Classification of fully characteristic subgroups in finitely generated Abelian groups

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 $$G_i$$.

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
This is direct.