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I have a question asking me to classify all groups of size 6 and below up to isomorphism. I can easily find out the answer, but what I'm interested in is how would one go about classifying all groups of a certain size up to isomorphism? The question then goes on to ask which of these are isomorphic to subgroups of $S_4$, which I'm stuck on, any help will be greatly appreciated!

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    $\begingroup$ General strategy is to use the Sylow theorems in some way. You can probably find some worked examples through Google. $\endgroup$ – Trevor Gunn Jun 10 '17 at 16:36
  • $\begingroup$ As another heuristic, the more complicated the prime factorization of a finite group, the more difficult it will be to classify groups of that order. $\endgroup$ – Duncan Ramage Jun 10 '17 at 17:06
  • $\begingroup$ @DuncanRamage I don't think I agree with that. The most difficult order are powers of primes, mainly because there are so many groups. Order $1024$ is the only order up to $2000$ for which the groups have not been completely classified. $\endgroup$ – Derek Holt Jun 10 '17 at 18:53
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    $\begingroup$ @DerekHolt I could have worded it more clearly, consider the powers of the primes to have a greater weight of difficulty in the prime factorization than the number of distinct primes. $\endgroup$ – Duncan Ramage Jun 10 '17 at 19:10
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Well, for the general question you can find (in some sense) your answer here https://en.wikipedia.org/wiki/Cayley's_theorem; every group of a certain size $n$ is a subgroup of $S_n$, and in the symmetric group you can find every finite group; this is not very practical actually, because classifying the subgroups of the symmetric group is certainly not a trivial task. In your case, you can take an $n-cycle \ \sigma$ in $S_4$ and the subgroups generated by $\sigma$ to obtain that $\mathbb Z/2\mathbb Z, \ \mathbb Z/3\mathbb Z \ $ and $ \mathbb Z/4\mathbb Z$ are subgroups of $S_4$; also $S_3$ is a subgroup of $S_4$ in a trivial way. $\mathbb Z/2\mathbb Z \times\mathbb Z/2\mathbb Z$ is the subgroup generated by $(1 2 )$ and $(34)$. On the contrary, $\mathbb Z/6\mathbb Z$ is not a subgroup of $S_4$; in fact it has an element of order 6, but no element with such order exists in $S_4$.

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