Are isomorphic quotients of abelian groups induced by automorphisms? [closed]

Question

If I have an (abelian) group $G$ and an automorphism $\sigma: G \to G$ then for any subgroup $H$ of $G$ there is an induced isomorphism $G/H \cong G/\sigma(H)$ given by the map $gH \mapsto \sigma(g)\sigma(H)$.

I want to know if the converse statement is true. Suppose I have an (abelian) group $G$ and two subgroups $H_1$ and $H_2$ such that $G/H_1 \cong G/H_2$ via some isomorphism. Then is there an automorphism $\sigma: G \to G$ which induces the automorphism on the quotient? I'm also interested in the slightly weaker question of if there is always an automorphism so that $\sigma(H_1) = H_2$ (ie an automorphism which induces some potentially different isomorphism on the quotients).

This feels plausible to me (I can't find a counterexample), but I haven't been able to construct the automorphism in general.

Answer 1

The answer is rather obviously no for abelian groups. An infinite series of counterexamples can be found by taking $G=\mathbb{Z}_4 \times (\mathbb{Z}_2)^n,$ with $$H_1 = \mathbb{Z}_4 \times (\mathbb{Z}_2)^{n-1}, \quad H_2 = \mathbb{Z}_2 \times (\mathbb{Z}_2)^{n}.$$

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Since you put "abelian" into parentheses, let me show that the answer is no also for non-abelian groups, providing a counterexample in which $H_1$ and $H_2$ are isomorphic as abstract groups. Consider the group of order $16$ $$G := \langle a,\,x \mid a^4 = x^4 = 1, \, xax^{-1} = a^{-1} \rangle,$$ namely, the non-trivial semidirect product of $\mathbb{Z}_4$ with itself, whose GAP label is $G(16, \, 4)$. Now, take the two subgroups $$H_1= \langle a^2, \, x \rangle, \quad H_2= \langle a, \, x^2 \rangle.$$ They are both isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_2$, so they are normal and the quotients are both isomorphic to $\mathbb{Z}_2$. However, $H_2$ is characteristic, so every automorphism of $G$ sends $H_2$ to itself.