# 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 characteristic subgroup of $G$ if and only if $H$ 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 finite groups

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

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

where each $G_i$ is an Abelian group of prime power order for different primes. Then, a subgroup $H$ of $G$ is a characteristic subgroup if and only if it 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 $G_i$.

### Case of finitely generated groups

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

$G = A \oplus B$

where $A$ is a torsion-free group. A subgroup $H$ of $G$ is a fully characteristic subgroup of $G$ if and only if $H = (H \cap A) \oplus (H \cap B)$ and we have that $H \cap A$ is fully characteristic in $A$ and $H \cap B$ is characteristic in $B$.

## 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 $a < b$, 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.

Define $\sigma_{i,j}$ as the sum of the endomorphism $\alpha_{i,j}$ and the identity map on $G$. Clearly, $\sigma_{i,j}$ is an automorphism.

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 $\varphi_{j,i}$ as the automorphism obtained as the sum of $\beta_{j,i}$ and the identity map.

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

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

$(g_1,g_2, \dots, g_r) \in H$.

We want to show that the elements $(g_1,0,0, \dots, 0), (0,g_2,0,0, \dots 0), \dots, (0,0,\dots,0,g_r)$ are in $H$.

If all except one $g_i$ are zero, then we are done. Otherwise, there exist $i < j$ such that both $g_i$ and $g_j$ are nonzero.