Conjugacy classes of a subgroup of index 2. Let $H \leq G$ be a subgroup of index $2$ in $G$  and $K$ a conjugacy class of $G.$ 
How can one show that $K \cap H$ is either empty, a conjugacy class of $H$ or is split into two conjugacy classes of $H$ of equal order?
In particular why is it that $K \cap H$ gets split if and only if some $k \in K$ commutes with some $g \not \in H$?
 A: In this answer I denote conjugation by a dot: $g\cdot x$ means $gxg^{-1}$.
Since $H$ is a normal subgroup of $H$ with Abelian quotient, all orbits $H\cdot x$ in $G$ under conjugation by $H$ are contained in one of the (two) cosets in $G/H$. Now if $g_1\in G-H$ is arbitrary, it is easily seen that any conjugacy class $K=G\cdot x$ is given by $K= H\cdot x\cup H\cdot(g_1\cdot x)$, where the two orbits being united are either disjoint or equal. Depending on the number of distinct orbits among $H\cdot x$ and $H\cdot(g_1\cdot x)$ that lie inside $$g_1\in G-H$H$, which number can be $0$, $1$, or $2$, the intersection $K\cap H$ consists of $0$, $1$, or $2$ conjugacy classes in$~H$. (Also the number is $0$ iff $x\notin H$.)
When there are two of them, conjugation by $g_1$ defines a bijection between those two conjugacy classes.
As for the final statement, it should read "$K \cap H$ is a $H$-conjugacy class (so it does not get split) if and only if $K\subseteq H$ and some $k\in K$ commutes with some $g \notin H$". The condition that $k\in K\cap H$ commutes with $g \notin H$ means that for some $g_1\in G-H$ one has $H\cdot k=H\cdot(g_1\cdot k)$: if the condition holds, one can take $g_1=g$; conversely if $h\cdot k=g_1\cdot k$ then $h^{-1}g_1\notin H$ commutes with $k$. So in this case the union above is really just one $H$-orbit. Then either $K\cap H=\emptyset$ or $K\cap H=K$. In fact when $K\cap H=\emptyset$ we are always in this situation (since one can take $g=k$) but when $K\subseteq H$ it depends. The above classification is a bit misleading; more explicitely one of the following happens:


*

*$K\cap H=\emptyset$ and $K$ is one $H$-conjugation orbit (but not of course a conjugation class of$~H$).

*$K\subseteq H$ and some $k\in K$ commutes with some $g\notin H$; then $K$ is a single $H$ conjugation class.

*$K\subseteq H$ but no $k\in K$ commutes with any $g\notin H$; then $K$ splits into two $H$ conjugation classes.

