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I think it is an easy question, but for some reasons I am confused. Thank you for your help! Let $G$, $H_1$ and $H_2$ be finite groups. If

$G/H_1\cong G/H_2$, is it true that $H_1\cong H_2$?

If not, do you know a counterexample?

Thank you!

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1 Answer

No. Take $G=\mathbf{Z}_2 \oplus\mathbf{Z}_4$. Let $H_1 = \mathbf{Z}_2\times\langle 2\rangle$, and $H_2=\{0\}\times\mathbf{Z}_4$. Then $G/H_1 \cong \mathbf{Z}_2\cong G/H_2$, but $H_1$ is the Klein $4$-group and $H_2$ is cyclic of order $4$.

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Oh thanks! Do you know if it is true when they are cyclic groups? –  math430 May 17 '12 at 16:32
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@math430: When who is a cyclic group? If $G$ is cyclic, then if $G$ is finite it has a unique subgroup of any given order, so if $G/H_1$ has the same size as $G/H_2$, then $H_1=H_2$. If $G$ is infinite, then $H = \langle [G:H]\rangle$, so again $G/H_1\cong G/H_2$ implies $H_1=H_2$. If $H_1$ and $H_2$ are both cyclic, then $G$ finite implies $H_1$ and $H_2$ are of the same order, so they must be cyclic of the same order, hence isomorphic. But if $G$ is infinite, then no. Take $G=\mathbb{Z}\oplus\mathbb{Z}\oplus\cdots$, $H_1=\mathbb{Z}$, and $H_2=\{0\}$. –  Arturo Magidin May 17 '12 at 16:37
    
Ok thank you, I meant G cyclic. Thank you for your answer! –  math430 May 17 '12 at 16:40
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