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Rotman's book An Introduction to the Theory of Groups (Fourth Edition) asks, on page 22, Exercise 2.8, to show that S(n) cannot be embedded in A(n+1), where S(n) = the symmetric group on n elements, and A(n) = the alternating group on n elements. I have a proof but it uses Bertrand's Postulate, which seems a bit much for page 22 of an introductory text. Does anyone have a more appropriate (i.e., easier) proof?

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There is no such embedding for even $n$, just consider the orders of the respective groups: you don't have $|S{n}|$ dividing $|A{n+1}|$ by comparing the order of exponents of $2$. – Olivier Bégassat May 15 '11 at 21:55
By the way, you got your title wrong! – Olivier Bégassat May 15 '11 at 22:10
@Arturo: No, that's not really the rules of this forum. MathOverflow is mainly geared towards research-level mathematicians, and so any proof is fair game. That said, what you've highlighted is that this question probably isn't appropriate for MO; it could easily be closed as "too localized", which is our closest approximation to "homework-level". I would rather Len just accept Darij's answer below. Conversely, Len already says he has a proof. – Theo Johnson-Freyd May 15 '11 at 23:05
@Theo: I should have said "is not the intended answer by Rotman" rather than "cannot be used". – Arturo Magidin May 15 '11 at 23:07
Someone should add the restriction of n > 1. Gerhard "Ask Me About System Design" Paseman, 2011.05.15 – Gerhard Paseman May 16 '11 at 6:42

I think the following is sufficiently elementary: a transposition in $S_n$ is an element of order 2 commuting with at least $2(n-2)!$ elements of the group. But $A_{n+1}$ does not have such an element if $n$ is large enough. Indeed, if $\sigma\in A_{n+1}$ is of order 2, then it is a product of $k$ independent transpositions where $k$ is even and $2\le k\le(n+1)/2$. The number of elements of $A_{n+1}$ commuting with such $\sigma$ equals $2^{k-1}k!(n+1-2k)!$, and this is smaller than $2(n-2)!$ provided that $n\ge 6$.

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I think this is solved on .

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Thanks, that's a perfectly fine proof. And yet ... it uses concepts that Rotman has not yet introduced at page 22. – Len Schrieber May 15 '11 at 22:48
@Len: then you might want to be more precise about what concepts Rotman has introduced by page 22. – Qiaochu Yuan May 16 '11 at 4:39
I wouldn't be surprised if this is just another mistake in Rotman. – darij grinberg May 16 '11 at 9:31

One could ask Rotman. It may be that in a reorganization of the material in the book that problem ended up earlier than the material needed for the (intended) answer. On the other hand it is not a bad experience for students to see problems where the complete solution seems slightly out of reach. Here, one can prove several small cases and see various potential directions for a general proof. Which will work? which are in the spirit of the subject? Of course it is best to set up the expectation that there might be problems like this.

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I'd just like to add that it's true enough that this question is not quite appropriate for MO; but still, judging from the number of views it has received, even serious mathematicians like to relax once in a while with an elementary problem! – Len Schrieber May 18 '11 at 19:35

Of course this is not a research level question, and so is not appropriate for MO, but I remember being puzzled myself about what proof Rotman had in mind for this. I think we had better assume Lagrange's Theorem or it will be completely hopeless! Perhaps the proof using Bertrand's Postulate was intended, because students might expect to have heard of that, even if they have not read a proof?

Let's spell that out. As already noted, we can assume $n+1 = 2m$ is even by Lagrange. If $S_n$ embeds into $A_{n+1}$, then the index of the image of the embedding is $m$, so there is a nontrivial homomorphism (multiplicative action on cosets) $\phi: A_{n+1} \rightarrow S_m$.

By BP, there is a prime $p$ with $m < p < n+1$, so $p$ does not divide $|S_m|$. Hence all elements of order $p$ lie in ${\rm Ker}(\phi)$, including $g = (1,2,\ldots,p-1,p)$ and $h = (1,2,\ldots,p-1,p+1)$. Then $g^{-1}h$ is a 3-cycle ( $(1,p,p+1)$ if you multiply permutations left to right), so ${\rm Ker}(\phi)$ contains all 3-cycles, which generate $A_{n+1}$, contradicting the nontriviality of $\phi$.

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You don't need Betran's Postulat for that argument: Small n can be done by hand and for $n\geq 4$ the group $A_{n+1}$ is simple, so that the homomorphism $A_{n+1}\to S_m$ must be injective. Now $m! < \frac{(n+1)!}{2}$ and that's the contradiction. – Johannes Hahn May 16 '11 at 10:27
Yes but the whole point was to avoid using the simplicity of $A_{n+1}$, which has certainly not been covered by Page 22 of Rotman's book. – Derek Holt May 16 '11 at 10:34
That $A_{n+1}$ is generated by the 3-cycles - has that been established by page 22? – Gerry Myerson May 16 '11 at 12:29
@Gerry: Actually, that is the immediately previous exercise to the one in question. (The one Len is asking about is 2.8; proving that $A_n$ is generated by 3-cycles for $n\gt 2$ is Exercise 2.7). – Arturo Magidin May 16 '11 at 16:31
@Derek: For that matter, cosets are not defined until page 24, right after this set of exercises, and Lagrange appears in page 26. The immediately previous exercise consists of showing that the 3-cycles generate $A_n$ for $n\gt 2$, so perhaps the argument Rotman has in mind consists of showing somehow that any putative image of $S_n$ in $A_{n+1}$ would necessarily contain all $3$-cycles. – Arturo Magidin May 16 '11 at 16:53

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