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1) Suppose $G_{1}$ and $G_{2}$ are groups with respective normal subgroups $N_{1}$ and $N_{2}$. Suppose $G_{1}/N_{1} \cong G_{2}/N_{2}$ and $N_{1} \cong N_{2}$. Does this imply that $G_{1} \cong G_{2}$?

2) Suppose $G/N \cong H$ and it is known that $N$ and $H$ are both finite. Does this imply that $G$ is finite?

Can't think of any counter examples. I'm trying to get some information about a group with only knowledge about it's subgroups and quotients.

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    @BabakSorouh Yea, I'm really tired; I figured at least one of them was obvious.2012-06-21

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The first question is false. Consider $G_1=\mathbb{Z/9 Z}$ and $G_2=\mathbb{(Z/3Z)}^2$. Take $N_1=N_2=\mathbb{Z/3Z}$.

The second question is indeed true for set theoretical considerations, $G$ is a finite union of $|N|$ many copies of a set of size $|H|$ (to see this fact note that if $\varphi$ is the surjective homomorphism from $G$ onto $H$ then all the fibers have the same size).

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Take $G_{1}=\mathbb{Z}_{2}\times\mathbb{Z}_{2},G_{2}=\mathbb{Z}_{4}$ and $N_{1}=N_{2}=\mathbb{Z}_{2}$ then $G_{i}/N_{i}\cong\mathbb{Z}_{2}$but $G_{1}\not\cong G_{2}$since $G_{1}$is not cyclic and $G_{2}$ is cyclic.

For your second question: $|G/N|=|H|$ but form lagrange $|G/N||N|=|G|\implies|G|=|N||H|$ hence it is also finite.

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    @BabakSorouh - just check out the proof. I don't have a reference2012-06-21