If $G$ has no subgroup of index $2$ and $G\leq S_n$, then $G \leq A_n$.

Question

I am currently reading Abstract Algebra by Dummit & Foote. Discussing some techniques about the Sylow theorems they prove the proposition mentioned in the title (Proposition 12 (1), p. 204, 3rd Edition) by contraposition.

In order to prove the claim, the book starts by saying: If $G$ is not contained in $A_n$, then $A_n < GA_n$ so we must have $GA_n = S_n$.

I have two problems with this sentence:

Firstly, using contraposition I would start by saying: If $G$ is no subgroup of $A_n$. Why is this equal to "$G$ is not contained in $A_n$"?

Secondly, I do not understand why/how I can conclude the second part of the sentence ("then $A_n < GA_n$ so we must have $GA_n = S_n$").

Could someone please explain this sentence in a bit more detail?

Answer 1

1. Since $G$ is a subgroup of $S_n$, asserting that it is a subgroup of $A_n$ is the same thing as asserting that it is a subset of $A_n$. So, denyin that $G$ is a subgroup of $A_n$ is the same thing as denying that it is a subset.
2. The only subgroup $G$ of $S_n$ such that $A_n\varsubsetneq G$ is $S_n$ it self. So, since $GA_n$ is a subgroup of $S_n$ which strictly contains $A_n$, it is equal to $S_n$.

Answer 2

If $G\subsetneq A_n$ then $G\cap A_n$ is a proper normal subgroup of $G$. Then $G/(G\cap A_n)\cong GA_n/A_n\le S_n/A_n$. So $G/(G\cap A_n)$ is a subgroup of a group of order $2$,with fewer than $2$ elements, so $G\cap A_n$ has index $2$ in $G$, a contradiction.