$H \le G , h \in H \implies h^{-1}H = H$ Suppose $H$ is a subgroup of $G$
In a proof, I come across this:
If $h \in H$, then  $h^{-1}H = \{h^{-1}k | k \in H\} = H$
why is $h \in H$ required or is this unnecessary since it should be implies by $H \le G$?
I think I see intuitively why this holds but am not sure how to clearly explain it. My thinking is that since $h^{-1}h_1 = h^{-1}h_2 \implies h_1 = h_2$ and $H$ is closed under multiplication, $h^{-1}H$ must preserve all the values of $H$ so that the implication must follow but this seems inexact.
How might this be more clearly expressed?
 A: In general, $\;xH=H\iff x\in H\;$ , so if $\;x\notin H\;$ then $\;x,\,x^{-1}\notin H\;$ and then $\;x^{-1}H\neq H\;$
A: $H$ is closed under multiplication means that if you multiply two elements that lie in $H$, then their product will lie in $H$.
But of course $H$ cannot be closed under multiplication by some element outside $G$, as if $g \notin H$, then $gH\ni g\circ e=g\notin H$ thus $gH$ cannot be a subset of $H$.
In Ring theory there exists objects satisfying this closedness under multiplication, for example ideals.
A: Prove $h^{-1}H \subseteq H$ and $H \subseteq h^{-1}H$.
$h^{-1}H \subseteq H$ is direct because $h^{-1} \in H$ and $H$ is a subgroup.
Now, to prove $H \subseteq h^{-1}H$, take $h' \in H$ and notice that $h'=(h^{-1}h)h'= h^{-1}(hh')$ and $hh'$ is in $H$, so $h' \in h^{-1}H$
A: For $h\in H$, define the map $\mu_h\colon H\to H$ by $\mu_h(x)=hx$. This is indeed a map into $H$ (here we use the fact that $a,b\in H$ implies $ab\in H$).
By definition, $hH$ is the image of $\mu_h$. If we prove that $\mu_h$ is bijective, we are done. Indeed, the map $\mu_{h^{-1}}$ is the inverse of $\mu_h$ (here we use the fact that $a\in H$ implies $a^{-1}\in H$): if $x\in H$,
$$
\mu_{h}\circ \mu_{h^{-1}}(x)=h(h^{-1}x)=x
\qquad
\mu_{h^{-1}}\circ \mu_{h}(x)=h^{-1}(hx)=x
$$
Therefore $hH=H$ and $h^{-1}H=H$ for all $h\in H$.
