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 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_is.

Finally, we need to show the conditions on the l_is. 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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