Dgdrxrt

I'm working through Rotman (1994). It's the 2nd chapter where my curiosity lead me to a question that is not from the book. Here's the problem that inspired the question:

2.8. Imbed $S_n$ as a subgroup of $A_n+2$, but show, for $nge2$, that $S_n$ cannot be imbedded in $A_n+1.$

Since the alternate subgroup has the property of even parity, I thought that a little investigation of parity for the second and third symmetric groups was in order. So I constructed a table for each of them, grouped by even and odd number of disjoint products.

I'll forego writing the actual table and simply provide the answers. In symmetric groups with an even idegree, two even permutations come together under composition to produce an even permutation. Two odds also make an even. Two different parities produce odd elements.

When the symmetric group has an odd degree, e.g. $S_3$, two odds make an odd, two evens make an odd, and two different parities make an even element. Exactly the opposite results from the symmetric groups with even index.

Two questions: is this enough to prove the $A_n+1$ case? And can this be generalized modulo 2 e.g. $A_2n+1$?

I don't think you understand even and odd permutations. Odd * odd or even * even is even, odd * even or even * odd is odd. It doesn't matter what $n$ is.

You can embed $S_n$ in $A_n+2$ by the map $f$ where $f(sigma) = sigma$ if $sigma$ is even and $f(sigma) = sigma cdot (n+1,n+2)$ if $sigma$ is odd.

You are not computing the parity correctly. In any symmetric group, the product of two even permutations is an even permutation (otherwise, $A_n$ would not be a subgroup!), the product of two odd permutations is an even permutation; and the product of two permutations of different parity is an odd permutation.

Instead, let's think about $S_3$, and how to embed it (or be unable to embed it) in $A_4$ and $A_5$.

$$S_3 = mathrmId, (1,2), (1,3), (2,3), (1,2,3), (1,3,2).$$
Now, $A_4$ doesn't have any subgroups of index $2$: they would be normal, and the only normal subgroups of $A_4$ are the trivial group, they whole group, and the subgroup $mathrmId, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)$ of index $3$. So $S_3$ cannot possibly embed into $A_4$.

What about $A_5$? There we have more leeway: there is no problem in mapping $(1,2,3)$ and $(1,3,2)$ to $A_5$: they have to map to $3$-cycles, so up to an automorphism of $A_5$ we may assume that we send $(1,2,3)$ to itself, and therefore $(1,3,2)$ to itself as well. We cannot just map $(1,2)$ to itself, though: $(1,2)notin A_5$. However, we can "fix" the parity by multiplying by a single that won't mess up the action on $1,2,3$ (so that we still get that the product of the image of $(1,2)$ and the image of $(1,3)$ is $(1,3,2)$, as needed, since $(1,2)(1,3)=(1,3,2)$). Just take $(1,2)(4,5)$! Similarly, try sending $(1,3)$ to $(1,3)(4,5)$, and $(2,3)$ to $(2,3)(4,5)$. Check that everything works out.

You may have to do the small $n$ cases separately, but after $n=3$, it should be simple to see how to get this done.

Added. Moreover, since $S_nhookrightarrow S_n+1$ (as the subgroup that fixes $n$), and this embedding necessarily respects parity, you have that $S_n$ embedds into $S_k$ for all $kgeq n$, and $A_n$ embeds into $A_k$ for all $kgeq n$. Added to the exercise, this shows that:

Proposition. Let $n$ be a positive integer. Then $S_n$ embeds into $A_k$ if and only if $n=1$ and $k$ is arbitrary; or $ngt 1$ and $kgeq n+2$.

For your title question, the answer is "yes" (if $ngt 1$, then $2n+1gt n+2$; if $n=1$, then $S_n$ is trivial and embeds into every $A_n$, in particular into $A_2$, which is also trivial).

More interesting is the question of when $S_n$ embeds into $A_k$ as a maximal subgroup. The pairs $(n,k)$ with this property were completely characterized by Bret Benesh in A classification of certain maximal subgroups of alternating groups, in Computational Group Theory and the Theory of Groups, Contemporary Mathematics vol. 470, American Mathematical Society, pages 21-26.

That $S_n$ does not embed into $A_n+1$ for $nge 2$ is not proved in this thread (at this time), but a proof can be found at this MO post. Since this post has been used to close subsequent occurrences of this question as duplicate, I'm adding this cw answer for completeness.

For $n$ even it's clear since then $|S_n|$ does not divide $|A_n+1|$. In general, one proof essentially consists in showing that, for $nge 4$ the centralizer of any element of order $2$ in $A_n+1$ is smaller than that of a transposition in $S_n$, which requires a little argument. Alternatively one uses that the image would have index $(n+1)!/2<n+1$, which contradicts simplicity of $A_n+1$ when $nge 4$. For $n=3$, using that $mathrmHom(S_3,C_3)=1$ while $mathrmHom(A_4,C_3)neq1$, one sees that the image of such an embedding would be contained in the normal subgroup of index 3 of $A_4$, of order 4, to get a contradiction.