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If I want to quotient $V_{4}$ by $C_{2}$ then I've got three isomorphic copies of $C_{2}$ in $V_{4}$ to choose from, but whichever one I choose to quotient out by, I get $C_{2}$.

So I feel happy writing $V_{4} / C_{2} = C_{2}$

But is that always the case?

Is there a group G with two isomorphic normal subgroups $H_{1}$, $H_{2}$ s.t. $G/H_{1} \ncong G/H_{2}$

Obviously they'll always have the same number of elements, but why would they have the same structure?

Shaun
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John Lawrence Aspden
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  • errr.., very probably!, I was trying to make it (not) work with Z2XZ4 earlier. I'll try again.... – John Lawrence Aspden Mar 06 '22 at 18:34
  • Yes it does, I am shame. I'd somehow convinced myself that if it broke it would break on 12 and completely missed 02.... Thank you! I leave it to wiser persons to decide whether to delete this and spare my blushes or leave it as a different way to ask the (now linked) question. – John Lawrence Aspden Mar 06 '22 at 18:42

1 Answers1

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Let $G=S_3\times\Bbb Z_3$. Note that if $H=\{e,(1\ \ 2\ \ 3),(1\ \ 3\ \ 2)\}\subset S_3$, then $H$ is a normal subgroup of $S_3$ and $H\simeq\Bbb Z_3$. But $G/(H\times\{0\})\simeq\Bbb Z_2\times\Bbb Z_3$, whereas $G/(\{e\}\times\Bbb Z_3)\simeq S_3$.

José Carlos Santos
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