TITLE: Given $H\subseteq G$ with $G$ a group, then $HH^{-1} =H \implies aH = H \ \forall a \in H$.
QUESTION [5 upvotes]: I'm solving a problem which deals with equivalent definitions of a subgroup. The problem states that:

Given a subset $H \neq \emptyset$ of a group $G$, prove that the following are equivalent:

$H$ is a subgroup of $G$.
$H = H^2$ and $H^{-1}=H$
$HH^{-1}=H$
$aH = H, \quad  \forall a \in H$


where $XY = \left\{xy \ \vert \ x \in X, y \in Y\right\}$, $H^2 =HH$, $H^{-1} = \{h^{-1} \ |\  h \in h\}$ and $aH = \{ah \ | \ h \in H\}$.

I've seen that in problems of this type a common strategy is to make an "implication chain", i.e., show that $1 \implies 2, 2 \implies 3, 3 \implies 4, 4\implies 1$ and this proves equivalence between all the statements. I attempted to follow this idea in my proof. I've managed to show that $1 \implies 2$ and $2 \implies 3$, however, I got stuck on the proof that $3 \implies 4$.
My attempt went something like this. We need to show that given any $x \in H$ and $y \in H^{-1}$ we can write $xy = ah$ for some $h \in H$. The problem is that I couldn't find a way to relate $y \in H$ without using that $H$ is a subgroup (which is what I want to prove, so I can't assume this).
I've looked at proofs of $1 \implies 4$ like the one here, but every proof I've seen explicitly uses the fact that $H$ is a subgroup to assert that $h^{-1} \in H$, but in $3 \implies 4$ I cannot assume that, only that $h^{-1} \in G$.
Does anyone know how I could prove $3 \implies 4$?
Thank you!

REPLY [3 votes]: You are assuming $HH^{-1}=H$. We want to show that if $a\in H$ then $aH=H$.
First, we note that $H$ is closed under inverses: let $h\in H$. Then $h$ can be written as $xy^{-1}$, with $x,y\in H$. Therefore,
$h^{-1} = (xy^{-1})^{-1} = yx^{-1} \in HH^{-1}=H$. Thus, $H^{-1}\subseteq H$.
So, let $h\in H$. Then $ah = a(h^{-1})^{-1}\in HH^{-1}=H$, so $aH\subseteq H$.
Finally, to prove $H\subseteq aH$, we note that $a^{-1}h = a^{-1}(h^{-1})^{-1}\in HH^{-1}=H$, so $h = a(a^{-1}h) \in aH$. This proves the equality.