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

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


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

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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.