$h\in H$ group $\rightarrow hH=Hh$?

Question

Probably this is pretty simple (or even trivial), but I'm stucked.

If $H\leq G$ is a subgroup, does it follow that $hH=Hh$, if $h\in H$ ? I can't prove or find a counter-example. If anyone could help me, I'd be grateful!

Answer 1

They are Always equal to $H$, then ir Always holds

Answer 2

$$hH=\{hh' \mid h' \in H\}, \color{red}{H}h = \{h'h \mid h' \in H\}$$

Note that $hh' = \color{red}{hh'h^{-1}}\cdot h$, so $hH \subseteq Hh$. A similar trick shows the reverse inclusion.

Answer 3

If $h\in H$, $hH=Hh$ because both are equal to $H$.

Indeed, by definition of a subgroup, $hH\subset H$. Conversely, if $k\in H$, we can write $\; k=h(h^{-1}k)\in hH$, so $H\subset hH$. Similarly, one checks $H=Hh$.

Answer 4

If $H\leq G$ is a subgroup, then for any $h,h' \in H$ you have that $hh'\in H$.

Therefore $hH=H=Hh$. To see this, let's see the double inclusion:

An element of $hH$ is of the form $hg$ with $g\in H$. Since $H$ is a subgroup, then $hg \in H$.

The other way around, take $h' \in H$ you want to see that $h' \in hH$. Consider $h'=h(h^{-1} h') \in hH$