Classification of fully characteristic subgroups in finitely generated Abelian groups

Statement

Case of groups of prime power order

Suppose $G$ is an Abelian group whose order is a power of a prime $p$ . Then, we can write, by the structure theorem for finitely generated Abelian groups:

$G=\bigoplus _{i=1}^{r}G_{i}$

where:

$G_{i}\cong \mathbb {Z} /p^{k_{i}}\mathbb {Z}$ .

where $k_{1}\leq k_{2}\leq \dots \leq k_{r}$ .

Then a subgroup $H$ of $G$ is a fully characteristic subgroup of $G$ , and this happens if and only if the following are satisfied:

$H$ is the direct sum of the intersections $H\cap G_{i}$ .
If the orders of $H\cap G_{i}$ are $p^{l_{i}}$ , then $l_{1}\leq l_{2}\leq \dots \leq l_{r}$ and $k_{1}-l_{1}\leq k_{2}-l_{2}\leq \dots \leq k_{r}-l_{r}$ .

Case of finitely generated groups

Suppose $G$ is a finitely generated Abelian group. Then, we can write:

$G=\bigoplus _{i=0}^{r}G_{i}$

where $G_{0}$ is a torsion-free Abelian group, and each $G_{i}$ is an Abelian group of prime power order for different primes. Then, a subgroup $H$ of $G$ is a fully characteristic subgroup, if and only if the following are satisfied:

$H$ is the direct sum of the intersections $H\cap G_{i}$ .
Each $H\cap G_{i}$ is characteristic (equivalently, fully characteristic) in $G_{i}$ .

Proof

Case of prime power order

Given: $G$ is an Abelian group whose order is a power of a prime $p$ . Then, we can write, by the structure theorem for finitely generated Abelian groups:

$G=\bigoplus _{i=1}^{r}G_{i}$

where:

$G_{i}\cong \mathbb {Z} /p^{k_{i}}\mathbb {Z}$ .

where $k_{1}\leq k_{2}\leq \dots \leq k_{r}$ .

A subgroup $H$ of $G$ .

To prove: $H$ is characteristic if and only if it is fully characteristic in $G$ , if and only if $H$ is the direct sum of $H\cap G_{i}$ , and if the orders of the intersections are $p^{l_{i}}$ , then $l_{1}\leq l_{2}\leq \dots \leq l_{r}$ and $k_{1}-l_{1}\leq k_{2}-l_{2}\leq \dots \leq k_{r}-l_{r}$ .

Proof: The proof relies on two important homomorphisms. For $aleb$ , 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:

$G=\bigoplus _{i=1}^{r}G_{i}$ .

For $1\leq i<j\leq r$ , 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.

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.

We are now in a position to prove the main result.

Suppose $H$ is a fully characteristic subgroup of $G$ . We first show that $H$ is a direct sum of $H\cap G_{i}$ . For this, suppose the element:

$(g_{1},g_{2},\dots ,g_{r})\in H$ .

Since $H$ is fully characteristic, it is invariant under the projections to the direct factors, which are endomorphisms. Thus, each $(0,0,\dots ,0,g_{i},0,\dots ,0)$ is in $H$ . Thus, every element of $H$ can be expressed as a sum of elements in $H\cap G_{i}$ , and we get that $H$ is a direct sum of the $H\cap G_{i}$ s.

Finally, we need to show the conditions on the $l_{i}$ s. For this, observe that for $i<j$ , we have:

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 $l_{i}\leq l_{j}$ .
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 $k_{i}-l_{i}\leq k_{j}-l_{j}$ .

Now, we show that if $H$ satisfies the conditions described above, then $H$ is fully characteristic 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 $\rho (H\cap G_{i})\leq H$ . 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 characteristic subgroup of the cyclic group $G_{i}$ (all subgroups of cyclic groups are fully characteristic).
$j>i$ : In this case, $k_{i}\leq k_{j}$ . 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 $l_{i}\leq l_{j}$ , this subgroup of size $p^{l_{i}}$ is contained in the subgroup $H\cap G_{j}$ that has order $p^{j}$ .
$j<i$ : In this case, $k_{j}\leq k_{i}$ . 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 $p^{k_{j}-(k_{i}-l_{i})}\leq p^{l_{j}}$ because $k_{j}-l_{j}\leq k_{i}-l_{i}$ . Thus, it is in $H\cap G_{j}$ .

General case

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