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Let $G$ be a group such that $|G| = 2012$, how would you classify, up to isomorphism, all groups $G$?

Clearly $2012 = 503 \times 2 \times 2$ and so $G \cong C_{503} \times C_2 \times C_2$ but how would you find the others?

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I would begin by figuring out how many Sylow 503-subgroups there are :-) Or, if you haven't covered the theory of Sylow subgroups yet, I would show that any such group has a normal subgroup of order 503. – Jyrki Lahtonen Jun 10 '12 at 9:20
Don't forget $C_{503}\times C_4$... – Chris Gerig Jun 10 '12 at 9:39
Jyrki Lahtonen, how do you figure that out? We have that if $p^nq = |G|$ where $p$ doesn't divide $q$ and $n$ is as large as possible then the number of Sylow $p$ subgroups satisfy $1\mod p$ so there are $1\mod 503$ subgroups? – user26069 Jun 10 '12 at 9:55
But doesn't $1 \mod 503$ and $2\mod 503$ both divide 4? – user26069 Jun 10 '12 at 11:07
@morphism, the number of Sylow 503-subgroups, call it $n_{503}$ must satisfy $$n_{503}\equiv 1\pmod{503}$$ and $$n_{503}\mid 2012.$$ The first congruence shows that $503\nmid n_{503}$, so the second criterion simplifies to $n_{503}\mid 4$ (often the Sylow theorems are stated to say this, because we can always make this same deduction). So $n_{503}$ is either 1, 2 or 4. Only the choice $n_{503}=1$ satisfies the first congruence. – Jyrki Lahtonen Jun 10 '12 at 12:21

First: there are $\,2\,$ non-isomorphic abelian groups of order $\,2012\,$, which have already been mentioned by you and Chris.

Second: as Jyrki mentioned, if $\,|G|=2012\,$ then $\,G\,$ always has a normal sbgp. $\,P\,$ of order $\,503\,$, so $\,G\,$ is always an extension of such a sbgp. Since $\,|\operatorname{Aut}(P)|=502=2\cdot 251\,$ , we have at least two possible such extensions. Putting $\,P:=\langle p\rangle\,\,,\,C_4:=\langle c\rangle\,\,,\,C_2\times C_2:=\langle a,b\rangle$:$$(i) P\rtimes C_4\,\,,\,\text{with homomorphism}\,\,\,C_4\to\operatorname{Aut}(P)\,\,\,\text{defined by}\,\,\,p^c:=p^{-1}$$ $$(ii)P\rtimes\left(C_2\times C_2\right)\,\,,\,\text{with hom.}\,\, C_2\times C_2\to\operatorname{Aut}(P)\,\text{defined by}\,\,p^a:=p^{-1}\,,\,p^b:=p$$

The other "obvious" action of $\,C_2\times C_2\,$ on $\,P\,\,:p^b=p^{-1}\,,\,p^a=p\,$ gives us a semidirect product isomorphic with (ii) above, as we've an automorphism of the Klein group mapping each generator into the other one.

Thus, we've $\,4\,$ non-isomorphic groups of order $\,2012$

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You should mention why every such extension is split. – user641 Jun 11 '12 at 16:32
I don't think it is necessary at all in this particular kind of problems, and even if it were: let the OP deal with after getting an answer to his/her question – DonAntonio Jun 11 '12 at 16:46
How is it not necessary? You can't claim you've found every group of order 2012 without it! – user641 Jun 11 '12 at 17:26
Of course I can when I point out that every such group has a normal sbgp. of order 503. Whether I want to remark or not that then I can always form the semidirect product of this sbgp. with a Sylow $2$sbgp. of order $4$ is a matter of discussion and/of taste: for a first approach I think this is unnecessary in a not-too-long-and-sketchy answer in this site. If I were in the university I'd either show the whole thing or else leave to the instructor to write down some exercises on this...which, in fact, is what happens. – DonAntonio Jun 11 '12 at 18:07

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