Show that $H=\{x\in G:ax=xa\}$ has index 2 ($G$ contains 2 elements of order 3) 
Let $G$ be a group with exactly two elements $a,b$ that have order 3. Show that $H=\{x\in G:ax=xa\}$ is a subgroup of index 2. (Assume that $H\neq G$)

Since $a$ has order 3, we know that $a^2\neq e$ has order 3, so it follows that $b=a^2$. In a previous exercise, I showed that for each $x\in G$, we have $ax=xa$ or $ax=xb$. I also managed to show that $H$ is indeed a subgroup, so all that remains is to find $y\in G$ such that $G=H\cup yH$.
It holds that $ax=xa$ and $ax=xb$ cannot hold simultaneously. So we can split $G$ into two disjoint sets $H$ and $X$. Let $H'=X\cup\{e\}$. It's easy to show that $H'$ is a subgroup. How can I show now that for each $y\in G\setminus H$, we have $yH=H'$? It's clear to me that $yH\subset H'$. Or maybe I could show that $\vert H'\vert=\vert H\vert$. Any help would be appreciated.
 A: Let $N=\langle a\rangle$. Since $N$ is the unique subgroup of order 3 of $G$, it must be normal in $G$. Note that $H$ is exactly the centraliser of $N$ in $G$, so $H$ is normal in $G$ and by the N/C theorem, $G/H$ is isomorphic to a subgroup of $Aut(N)=Aut(C_3)\cong C_2$, so $|G/H|\leq 2$.
A: Since $H\ne G$ by assumption, fix $y\in G\setminus H$. Then $ay=yb$, as you showed; also $y^{-1}\notin H$, hence $ay^{-1}=y^{-1}b$, which implies $ya=by$.
Let $z\in G\setminus H$; then you have $az=zb$ and $za=bz$ as before, so
$$
ax=ay^{-1}z=y^{-1}bz=y^{-1}za=xa
$$
hence $x\in H$ and $z=yx\in yH$.
Thus $G=H\cup yH$.
A: Let $N = \langle a \rangle$. From another answer, we see that $N$ is normal, so $G$ acts on $N$ by conjugation.
We see that $H$ is the stabilizer of $a$. By the orbit-stabilizer theorem, $|G:H| = |\operatorname{Orb}(a)|$, where $\operatorname{Orb}(a) := \{gag^{-1} \mid g \in G\}$ is the orbit of $a$.
It is assumed that $H \ne G$, i.e. $|G:H| > 1$, i.e. $|\operatorname{Orb}(a)| > 1$.
We see that $e \notin \operatorname{Orb}(a)$: otherwise, if $gag^{-1} = e$, then $ga=g$, i.e. $a=e$, contradiction.
Well, $\operatorname{Orb}(a) \subseteq N \setminus \{e\}$, so $|\operatorname{Orb}(a)| \le |N \setminus \{e\}| = 2$.
In conclusion, $1 < |\operatorname{Orb}(a)| \le 2$, so $|\operatorname{Orb}(a)| = 2$, i.e. $|G:H| = 2$.
