Proving that the inverse in subgroup $H \leq G$ is the same as the inverse in $G$ : does it follow from uniqueness? Let $H$ be a subgroup of $G$. I am trying to show that the inverse element of some $h \in H$ is the same as its inverse in $G$. I know how to prove it without uniqueness, but I am trying to understand why this proof method would fail, assuming that it would. I am taking as proved that the identity in $H$ is the same as the identity in $G$, and am just calling it $e$.

Let $h \in H$. Then there exists an inverse $x \in H$ such that $hx = xh = e$. Furthermore, $H \subset G$, so there exists an inverse $y \in G$ such that $hy = yh = e$. By uniqueness of the inverse element, $x = y$.

Is the reason that uniqueness fails because the first equation $hx = e$ holds only for elements in $H$, and the second equation, $hy$, holds for elements in $G \setminus H$? If $H$ is the improper subgroup, this should work, but that would really boil down to the standard proof of uniqueness.

Take $h \in H$. Let $x$ be the inverse of $h$ in $H$. Let $y$ be the inverse of $h$ in $G$. We will prove that $x = y$ by contradiction.
Assume that $x \neq y$. By definition of the inverse, $x \in H$. Since $H \subseteq G$, then we also have $x \in G$. Hence, $x$ is an element of $G$ that satisfies $hx = xh = e$, making it also an inverse of $h$ in $G$. Since $x \neq y$, then we have two distinct inverses of $h$ in $G$. This contradicts with the uniqueness property of the inverse in a group.
Hence, it must be the case that $x = y$. That is, the inverse of $h$ in $H$ is just equal to the inverse of $h$ in $G$. In other words, the uniqueness argument in your proof is just fine.
