TITLE: Prove that $H$ is a normal subgroup of $G$
QUESTION [9 upvotes]: This is a problem from the book "Berkeley Problems in Mathematics":

Let $G$ be a group of order $120$, let $H$ be a subgroup of order $24$, and assume that there is at least one coset of $H$ (other than $H$ itself) which is equal to some right coset of $H$. Prove that $H$ is a normal subgroup of $G$.

To be honest, I couldn't make much progress. I would appreciate any kind of help. Thanks.

REPLY [2 votes]: As Daniel suggested in comments, if $H \neq aH = Hb$, then $aHb^{-1}=H \implies a.e.b^{-1}=h$ for some $h\in H\implies  ab^{-1}\in H \implies Hab^{-1}=H \implies Ha=Hb $.
Now  as $|G:H|=5$, we can make two sets of left and right cosets and name them as $L=\{H,a_1H,a_2H,a_3H,a_4H\}$ and $R=\{H,Hb_1,Hb_2,Hb_3,Hb_4\}$.
Now (WLG) Let $a_1H=Hb_1$, ($b_1$ may be identity too, easy case) but by above justification, $a_1H=Hb_1 \implies Ha_1=Hb_1 \implies a_1H=Ha_1 \implies a_1Ha_1^{-1}=H$.
Now this implies $a_1 \notin H$, lies in $N_G(H)$, so by lagranges theorem $N_G(H)=G$, and hence $H$ is normal in $G$