Characteristic equals fully invariant in odd-order abelian group

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.
View all subgroup property implications in odd-order abelian groups $|$ View all subgroup property non-implications in odd-order abelian groups $|$ View all subgroup property implications $|$ View all subgroup property non-implications

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 $k_{1}\leq k_{2}\leq \dots \leq k_{r}$.

$H$ 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 $a\leq b$, 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 $1\leq i<j\leq r$, 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 $\!l_{i}\leq l_{j}$.
• 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 $\!k_{i}-l_{i}\leq k_{j}-l_{j}$.

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 $\rho (H\cap G_{i})\leq H$. 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, $\!k_{i}\leq k_{j}$. 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 $l_{i}\leq l_{j}$, 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, $\!k_{j}\leq k_{i}$. 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 $\!p^{k_{j}-(k_{i}-l_{i})}\leq p^{l_{j}}$ because $\!k_{j}-l_{j}\leq k_{i}-l_{i}$. 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.