An easy proof that $S(n)$ does not embed into $A(n+1)$? 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?
 A: 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 $\phi$.
A: 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.
A: 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 $n\ge 6$.
A: Here is a short proof that follows from Rotman's material: Automorphisms of $S_m$ are all inner unless $m=2 \text{ or } 6$. It is not hard to use this to show that for $m\neq 2, 6$ subgroups of $S_m$ of index $m$  are in the form of $\{\sigma\mid \sigma(i)=i\}$ where $i\in\{1,\dots,m\}$ (see this stackexchange post).
Now a subgroup $H$ of $A_{n+1}$ isomorphic to $S_n$ yields an index-$(n+1)$ subgroup of $S_{n+1}$. Hence, except for $n=5$, $H$ must coincide with a stabilizer of the action $S_{n+1}\curvearrowright\{1,\dots,n+1\}$. But all such subgroups contain odd permutations, a contradiction. It remains to show that $S_5$ cannot be embedded in $A_6$. For this, notice that the former has elements of order six while the latter doesn't.
A: I think this is solved on http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=333049 .
A: Using concepts that Rotman have introduced before that exercise, I think we could provide the following counterexample: if $S_2$ could be embedded in $A_3$, then in $A_3$ there would be a subgroup $H$ with two elements. Such subgroup must contain $1$, so it must contain only one among $(1\ 2\ 3)$ and $(1\ 3\ 2)$; but if it contains one of them, it must contains the other one, as a power of the first element. So $A_3$ can not contain any subgroup with two elements, and $S_2$ can not be embedded in $A_3$.
