Characteristic equals fully invariant in odd-order abelian group

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

Related facts

 * Fully invariant implies characteristic

Opposite facts

 * Characteristic not implies fully invariant
 * Characteristic not implies fully invariant in finitely generated abelian group
 * Characteristic not implies fully invariant in finite abelian group
 * Characteristic not implies fully invariant in class three maximal class p-group
 * Characteristic not implies fully invariant in odd-order class two p-group

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.