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Cayley's theorem tells us that every group is isomorphic to a permutation group.

I am trying to study the gap(s) in this theorem and I found that the drawback of this classical theorem is that even for small groups the size of the permutation group is huge.

Are there any more drawbacks of this theorem?

Please let me know if I am wrong. Thanks in advance!

Shaun
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Learning
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  • It's a little unclear to me, but I think you might be misunderstanding the phrase "is isomorphic to a permutation group". That phrase should be read as an abbreviation of the longer phrase "is isomorphic to some subgroup of the full permutation group of some set". One way or another, it's very unclear what you mean by a "drawback", and the highlighted question is very vague. – Lee Mosher Nov 16 '19 at 17:17
  • @LeeMosher by drawback, I meant that we will get very huge size of permutation groups to whose subgroup the given group will be isomorphic when the size of the given group will be large. – Learning Nov 16 '19 at 17:39
  • @освящение, as said in my answer, that shouldn't be considered a drawback as such, but I agree that it pushes to look to "sharper" embeddings, namely into some $S_{m<n}$. –  Nov 16 '19 at 17:55

2 Answers2

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[E]ven for small groups the size of the permutation group is huge.

I think you've made a mistake.

Isomorphisms preserve the cardinality of the underlying sets of the groups in question; that is, if $\varphi: G\to H$ is an isomorphism, then $$\lvert G\rvert=\lvert H\rvert$$ because $\varphi$ is a bijection.

Shaun
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    I think there may be a confusion as to what "permutation group" means - all of $S_n$ or a subgroup of it. – Wojowu Nov 16 '19 at 15:24
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    @Shaun, I guess the OP refers to the fact that the isomorphic image of $G$ is "lost" in such a "wide" container as $\operatorname{Sym}(G)$ actually is as soon as the order of $G$ rise up. –  Nov 16 '19 at 15:27
  • @Wojowu I am talking about the groups $S_n$. – Learning Nov 16 '19 at 16:13
  • @Luca Could you please elaborate this fact about isomorphism and write an answer? – Learning Nov 16 '19 at 16:14
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    @освящение In that case, the Cayley's theorem as you wrote it ("every group is isomorphic to a permutation group") is not true. – Wojowu Nov 16 '19 at 16:20
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As pointed out in the comments, Cayley's Theorem states that any finite group $G$, of order $n$, is isomorphic to a subgroup of $S_n$ (not to $S_n$, if only for the fact that $n<n!$ as soon as $n>2$). In fact, $G$ embeds (=injective homomorphism) into $S_n$ via left multiplication, so $G$ is isomorphic to its image via such homomorphism, which in turn is a subgroup of $S_n$.

So, though it's not a problem that the codomain of such embedding ($S_n$) actually "inflates" as $n=|G|$ rises up, I think it's legitimate to ask whether embeddings of $G$ into $S_{m<n}$ (say "sharper") do exists. Have a look at this (it's a duplicate, but in its OP there is already an example of such a "sharpening"; therein linked post works out thoroughly the matter).