# Subgroups of finite abelian groups and automorphisms

Let $A$ a finite abelian group, and $B$, $C$ finite subgroups of $A$. If $B$, $C$ are isomorphic, and their quotient groups $A/B$, and $A/C$ are isomorphic, it may still happen that there does not exist an automorphism of $A$ that takes $B$ to $C$. See a question on MSE at this link, and the solution at this page. Note that the implication holds if $B$ (or $A/B$) has prime order (not that hard to prove).

I wonder if the following is true, or if anyone has a counterexample:

Let $A$ be a finite abelian group, $B$, $C$ isomorphic subgroups. Assume that for every(many) characteristic subgroup $B'$ of $B$, and corresponding $C'$ of $C$, the quotient groups $A/B'$, $A/C'$ are isomorphic. Then there exists an automorphism of $A$ that takes $B$ to $C$.

There is a counterexample. Given any prime $p$, let \begin{align} A &= \begin{pmatrix} \mathbb Z/p^5 & \mathbb Z/p^2 \\ \mathbb Z/p & \mathbb Z/p^4 \end{pmatrix} \\ &:= \left\{ \begin{pmatrix}x & y \\ z & w\end{pmatrix} \middle|\, x \in \mathbb Z/p^5, y \in \mathbb Z/p^2, z \in \mathbb Z/p, w \in \mathbb Z/p^4 \right\} \\ &\cong \mathbb Z/p^5 \times \mathbb Z/p^2 \times \mathbb Z/p \times \mathbb Z/p^4, \end{align}

let $B$ be generated by $$b_1 = \begin{pmatrix} p^3 & 0 \\ 1 & 0\end{pmatrix}\quad \text{and}\quad b_2 = \begin{pmatrix}0 & p \\ 0 & p^2 \end{pmatrix},$$ and let $C$ be generated by $$\begin{pmatrix}p^3 & p \\ 0 & 0 \end{pmatrix} \quad \text{and}\quad \begin{pmatrix}0 & 0 \\ 1 & p^2 \end{pmatrix}.$$

Thinking of $A$ columnwise as the product

$$A = A_1 \times A_2 = (\mathbb Z/p^5 \times \mathbb Z/p) \times ( \mathbb Z/p^2 \times \mathbb Z/p^4),$$

we have $B = \langle b_1 \rangle \times \langle b_2 \rangle \cong (\mathbb Z/p^2) \times (\mathbb Z/p^2),$ so the only characteristic subgroup of $B$ other than $B$ itself and the trivial group is $$pB = \langle p b_1 \rangle \times \langle p b_2 \rangle \cong \mathbb Z/p \times \mathbb Z/p.$$

It's straightforward to check that

$$A/B = A_1/\langle b_1 \rangle \times A_2/\langle b_2 \rangle \cong (\mathbb Z/p^4) \times (\mathbb Z/p \times \mathbb Z/p^3)$$

Similarly, thinking of $A$ rowwise as

$$A = (\mathbb Z/p^5 \times \mathbb Z/p^2) \times (\mathbb Z/p \times \mathbb Z/p^4)$$

makes it straightforward to check that

$$C \cong (\mathbb Z/p^2) \times (\mathbb Z/p^2),$$

with

$$A/C \cong (\mathbb Z/p^4 \times \mathbb Z/p) \times (\mathbb Z/p^3).$$

Moreover, $pB = pC$. So we have $B \cong C$, $A/B \cong A/C$, and $A/pB \cong A/pC$.

However, there is no automorphism of $A$ which sends $B$ to $C$. For example,

\begin{align} (pA\cap B)/(p^4A \cap B) &= (pA_1 \cap \langle b_1 \rangle)/(p^4A_1 \cap \langle b_1 \rangle) \times (pA_2 \cap \langle b_2 \rangle)/(p^4A_2 \cap \langle b_2 \rangle)\\ &\cong 1 \times (\mathbb Z/p^2), \end{align}

but similarly

$$(pA\cap C)/(p^4A \cap C) \cong (\mathbb Z/p) \times (\mathbb Z/p).$$

• Thank you! That looks very nice. – Orest Bucicovschi Oct 30 '17 at 0:39