Characteristic equals fully invariant in odd-order abelian group

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a odd-order abelian group. That is, it states that in a Odd-order abelian group (?), every subgroup satisfying the first subgroup property (i.e., Characteristic subgroup (?)) must also satisfy the second subgroup property (i.e., Fully invariant subgroup (?)). In other words, every characteristic subgroup of odd-order abelian group is a fully invariant subgroup of odd-order abelian group.
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Statement

In an odd-order abelian group, a subgroup is characteristic if and only if it is fully invariant.

Related facts

• Fully invariant implies characteristic
• Characteristic not implies fully invariant
• Characteristic not implies fully invariant in finitely generated abelian group
• Characteristic not implies fully invariant in finite abelian group
• Classification of fully invariant subgroups in finitely generated abelian groups

Proof for a group of prime power order

Given: An Abelian group $G$ of odd prime power order.

$G=\underset{i=1}{\overset{r}{⨁}}{G}_i$.

where we have:

$G_i\cong \mathbb{Z}/p^{k_i}\mathbb{Z}$.

such that $k_1\le k_2\le \dots \le k_r$.

$H$ is a characteristic subgroup of $G$.

To prove: $H$ is fully invariant in $G$.

Proof:

It is a direct sum of its intersections with the direct summands

We first prove that $H$ is a direct sum of the intersections $H\cap G_i$.

For this, consider the projection map ${\rho }_i:G\to G$ that sends an element $(g_1,g_2,\dots ,g_r)$ to the element $(0,0,\dots ,g_i,0,0,\dots ,0)$. ${\rho }_i$ is an endomorphism of $G$. Consider the map:

${\sigma }_i=\mathrm{I}\mathrm{d}+{\rho }_i$.

We claim that ${\sigma }_i$ is an automorphism. For this, note that ${\sigma }_i$ acts as the identity map on all coordinates other than $i$, and as the doubling, or square, map on the $i^{th}$ coordinate. Since the group has odd order, the doubling map is an automorphism of the $i^{th}$ coordinate, so ${\sigma }_i$ is an automorphism. Thus, we have:

${\sigma }_i(H)=H$.

Since $H$ is a subgroup, we get:

${\rho }_i(H)=H$.

Thus, the projections of any element of $H$ are in $H$. Thus, $H$ is a direct sum of its projections.

A characteristic subgroup satisfies the conditions for being fully invariant

For $a\le b$, there is an injective homomorphism:

${\nu }_{a,b}:\mathbb{Z}/p^a\mathbb{Z}\to \mathbb{Z}/p^b\mathbb{Z}$

that sends the generator on the left to $p^{b-a}$ times the generator on the right.

There is also a surjective homomorphism:

${\pi }_{b,a}:\mathbb{Z}/p^b\mathbb{Z}\to \mathbb{Z}/p^a\mathbb{Z}$

that sends the generator to the generator.

Recall that we have:

$G=\underset{i=1}{\overset{r}{⨁}}{G}_i$.

For $1\le i<j\le r$, define ${\alpha }_{i,j}$ as the endomorphism of $G$ that sends $G_i$ to $G_j$ via the map:

${\nu }_{p^{k_i},p^{k_j}}:G_i\to G_j$.

and is zero elsewhere. Define ${\gamma }_{i,j}$ as:

${\gamma }_{i,j}=\mathrm{I}\mathrm{d}+{\alpha }_{i,j}$.

Clearly, ${\gamma }_{i,j}$ is an automorphism of $G$.

Similarly, define ${\beta }_{j,i}$ as the endomorphism of $G$ that sends $G_j$ to $G_i$ via the map ${\pi }_{p^{k_j},p^{k_i}}$ and is zero elsewhere.

Define ${\phi }_{j,i}$ as:

${\phi }_{j,i}=\mathrm{I}\mathrm{d}+{\beta }_{j,i}$.

Clearly ${\phi }_{j,i}$ is an automorphism of $G$.

Thus, for $H$ to be characteristic, it must be invariant under both ${\gamma }_{i,j}$ and ${\phi }_{j,i}$. This forces $H$ to be invariant under ${\alpha }_{i,j}$ and ${\beta }_{j,i}$. If the order of $H\cap G_i$ is $p^{l_i}$, we obtain the following:

• The endomorphism ${\alpha }_{i,j}$ sends $H$ to itself: Thus, $H\cap G_i$ injects into $H\cap G_j$ via ${\alpha }_{i,j}$, so $l_i\le l_j$.
• The endomorphism ${\beta }_{j,i}$ sends $H$ to itself: Thus, ${\beta }_{j,i}$ induces a surjective endomorphism from $G_j/(H\cap G_j)$ to $G_i/(H\cap G_i)$, forcing $k_i-l_i\le k_j-l_j$.

Now, we show that if $H$ satisfies the conditions described above, then $H$ is fully invarant in $G$.

Suppose $\rho :G\to G$ is an endomorphism. Since $H$ is a direct sum of $H\cap G_i$, it suffices to show that $\rho (H\cap G_i)\le H$. For this, in turn, it suffices to show that the $j^{th}$ coordinate of $\rho (H\cap G_i)$ is contained in $H\cap G_j$. This is easily done in three cases:

• $j=i$: In this case, it is direct since $H\cap G_i$ is a fully invariant subgroup of the cyclic group $G_i$ (all subgroups of cyclic groups are fully invariant).
• $j>i$: In this case, $k_i\le k_j$. Consider the homomorphism from $H\cap G_i$ to $G_j$ obtained by composing the $j^{th}$ projection with $\rho$. This map must send $H\cap G_i$ to a subgroup of size at most $p^{l_i}$ in $G_j$. But since $l_i\le l_j$, this subgroup of size $p^{l_i}$ is contained in the subgroup $H\cap G_j$ that has order $p^{l_j}$.
• $j<i$: In this case, $k_j\le k_i$. COnsider the homomorphism from $G_i$ to $G_j$ obtained by composing the $j^{th}$ projection with $\rho$. The index of the image of $H\cap G_i$ is at least the index of $H\cap G_i$ in $G_i$, which is $p^{k_i-l_i}$. Thus, the image has size at most $p^{k_j-(k_i-l_i)}\le p^{l_j}$ because $k_j-l_j\le k_i-l_i$. Thus, it is in $H\cap G_j$.

Proof for an odd-order Abelian group

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