Classification of fully characteristic subgroups in finitely generated Abelian groups

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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.