TITLE: Suppose that $[G:H]$ is a prime integer, and that $g \notin H$. Prove that H is normal in G.
QUESTION [5 upvotes]: Let H be a subgroup of a group G. Let $k,g \in G$ such that $gH = Hk$.
Suppose further that $[G:H]$ is a prime integer, and that $g \notin H$. Prove that H is normal in G.
I have totally no idea at all how to do this question. Can someone tell me how to start?

REPLY [4 votes]: If there are $g,k\in G$ with $g\notin H$ and $gH = Hk$, then for this particular $g$ we have $gH = Hg$, or equivalently $gHg^{-1} = H$. This follows since under these hypotheses $g \in Hg \cap Hk = Hg \cap gH$, and two right cosets are either disjoint or equal.
Thus for the normalizer
$$N_G(H) = \{x\in G : xHx^{-1} = H\} \tag{$\ast$}$$
we know $g \in N_G(H)$, and consequently $H \subsetneqq N_G(H)$. Since
$$[G:H] = [G : N_G(H)] \cdot [N_G(H) : H]$$
is prime and $[N_G(H) : H] > 1$ by $(\ast)$ it follows that $[G : N_G(H)] = 1$, i.e. $N_G(H) = G$, in other words $H$ is normal.