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I saw an exercise that goes:

Let $G$ be a group of order 120, and $H$ be a subgroup of order 24. If at least one left coset of $H$ in $G$ is a right coset apart from $H$ itself, show that $H$ is normal.

Somehow I need to show that the remaining 3 left cosets are also right cosets. But I can't yet find a way to argue about that.

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up vote 2 down vote accepted

By assumption there exists an element $g \in G$ be such that $gH$ is a right coset of $H$, i.e. there is an element $g' \in G$ such that $gH = Hg'$. Then we also have $Hg=Hg'$ and thus $gH=Hg$, i.e. $g$ lies in the normalizer of $H$. Since 120/24=5 is prime, the normalizer of $H$ can either be $H$ or $G$ and since by assumption $g$ can be chosen outside of $H$, the normalizer of $H$ must be $G$ itself.

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Hint: Let $g$ be an element of order 5 in your group. Show every left coset is of the form $g^kH$. So in particular if $hH$ ($h=g^k$) the non-trivial coset which is left and right, all left cosets are of the form $h^kH$.

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An element $g$ of order 5 lies outside $H$ so the cosets are $Hg^i$, since one of these is also a left coset say $k=g^j$, $Hk=kH$, then the other cosets are $Hk^i=k^iH$, hence $H$ is normal.

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