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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.

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    $\begingroup$ If $H \neq aH = Hb$, then note that $Hb = Ha$ follows. Now think of $N_G(H)$. $\endgroup$ – Daniel Fischer Jul 10 '15 at 21:36
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A subgroup of order $24$ is maximal in a group of order $120=5\cdot 24$, meaning that if we add even one more element we get the whole group. Thus it suffices to find an element not in $H$ that normalizes $H$, because in that case the normalizer properly contains $H$ and hence is the whole group.

Suppose $a\notin H$ and $aH=Hb$. Then $b\in aH$, hence $b=ah$ for some $h\in H$. Thus $bh^{-1}=a$, so $a\in bH$, hence $aH=bH$. Thus $b^{-1}a\in H$. Since $aH=Hb$, $H=b^{-1}aH=b^{-1}Hb$. Thus $b^{-1}$ is an element not in $H$ that normalizes $H$, so we are done.

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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 $G$

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