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Is every coset of a group closed under taking inverses?

What I mean is that if $G$ is a group and $H$ is a subgroup, and let $a$ be any element of $G$. Then the coset $aH$ is not necessary a group. But does every element of $aH$ have an inverse element also in $aH$?

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    $\begingroup$ No, just take the coset $\{1\}$ of $\{0\}$ in $\mathbf{Z}/n\mathbf{Z}$ for $n\ge 3$. In general, for a group $G$ and subgroup $H$, one has (exercise): every left coset $gH$ is closed under inversion iff $H$ is normal and $G/H$ is an elementary $2$-group. $\endgroup$
    – YCor
    Sep 4, 2019 at 14:28

2 Answers 2

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No, this does not have to hold.

Example: The symmetric group $S_3$ with 6 elements.

$U=\{\operatorname{id}, (12)\}$ is a subgroup.

Now view $(13)U=\{(13), (123)\}$. Then the element $(123)$ has no inverse.

We have $(123)(13)=(23)$ and $(123)(123)=(132)$

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No. $a\in aH$. But if $a^{-1}\in aH$, then $a^{-1}=ah\implies h=a^{-2}$. So $a^2\in H$.

So, for instance, consider the dihedral group, $D_4=\langle r,s\mid r^4,s^2, (rs)^2\rangle $.

Take $H\le D_4$ where $H=\{s,e\}$. Then $r^2\not\in H$.

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    $\begingroup$ Indeed, $a^{-1} \in aH$ iff $a^2 \in H$. $\endgroup$
    – lhf
    Sep 4, 2019 at 12:07

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