Characteristic equals fully invariant in odd-order abelian group

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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a odd-order abelian group. That is, it states that in a Odd-order abelian group (?), every subgroup satisfying the first subgroup property (i.e., Characteristic subgroup (?)) must also satisfy the second subgroup property (i.e., Fully invariant subgroup (?)). In other words, every characteristic subgroup of odd-order abelian group is a fully invariant subgroup of odd-order abelian group.
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Statement

In an odd-order abelian group, a subgroup is characteristic if and only if it is fully invariant.

Related facts

Opposite facts

Proof for a group of prime power order

Given: An abelian group G of odd prime power order.

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

where we have:

G_i \cong \mathbb{Z}/p^{k_i}\mathbb{Z}.

such that k_1 \le k_2 \le \dots \le k_r.

H is a characteristic subgroup of G.

To prove: H is fully invariant in G.

Proof:

It is a direct sum of its intersections with the direct summands

We first prove that H is a direct sum of the intersections H \cap G_i.

For this, consider the projection map \! \rho_i:G \to G that sends an element \! (g_1,g_2, \dots, g_r) to the element \! (0,0, \dots, g_i, 0, 0, \dots, 0). \! \rho_i is an endomorphism of G. Consider the map:

\sigma_i = \operatorname{Id} + \rho_i.

We claim that \! \sigma_i is an automorphism. For this, note that \! \sigma_i acts as the identity map on all coordinates other than \! i, and as the doubling, or square, map on the \! i^{th} coordinate. Since the group has odd order, the doubling map is an automorphism of the \! i^{th} coordinate, so \! \sigma_i is an automorphism. Thus, we have:

\! \sigma_i(H) = H.

Since \! H is a subgroup, we get:

\! \rho_i(H) = H.

Thus, the projections of any element of H are in H. Thus, H is a direct sum of its projections.

A characteristic subgroup satisfies the conditions for being fully invariant

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. Define \! \gamma_{i,j} as:

\! \gamma_{i,j} = \operatorname{Id} + \alpha_{i,j}.

Clearly, \! \gamma_{i,j} is an automorphism of G.

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:

\varphi_{j,i} = \operatorname{Id} + \beta_{j,i}.

Clearly \! \varphi_{j,i} is an automorphism of G.

Thus, for H to be characteristic, it must be invariant under both \! \gamma_{i,j} and \varphi_{j,i}. This forces H to be invariant under \alpha_{i,j} and \! \beta_{j,i}. If the order of H \cap G_i is p^{l_i}, we obtain the following:

  • 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 invarant 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 invariant subgroup of the cyclic group \! G_i (all subgroups of cyclic groups are fully invariant).
  • \! 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^{l_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.

Proof for an odd-order abelian group

This is direct because the group splits up as a direct sum of its p-Sylow subgroups, and characteristic or fully invariant subgroups also split up as direct sums.