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Assume that we have two groups $A$ and $B$ such that $C \subset A$ and $C \subset B$ where $C$ is a normal subgroup of both $A$ and $B$. If we have that $A/C \cong B/C$ is it true that $A \cong B$? I feel like this shouldn't be true but I can't seem to find any counter examples and my attempts to prove it thus far has been unsuccessful. It would be nice to find a finite counterexample if possible.

This is not the case where $A$ and $B$ have isomorphic subgroups $C$ and $C'$ such that $A/C \cong B/C'$. I have intentionally excluded these cases as uninteresting. Subgroups can be embedded in all sorts of strange ways into other groups. This forces the subgroup $C$ in $A$ to actually be the same set inside of $B$.

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    $\begingroup$ Like you say, subgroups can be embedded in all sorts of ways into other groups. So if $C$ and $C'$ are isomorphic but not exactly the same, it's a simple matter of modifying $A$ and $B$ and identifying elements in certain ways to make $C$ and $C'$ actually identical. See Ittay Weiss's answer for example. $\endgroup$ – PatrickR Sep 6 '13 at 4:31
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Consider $A=\{0,1,2,3\}$, where $0,1,2,3$ are just four distinct symbols. Turn $A$ into a group by interpreting these symbols as the elements of $\mathbb Z_4$, the usual group. Now let $B=\{0,2,2',2''\}$ where $0,2$ are the same symbols in $A$ and $2',2''$ are two new symbols. Turn $B$ into a group so that each element has order at most $2$ and so that $0$ is the identity element (this can be done in precisely one way). Now, $C=\{0,2\}$ is a subgroup of both $A$ and $B$, the quotients are obviously isomorphic, but $A$ and $B$ are not isomorphic.

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  • $\begingroup$ Is there a larger group into which $A$ and $B$ can be embedded as subgroups? Or are we basically just considering $\{0, 2\}\subset \mathbb{Z}_4$ and $\{(0,0), (1,1)\}\subset \mathbb{Z}_2\times \mathbb{Z}_2$ to be the same set via the obvious identification? $\endgroup$ – Josh Keneda Sep 6 '13 at 4:36
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For a counterexample, take $A = \mathbb Z_4$, the cyclic group of order $4$, and $B = \mathbb Z_2 \times\mathbb Z_2$, the product of two cyclic group of order $2$.

Both $A$ and $B$ contain a subgroup $C$ isomorphic to $\mathbb Z_2$ and $A/C\cong B/C\cong\mathbb Z_2$, but $A$ and $B$ are not isomorphic.

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  • $\begingroup$ I will admit I did not overly stress that I was excluding the obvious counterexamples. $C$ is a subset of $A$ and $B$ and so your counterexample doesn't apply because the two subgroups are embedded as different sets inside of $\mathbb{Z}_4$ and $\mathbb{Z}_2 \times \mathbb{Z}_2$. $\endgroup$ – Devin Murray Sep 6 '13 at 3:21
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If you're willing to weaken your condition to $C \subset A$ and $C' \subset B$, where $C \cong C'$, then the answer to your question is no.

Take $A = \mathbb{Z}_4$, $B = \mathbb{Z}_2 \times \mathbb{Z}_2$, $C = \{0, 2\}\subset A$, and $C' = \{(0,0), (1,1)\}\subset B$. Then $C \cong C'$ and $A/C \cong B/C'\cong \mathbb{Z}_2$, but $A \not\cong B$.

I'm not sure if this extends to an answer to the question as phrased, but this might be close enough to satisfy your curiosity.

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  • $\begingroup$ I'm aware that weakening my assumptions results in easy counterexamples. I'm interested in the much more restricted case. $\endgroup$ – Devin Murray Sep 6 '13 at 3:23
  • $\begingroup$ @DevinMurray I'd say that these two formulations are equivalent, so it is not really a weakening. You can always transfer group structure via a bijection. This is basically what the Patrick's comment under your question says. $\endgroup$ – Martin Sleziak Sep 6 '13 at 6:08

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