# Is every coset of a group closed under taking inverses?

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 $$aH$$?

• 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.
– YCor
Sep 4, 2019 at 14:28

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 $$(123)(123)=(132)$$

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 $$r^2\not\in H$$.

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