I'm trying to solve this question from Hungerford's book:
The lemma 6.11 says
Let $T$ be the subgroup of index $2$, if it generates $A_n$, then $A_n\subset T$, thus by Lagrange theorem: $[S_n:A_n]|A_n|=[S_n:T]|T|$, which implies $|A_n|=|T|$, so $A_n=T$ (why formally? pigeonhole principal?)
So the solution follows easily using the hint, but I couldn't prove why every subgroup of index $2$ must contain all $3$-cycles of $S_n$. What I know is every $3$-cycle has order $3$ and I'm trying to find something using this fact and the Lagrange theorem, without any success.
I need help.
Thanks in advance
Choose any $3$-cycle $\alpha \in S_n$. We want to show that $\alpha \in T$. To see this, we argue by contradiction. Suppose instead that $\alpha \notin T$. Then since $T$ has index $2$, we know that $S_n = T \cup \alpha T$ with $T \cap \alpha T = \emptyset$. Now consider the element $\alpha^2 \in S_n = T \cup \alpha T$. It can only land in two places:
Case 1: Suppose that $\alpha^2 \in T$. Then since $\alpha$ has order $3$, we know that $\alpha^{-1} = \alpha^2 \in T$. But groups are closed under inverses, so $\alpha = (\alpha^{-1})^{-1} \in T$, a contradiction.
Case 2: Suppose that $\alpha^2 \in \alpha T$. Then there is some $\beta \in T$ such that $\alpha^2 = \alpha\beta$. But then left cancellation by $\alpha$ gives us that $\alpha = \beta \in T$, a contradiction.
Thus, we conclude that $T$ contains all $3$-cycles of $S_n$, as desired. $~~~\blacksquare$