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I have the form of Cayley's theorem that doesn't say any more than : for every finite group G there exists n such that Sn has a subgroup which is isomorphic to G. Now I'd be interested in knowing whether or not we know ways of finding the smallest such n?

For example, I've tried G = D10 and found 5, also for G = C5 I get 5. But that only uses Lagrange's Theorem and finding the actual subgroup... What about more complicated examples like C2xC2xC2? or S3xS3? or Z9?

Could you show me how you lead your reasoning on these?

Thanks a lot!

Timmy
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  • Typically a stronger statement of Cayley's Theorem is that $|G| = n$. I think that is the smallest $n$ that is always guaranteed, though for some specific cases, you might be able to find a smaller $n$ (not sure). – Kaj Hansen May 15 '14 at 19:42
  • Related to http://math.stackexchange.com/questions/36846/minimal-embedding-of-a-group-into-the-group-s-n – Jack Schmidt May 15 '14 at 19:43
  • I've read it was more common to find some n smaller than |G|... Do you have an insight on the examples above? – Timmy May 15 '14 at 19:44
  • Short answer: finding the exact minimum is hard, but some general bounds are easy. $n \leq |G|$, if we let $\mu(G)$ be the smallest such $n$, then $\mu(G \times H) \leq \mu(G) + \mu(H)$ and $\mu(C_{p^n}) = p^n$. – Jack Schmidt May 15 '14 at 19:45
  • Oh these are really usefull, thank you! It might weird but that means there is a subgroup of S6 isomorphic to S3xS3, I don't see which one... – Timmy May 15 '14 at 19:49
  • $\langle (1,2),(1,2,3) \rangle \times \langle (4,5), (4,5,6) \rangle$ works because $6$ has two copies of $3$ in it. – Jack Schmidt May 15 '14 at 19:56
  • This is another question related to minimal embeddings. – Alexander Gruber May 16 '14 at 00:48

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Finding subgroups of $S_n$ isomorphic to $G$ is the same thing as finding injective homomorphisms from $G$ to $S_n$.

Finding homomorphisms from $G$ to $S_n$ is the same as choosing conjugacy classes of subgroups $H_i$ of $G$ so that $n = \sum [G:H_i]$. The kernel of the homomorphisms is the intersection of the conjugates of all of the $H_i$.

If $G=H\times K$, then take $H_1 = H \times 1$ and $H_2 = 1 \times K$, and we get $H_1 \cap H_2 = 1 \times 1$, so the homomorphism defined by $\{H_1, H_2\}$ is injective, and $G$ is a subgroup of $S_n$ for $n = [G:H_1] + [G:H_2] = |K| + |H|$.

Repeating this with 3 groups shows that $C_2 \times C_2 \times C_2$ is isomorphic to a subgroup $S_6$ since $6=2+2+2$. By considering a Sylow 2-subgroup of $S_5$, one can see that $6$ is in fact the smallest possible $n$. An explicit subgroup is $\langle (1,2) \rangle \times \langle (3,4) \rangle \times \langle (5,6) \rangle$.

Similarly $S_3 \times S_3$ is isomorphic to a subgroup of $S_6$ since $6=3+3$. It is not isomorphic to a subgroup of any smaller $S_n$ since $S_5$'s order is not divisible by 36. An explicit subgroup is $\langle (1,2), (1,2,3) \rangle \times \langle (4,5),(4,5,6) \rangle$.

$Z_9$ is isomorphic to a subgroup of $S_9$, and by checking a Sylow 3-subgroup of $S_8$, one can see no smaller $n$ will work. An explicit subgroup is $\langle (1,2,3,4,5,6,7,8,9) \rangle$.

See https://mathoverflow.net/questions/48928/smallest-n-for-which-g-embeds-in-s-n for some higher level ideas.

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