# Difference between revisions of "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 $k_1 \le k_2 \le \dots \le k_r$.

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 $l_1 \le l_2 \le \dots \le l_r$ and $k_1 - l_1 \le k_2 - l_2 \le \dots \le k_r - l_r$.

### 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 $k_1 \le k_2 \le \dots \le k_r$.

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 $l_1 \le l_2 \le \dots \le l_r$ and $k_1 - l_1 \le k_2 - l_2 \le \dots \le k_r - l_r$.

Proof: The proof relies on two important homomorphisms. 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:

$G = \bigoplus_{i=1}^r G_i$.

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.

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 $l_i \le l_j$.
• 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 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 $\rho(H \cap G_i) \le H$. 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 characteristic subgroup of the cyclic group $G_i$ (all subgroups of cyclic groups are fully characteristic).
• $j > i$: 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 $p^j$.
• $j < i$: 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 $H \cap G_j$.

### General case

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