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 $n\ge2$, 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. $A_{2n+1}$?
Best Answer
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.