Let H be a subgroup of G with index 2.
a. Prove that H is a normal subgroup of G.
b. Prove that g^(2) E H for all g E G.
There is a fairly well-known theorem which says that "if all the right cosets of a subgroup are left cosets of the subgroup then the subgroup is normal". So, notice that if $\displaystyle L_1,L_2$ are the left cosets and $\displaystyle R_1,R_2$ the right then $\displaystyle L_1\amalg L_2=R_1\amalg R_2$ (where that funny symbol just means union but the sets are pairwise disjoint). What now?
A more general theorem says that if $\displaystyle N\unlhd G$ then $\displaystyle g^{\left[G:N\right]}\in N$ for all $\displaystyle g\in G$. To see this think about $\displaystyle G/N$. What's it's order? what is $\displaystyle \left(gN\right)^{[G:N]}=?$b. Prove that g^(2) E H for all g E G.
For the first part:
for g E G
if g E H , then gH = H = Hg. Therefore H is a normal subgroup.
if g is not an element of H, G = eH U gH because there are only 2 cosets.
also G = He U Hg.
So G = H U Hg and G = H U gH (H = eH = He)
Therefore, gH = Hg. H is a normal subgroup.
I don't know how to prove part b. Sorry
That's lame. I wrote up a proof. Here's the basic idea. Break it into two cases
1. If $\displaystyle g\in N$ it's obvious what to do.
2. If $\displaystyle g\notin N$ then it has to be in the other coset, call it $\displaystyle hN$. And, so if $\displaystyle g^2\notin N\implies\cdots g\notin g^{-1}N\implies\cdots$