive at the following consequence.

######  Corollary 3.3.

There is a group $A$ with minimal characteristic inertia, but the square
$G=A\oplus A$ does not have minimal characteristic inertia.

###### Proof.

Let $A$ be any group with minimal characteristic inertia, but not minimal full
inertia. ∎

As already noticed above, many examples of groups without minimal full inertia
have been given, i.e., groups $G$ with a fully inert subgroup $X$ that is not
commensurable with a fully invariant subgroup (for more details, we refer to
[9]). The following, however, gives an example where $X$ is actually
characteristic.

######  Example 3.4.

There is a $2$-group $G$ that has a characteristic subgroup $X$ which is fully
inert, but not commensurable with a fully invariant subgroup.

###### Proof.

Suppose $U=\mathbb{Z}_{2}\oplus\mathbb{Z}_{8}$ and $S$ is the subgroup
$\langle(1,2)\rangle=\\{(0,0)$, $(1,2)$, $(0,4)$, $(1,6)\\}$; actually,
Kaplansky noted in [13] that $S$ is characteristic in $U$, but not fully
invariant.

We now let $U^{\mathbb{N}}=\oplus_{i\in\mathbb{N}}U_{i}$, where each $U_{i}$
is an isomorphic copy of $U$. Consider the collection $R$ of endomorphisms of
$U^{\mathbb{N}}$ of the form $\phi:=(\phi_{i})_{i\in\mathbb{N}}$, where
$\phi_{i}:U_{i}\to U_{i}$ and such that, for some $N\in\mathbb{N}$ and
$m\in\mathbb{Z}$, the map $\phi$ restricted to $\oplus_{i\geq N}U_{i}\subseteq
U^{\mathbb{N}}$ is simply multiplication by $m$.

Using again the classical result of Corner ([4], Theorem 10.2), we can
construct a group $G$ with $2^{\omega}G=U^{\mathbb{N}}$ such that the
endomorphisms of $G$, when restricted to $U^{\mathbb{N}}$, are precisely the
ring $R$ and whose automorphisms, again when restricted to $U^{\mathbb{N}}$,
are precisely the units of $R$.

Let $X:=S^{\mathbb{N}}=\oplus_{i\in\mathbb{N}}S_{i}$, where $S_{i}$ in $U_{i}$
is simply $S$ in this copy of $U$.

Since $S$ is characteristic in $U$ and all of the endomorphism in $R$ respect
the given coordinate structure, it immediately follows that $X$ is actually
characteristic in $G$.

We now observe that $X$ is fully inert. If $\phi\in R$ and we restrict $\phi$
to $U^{\mathbb{N}}$ (using the same letter), we can by construction find an
$N\in\mathbb{N}$ and $m\in\mathbb{Z}$ such that $\phi$ restricted to
$\oplus_{i\geq N}U_{i}\subseteq U^{\mathbb{N}}$ is simply multiplication by
$m$. It follows that $\phi(\oplus_{i\geq N}S_{i})\subseteq X$. Therefore,

$\hat{\phi}(X)=[\phi(\oplus_{i<N}S_{i})+X]/X.$

Since $\phi(\oplus_{i<N}S_{i})$ is finite, it readily follows that
$\hat{\phi}(X)$ is, as well, so that $X$ is fully inert.

On the other hand, consider the idempotent homomorphism $\kappa:U\to U$ given
by $\kappa((a,b))=(a,0)$. So, $(1,2)\in S$, but $\kappa((1,2))=(1,0)\not\in
S$. For each $n\in\mathbb{N}$, consider the endomorphism $\phi_{n}$ of $G$
that restricts to $\kappa$ on each $U_{i}$ for $i\leq n$, and is 0 on each
$U_{i}$ for $i>N$. It is easy to see that $|\hat{\phi}_{n}(X)|=2^{n}$ for all
$n\in\mathbb{N}$. Consequently, $X$ is not uniformly fully inert (see [9]).
But this means that it is not commensurable with a fully invariant subgroup,
either. ∎

In contrast to the statement of the previous example, we may now a little
refine it in order to exhibit a $2$-group having a characteristic subgroup
that is not fully inert, thus somewhat extending the ingenious example from
[4].

######  Example 3.5.

There is a $2$-group $G$ having a characteristic subgroup which is not a fully
inert subgroup.

###### Proof.

Retain the group $U$ and its characteristic subgroup $S$ from the last
example. If $R$ is the endomorphism ring of $U$, then associating each
$\phi\in R$ with the endomorphism $U^{\mathbb{N}}\to U^{\mathbb{N}}$ given by
applying $\phi$ to each coordinate individually, we can view $R$ as acting on
$U^{\mathbb{N}}$.

We again use Corner’s Theorem to produce a group $G$ with
$2^{\omega}G=U^{\mathbb{N}}$ as above. Again, the same $X$ as defined in our
previous example is easily seen to be characteristic, but not fully inert. ∎

In [4], Corner produced an example of group that was transitive, but not fully
transitive. For $p\not=2$, any transitive group is always fully transitive, so
his example was necessarily a $2$-group. In our final result, we amend his
construction to show there is such a group with the property that all of its
characteristic subgroups are fully invariant.

######  Example 3.6.

There is a transitive, non-fully transitive $2$-group, all of whose
characteristic subgroups are fully invariant.

###### Proof.

Let $G$ be Corner’s original construction of a transitive, but not fully
transitive, group. It has the specific property that $2^{\omega}G$ is finite;
in fact, it follows from our previous examples that
$2^{\omega}G=\mathbb{Z}_{2}\oplus\mathbb{Z}_{8}=U$. In addition, if $R$ is the
collection of endomorphisms of $G$ in Corner’s example restricted to
$2^{\omega}G$, then $R$ is the subring of the endomorphisms of $2^{\omega}G$,
generated by the collection of automorphism of $2^{\omega}G$, and any such
automorphism of $2^{\omega}G$ extends to an automorphism of $G$.

Since the transitivity or full transitivity of $G$ depends only on how the
endomorphism ring acts on $2^{\omega}G$, adding a separable summand to $G$
will always give another group with the same properties. Therefore, without
loss of generality, we may assume that for every $n\in\mathbb{N}$, the $n$th
Ulm invariant of $G$ is at least $2$.

We need to show that an arbitrary characteristic subgroup $X$ is fully
invariant. So, it suffices to show that, if $\gamma$ is any endomorphism of
$G$ and $x$ is any element of $X$, then $\gamma(x)\in X$.

To that aim, suppose first that $2^{k}$ is the order of $x+2^{\omega}G\in
G/2^{\omega}G=G_{1}$; so $|2^{j}x|<\omega$ if, and only if, $j<k$.

We know that $\gamma$ restricted to $2^{\omega}G$ is a sum of automorphisms of
$2^{\omega}G$, say $\phi_{1}+\cdots+\phi_{m}$, and that all of these maps can
be extended to automorphisms of $G$. Since $2^{k}x\in 2^{\omega}G$, we have

$\gamma(2^{k}x)=\phi_{1}(2^{k}x)+\cdots+\phi_{m}(2^{k}x).$

If $\gamma^{\prime}=\gamma-(\phi_{1}+\cdots+\phi_{m})$, then since $X$ is
characteristic, $\gamma(x)\in X$ if, and only if, $\gamma^{\prime}(x)\in X$.
Replacing $\gamma$ by $\gamma^{\prime}$, there is no loss of generality in
assuming that $\gamma(2^{k}x)=0$.

Let $z=\gamma(x)$; so $2^{k}z=0$ and we clearly have an inequality of height
sequences $\|x\|\leq\|z\|$. Set $y_{j}=2^{j}x$ and $y^{\prime}_{j}=-2^{j}x$
for all $j\geq k$. By a technique that goes back to Theorem 26 of [13] (which
also appears in Theorem 2.13 of [3] and was mentioned earlier in this work),
we can induct backwards from $j=k$ down to $j=0$ to construct
$y_{k-1},\dots,y_{1},y_{0}$ and
$y^{\prime}_{k-1},\dots,y^{\prime}_{1},y^{\prime}_{0}$ such that, for each
$0\leq j$, we have:

(1) $2^{j}z=y_{j}+y^{\prime}_{j}$, i.e., $2^{j}z-y_{j}=y^{\prime}_{j}$;

(2) $2y_{j}=y_{j+1}$ and $2y_{j}^{\prime}=y_{j+1}^{\prime}$;

(3) $|y_{j}|=|y^{\prime}_{j}|=|2^{j}x|$.

In fact, these conditions clearly hold for all $j\geq k$, so assume they hold
down to $j+1$ and we need to construct $y_{j}$, which by (1) will define the
wanted $y_{j}^{\prime}$.

Case 1: $|2^{j}x|+1=|2^{j+1}x|=|y_{j+1}|$: Let $y_{j}$ be any element of $G$
satisfying $2y_{j}=y_{j+1}$ and $|y_{j}|=|2^{j}x|$. By (1), we must let
$y_{j}^{\prime}=2^{j}z-y_{j}$ and (2) follows easily.

Certainly, $|y_{j}^{\prime}|\geq\min\\{|y_{j}|,|2^{j}z|\\}=|2^{j}x|$. And if
$|y_{j}^{\prime}|>|2^{j}x|$, then

$|2^{j}x|+1=|y_{j+1}|=|y_{j+1}^{\prime}|=|2y_{j}^{\prime}|\geq|y_{j}^{\prime}|+1>|2^{j}x|+1.$

This contradiction verifies (3) for $j$.

Case 2: $|2^{j}x|+1<|2^{j+1}x|=|y_{j+1}|$: Let $n=|2^{j}x|$. We start by
finding $s\in G$ such that $|s|\geq n+1$ and $2s=y_{j+1}$.

Recall that the $n$th Ulm factor of $G$ is isomorphic to

$U_{n}:=\\{g\in 2^{n}G:2g\in 2^{n+2}G\\}/2^{n+1}G.$

Since $|2^{j}z|\geq|2^{j}x|=n$ and $|2^{j+1}z|\geq|2^{j+1}x|\geq n+2$, we see
that $v:=2^{j}z+2^{n+1}G$ represents a (possibly zero) element of $U_{n}$.

Since $U_{n}$ has at least $3$ elements, it has a non-zero element $w\neq v$.
Using the usual way to think of Ulm factors, we can find $t\in(2^{n}G)[2]$
such that $t+2^{n+1}G=w$; since $w\neq 0$, we have $|t|=n$.

Let $y_{j}=s+t$. So, $y_{j}+2^{n+1}G=t+2^{n+1}G=v$ and
$2y_{j}=2s+2t=2s=y_{j+1}$. To make sure (1) continues to hold, we must define
$y_{j}^{\prime}=2^{j}z-y_{j}$, which will again imply that
$2y_{j}^{\prime}=y_{j+1}^{\prime}$, so that (2) will hold as well.

Regarding (3), since $|s|\geq n+1$, we can conclude that
$|y_{j}|=|s+t|=n=|2^{j}x|$, which is half the battle. In addition, since
$|y_{j}|=n$ and $|2^{j}z|\geq|2^{j}x|=n$, we can deduce that
$|y_{j}^{\prime}|\geq n$. Finally, since

$y^{\prime}_{j}+2^{n+1}G=(2^{j}z+2^{n+1}G)-(y_{j}+2^{n+1}G)=v-w\in U_{n}$

is non-zero, we can conclude that $|y_{j}^{\prime}|=n=|2^{j}x|$, as required.

Setting $j=0$, we can infer from (1) that $z=y_{0}+y^{\prime}_{0}$. And from
(2) and (3), the height sequences must satisfy
$\|y_{0}\|=\|y_{0}^{\prime}\|=\|x\|$. So, since $G$ is transitive, there are
automorphisms $\alpha$ and $\alpha^{\prime}$ such that $\alpha(x)=y_{0}$ and
$\alpha^{\prime}(x)=y_{0}^{\prime}$. Therefore, since $X$ is characteristic,
one finds that

$z=y_{0}+y_{0}^{\prime}=\alpha(x)+\alpha^{\prime}(x)\in X,$

as required. ∎

A query which immediately arises is of whether or not the Krylov transitive
$2$-group as constructed in [1] to be neither transitive nor fully transitive
has the same property as in the preceding example, that is, are all its
characteristic subgroups fully invariant, or even commensurable with fully
invariant subgroups? Moreover, it is rather logical to have true the assertion
that, for all primes $p$, any Krylov transitive $p$-group whose characteristic
subgroups are (commensurable with) fully invariant subgroups is necessarily
transitive.

We end our work with two problems of some interest. The following question is
clearly important and possibly difficult (compare with Example 3.2).

Problem 1. Does every group with minimal full inertia also have minimal
characteristic inertia?

In case this is not true, that is these two properties are independent each
other, we proceed with the next possibly challenging question, which is
relevant to Corollary 2.5.

Problem 2. Do totally projective $2$-groups have the minimal characteristic
inertia?

It is worthwhile noticing that, in view of [14], they always have the minimal
full inertia.

Funding: The work of the first-named author, P.V. Danchev, is partially
supported by the Bulgarian National Science Fund under Grant KP-06 No. 32/1 of
December 07, 2019.

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