# If $G/N$ and $H/N$ are isomorphic, does it imply $G$ and $H$ are isomorphic?

If $N$ is a subgroup of both $G$ and $H$, and $G/N \cong H/N$, then is $G \cong H$? (Or more generally, if $N$ and $M$ are subgroups of $G$ and $H$ respectively, and $N \cong M$ and $G/N \cong H/M$, then is $G \cong H$?)

I suspect the answer is no, but if the answer is yes, is there a natural isomorphism between $G$ and $H$ given an isomorphism $\phi:G/N \rightarrow H/N$?

Also, just making sure, is $(G \times H)/G \cong H$ and $(G \times H)/H \cong G$?

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If $G$ and $H$ are different groups, how can $N$ be a subgroup of each? It must live in one but not the other. Do you mean that $G$ and $H$ and each contained in some common larger group? –  Potato Mar 13 '13 at 7:51
@Potato: Think categorically. $N,G,H$ are groups and $N \to G$ and $N \to H$ are given regular monomorphisms. The question makes sense, even if we don't use the pushout $G \sqcup_N H$. –  Martin Brandenburg Mar 14 '13 at 17:58
@Potato, there is no obstacle for a group to be a subgroup of two different groups, even if the latter are not both contained in a larger group! –  Mariano Suárez-Alvarez Mar 14 '13 at 18:01
@MartinBrandenburg: Funny, my interpretation of "think categorically" would definitely consider "N is a subgroup of G" to be different from "N is isomorphic to a subgroup of G." –  Cam McLeman Mar 14 '13 at 19:20

$N = \mathbb Z/2$ is isomorphic to a subgroup of $G = \mathbb Z/2 \times \mathbb Z/2$ and of $H = \mathbb Z/4$. These groups are not isomorphic but $G/N \simeq \mathbb Z/2 \simeq H/N$.

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Your first/second question can be reformulated as asking whether the isomorphism class of the normal subgroup $N$ of the group $G$, and that of the quotient group $G/N$, together determine the isomorphism class of $G$.

The short answer is no, as shown by @Jim. A fuller answer is that you are actually inquiring about the intricate problem of group extensions.

As to your third question, I recommend a bit of care. If you write $$(\mathbf{Z} \times \mathbf{Z}) / \mathbf{Z},$$ the $\mathbf{Z}$ at the denominator refers to which of the many copies of $\mathbf{Z}$ in $\mathbf{Z} \times \mathbf{Z}$? If you choose for instance $$\{ (2x, 0) : x \in \mathbf{Z} \} \cong \mathbf{Z}$$ as your copy of $\mathbf{Z}$ at the denominator, then the quotient is definitely not isomorphic to $\mathbf{Z}$, but to $\mathbf{Z}/2 \mathbf{Z} \times \mathbf{Z}$.

So you should probably say something like $$1 \to H \to G \times H \to G \to 1$$ is an exact sequence, where $H \to G \times H$ is the map $h \mapsto (1, h)$ and $G \times H \to G$ is $(g, h) \mapsto g$.

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Let $G=S_3$, $H=\mathbb Z_6$, $N_1=\langle(1~2~3)\rangle$ and $N_2=\langle[2]\rangle$. We have $$N_1\leq G,~~~N_2\leq\ H,~~~N_1\cong N_2$$ and $G/N_1\cong H/N_2$ but $G\ncong H$.

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How are you writing $N_2$ as a subgroup of $H$? If you mean there's a copy of $\mathbb{Z}_2$ (or $\mathbb{Z}_3$) with isomorphic quotient in $G$ and in $H$, but that $G$ and $H$ aren't isomorphic, that's correct, but your notation is getting in the way of it... –  Steven Stadnicki Mar 14 '13 at 18:27
@StevenStadnicki: Yes! I mean there is a copy of $N_1$ in $G$ and the same story for $N_2$ and $H$ as exactly you kindly noted. Thanks Steven for your time. –  B. S. Mar 14 '13 at 18:31
the problem is that the generator given for $N_2$ makes it hard to see how $N_2$ is a subgroup of $H$ - you've written it as though it were a subgroup of $G(=S_3)$, in permutation notation, but $\mathbb{Z}_6$ isn't a permutation group per se. –  Steven Stadnicki Mar 14 '13 at 18:36
@StevenStadnicki: I see what I made the mistake. I fixed it –  B. S. Mar 14 '13 at 18:39
I like your counterexample! +1 –  amWhy Mar 15 '13 at 1:29