# $G_{1}/N_{1} \cong G_{2}/N_{2}$ and $N_{1} \cong N_{2} \Rightarrow G_{1} \cong G_{2}$?

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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I think you can find the counter examples for 1 among groups of order $4$. – Babak S. Jun 21 '12 at 6:24
For 1) see en.wikipedia.org/wiki/Group_extension . This is true if and only if a certain $\text{Ext}$ group is trivial. – Qiaochu Yuan Jun 21 '12 at 6:27
@BabakSorouh Yea, I'm really tired; I figured at least one of them was obvious. – Jackson Jun 21 '12 at 6:53

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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@Eugene it does not require that $G$ is finite – Belgi Jun 21 '12 at 6:32
@Eugene - Lagrange's theorem doe's not require the group to be finite, to put in words of "divides" you do need to say that the group is finite but if you will look at the proof you will see that (and it is written later in the link) that the bijection shows this is true for any cardinality. – Belgi Jun 21 '12 at 6:35
Give me the link I can see what you noted is true. Thanks – Babak S. Jun 21 '12 at 6:48
@BabakSorouh - just check out the proof. I don't have a reference – Belgi Jun 21 '12 at 6:49