# Rotman's exercise 2.8 “$S_n$ cannot be imbedded in $A_{n+1}$”

This question is about the (in)famous Rotman's exercise 2.8 in "An Introduction to the Theory of Groups." I've searched and found similar questions here and in MO, but none of them contains a valid proof. (Does $S_n$ belong as a subgroup to $A_{2n+1}$?)

According to Rotman, a valid proof can only use the concepts introduced up to this exercise: cycle permutations, factorization of permutations, odd and even permutations, semigroups, groups, homomorphism and subgroups. Cosets, Lagrange's theorem, normal subgroups, and so on are not yet introduced. I stress this point because all of the proofs I've seen use Lagrange or actions, on cosets.

Now my attempt is to use exercise 2.7 (solved) which is about a proof that $A_n$ ($n>2$) is generated by all the $3$-cycles and exercise 2.4 (solved) " if $S$ is a proper subgroup of $G$ then $\langle G \setminus S\rangle=G$ " in this way:

Suppose that for every $\phi : S_n \to A_{n+1}$ imbeddings, all the $3$-cycles are contained in $\operatorname{Im}\phi$, then the assertion is proved by absurd. But I can't find a way to prove if it is possible or either find a counterexample to this kind of approach.

If someone has another proof which use only basics concepts is well accepted of course, but I mainly need some hints about correctness or not of my reasoning and how to proceed if it is correct. Thank you in advance.

• I am not very sure about the way i am following as you said i can not use more group theory but, Suppose $S_n$ is in $A_{n+1}$, what is the order of $S_n$ and what is the order of $A_{n+1}$... does $n!$ divides $\frac{(n+1)!}{2}$ in general.... – user87543 Aug 18 '13 at 14:02
• Have you tried counting the number of elements of order $2$ in both $S_n$ and $A_{n+1}$? The computations are rather cumbersome, but it requires no knowledge of group theory. – Inactive - avoiding CoC Aug 18 '13 at 19:51
• @user84559: I thought counting elements of order 2 might work too, but it doesn't. To save doing the combinatorics here, oeis.org/A000085 and oeis.org/A000704 give the relevant counts and the only case where $S_n$ has more elements of order 2 than $A_{n+1}$ is when $n = 2$. – Rob Arthan Aug 19 '13 at 16:29
• Isn't there a result that the largest subgroup of $A_{n+1}$ is $A_n$? I may be mistaken. – alex.jordan Aug 19 '13 at 20:49
• Here is a link to an answer at MO: mathoverflow.net/questions/65083/… – alex.jordan Aug 19 '13 at 20:57

If $n$ is odd there exists a subset $S\subset S_n$ of pairwise commuting elements of order $2$ with $|S|=\tfrac{n-1}{2}$. For example $$\{(1\ 2),(3\ 4),\ldots,(n-2\ n-1)\}.$$ An element $\sigma\in A_{n+1}$ of order $2$ is necessarily a product of an even number of disjoint $2$-cycles. Therefore a subset $A\subset A_{n+1}$ of pairwise commuting elements of order $2$ cannot contain more than $\tfrac{n+1}{4}$ elements. Hence for an embedding $S_n\ \hookrightarrow\ A_{n+1}$ to exist we must have $$\frac{n-1}{2}\leq\frac{n+1}{4},$$ or equivalently $n\leq3$. At this point it should not be hard to verify that $S_3$ does not embed into $A_4$, and of course that $S_1$ does embed into $A_2$.
A similar argument can be made when $n$ is even, to find that for an embedding $S_n\ \hookrightarrow\ A_{n+1}$ to exist we must have $$\frac{n}{2}<\frac{n+1}{4},$$ or equivalently $n<1$, immediately yielding a contradiction.
• From your comment I conclude that you realized in meanwhile that $A_4$ has an elementary abelian subgroup of order $4$, i.e., two commuting elements of order $2$. – j.p. Jul 2 '15 at 10:52