TITLE: An easy proof that S(n) does not embed into A(n+1)?
QUESTION [15 upvotes]: 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?

REPLY [33 votes]: 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 $n\ge 6$.

REPLY [7 votes]: 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.