Prove that if $H \le G$, then the identity in $H$ is the same as the identity is $G$ My question is a generalization of a result I saw in Linear Algebra.

Prove that if $H \le G$, then the identity in $H$ is the same as the identity in $G$.

I would like to know if my justification is sound:
PROOF: Let $g \in H$. Since $H \le G$, we know that $g \in G$ also. Let $e_H$ be the identity in $H$ and $e_G$ be the identity in $G$. Then by definition of identity, $$ge_H = e_Hg = g \;\;\;\;\text{and}\;\;\;\;ge_G = e_Gg = g.$$ Transitively, since both of these equations equal $g$, we may equate them to receive $$ge_H = ge_G.$$ Because $H \le G$, we know that $g^{-1} \in H$ and $g^{-1} \in G$. So $$g^{-1}ge_H = g^{-1}ge_G$$ $$e_He_H = e_Ge_G$$ $$e_H = e_G$$ which means that the identities in both the subgroup and the group are equal.
 A: I think you have a vicious circle here, because you appear to be assuming $e_{H} = g^{-1} g = e_{G}$, or equivalently that inverses in $H$ and $G$ are the same.
I believe the argument should really be the following.
$$e_{H} =  e_{H} e_{H}$$ since $e_{H} \in H$, and $e_{H}$ is the identity in $H$.
Now multiply on the left, say, by the inverse $e_{H}^{-1}$ of $e_{H}$ in $G$, so that $e_{H}^{-1} e_{H} = e_{G}$.
You get 
$$
e_{G} = e_{H}^{-1} e_{H} = e_{H}^{-1} (e_{H} e_{H}) = (e_{H}^{-1} e_{H}) e_{H} = e_{G} e_{H} = e_{H},
$$
as $e_{G} x = x$ for all $x \in G$.
A: In every group $G$ for $g\in G$ we have:$$g^2=g\iff g=e_G$$
where $e_G$ denotes the identity of $G$.
Proving this is straightforward.
In words: every group has exactly one element that is idempotent, which is its identity.
As a subgroup of $G$ group $H$ contains $e_G$ which is idempotent so we conclude that $e_G=e_H$.

Sidenote:
By definition a grouphomomorphism $\phi:G\to G'$ is a map that respects multiplication. The demand that it sends identity to identity (as you would expect) is left out. Why? Because it turns out to be redundant. The fact that it respects multiplication implies that it respects idempotents (if $g^2=g$ then $\phi(g)^2=\phi(g^2)=\phi(g)$) and the uniqueness of identities as idempotents ensures that $\phi(e_G)=e_{G'}$.
