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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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    I think you can find the counter examples for 1 among groups of order $4$.2012-06-21
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    For 1) see http://en.wikipedia.org/wiki/Group_extension . This is true if and only if a certain $\text{Ext}$ group is trivial.2012-06-21
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    @BabakSorouh Yea, I'm really tired; I figured at least one of them was obvious.2012-06-21

2 Answers 2

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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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    @Eugene it does not require that $G$ is finite2012-06-21
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    @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.2012-06-21
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    Give me the link I can see what you noted is true. Thanks2012-06-21
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    @BabakSorouh - just check out the proof. I don't have a reference2012-06-21