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Let $G$ be a finite abelian group.

We know that the least size of a minimal generating set of a proper nontrivial subgroup $H \leqslant G$ might be the same as $G$, for example ($\Bbb Z_6 \cong \Bbb Z_3 \times \Bbb Z_2 $ has a subgroup isomorphic to $\Bbb Z_2$, and both $\Bbb Z_6$ and $\Bbb Z_2$ can be generated by $1$ element). Also, in general, given two isomorphic subgroups $H_1$ and $H_2$ of $G$, it is not necessarily true that $G/H_1$ and $G/H_2$ are isomorphic (Isomorphic quotient groups).

However, if the least size of a minimal generating set of each $H_i$ is the same as that of $G$, and $H_1 \cong H_2$, does this implies that $G/H_1 \cong G/H_2$?

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  • $\begingroup$ Do you really mean "same number of minimal generating sets", or do you mean "the least size of a minimal generating set of $H_i$ is the same as that of $G$"? $\endgroup$ Jul 24, 2023 at 14:10
  • $\begingroup$ (The reason I ask is that $\mathbb{Z}_6$ has $1$ as the smallest size of a minimal generating set, but has four "minimal generating sets", namely $\{1\}$, $\{5\}$, $\{2,3\}$, and $\{3,4\}$). $\endgroup$ Jul 24, 2023 at 14:13
  • $\begingroup$ yes, @ArturoMagidin, thank you! $\endgroup$
    – ghc1997
    Jul 24, 2023 at 14:17
  • $\begingroup$ @BrevanEllefsen unfortunately $\Bbb Z_3 $ and $\Bbb Z_2 $ are not isomorphic $\endgroup$
    – ghc1997
    Jul 24, 2023 at 14:20

1 Answer 1

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There are two subgroups isomorphic to $C_2 \times C_4$ in $C_4 \times C_8$ with non-isomorphic quotient groups.

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