Show identity of subgroup is same as identity of group Let $H$ be a subgroup of $G$. Let $1_H$ and $1_G$ be the identities of $H$ and $G$, respectively. Show that $1_H=1_G$.
My attempt is: since we know that the identity of a group is unique, and hence $1_H=1_G$.
Is my proof correct?
 A: Knowing that the identity of a group is unique is not sufficient, since being an identity of $H$ does not directly imply being an identity of $G$. Also, previous answers have overlooked how the cancellation law depends on the identity element to work. For instance, by $1_H^{-1}$ do we mean an element from $G$ such that $1_H^{-1}1_H=1_G$? Or an element from $H$ such that $1_H^{-1}1_H=1_H$?
Let $1_H^{-1}$ be an element from $G$ such that $1_H^{-1}1_H=1_G$, for then $1_H=1_G1_H=1_H^{-1}1_H1_H=1_H^{-1}1_H=1_G$, which is what we wanted to show. (We only used the operation of $G$. The first equality follows from $1_G$ being the identity of $G$; the third equality follows from $1_H$ being the identity of $H$.)
A: Hint: 
Start with $1_{H}^2 = 1_{H}$.
A: The identity of $H$ is an element of $G$ satisfying $x^2 = x$, and the only such element can be $1_G$.
A: Working in $G$, we have $1_H1_H=1_H=1_H1_G$. The first equality follows from the fact that $1_H$ is the identity of $H$ and $H$ inherits its operation from $G$. The second follows from the fact that $1_G$ is the identity of $G$. Now premultiply by $1_H^{-1}$ to obtain the result.
A: Let $h_1, h_2 \in H$ where $h_2$ is the inverse of $h_1$. Note that this is allowed since $H$ itself is a group. Then $h_1h_2 = 1_H$, but $h_1, h_2$ are both elements in $G$ as well, so $h_1h_2 = 1_G$. Therefore, $1_H = 1_G$.
A: I thought I would post about something related that might interest the readers.  Let $\Bbb{Z}/n$ be a ring and take its multiplicative monoid $M$, not its unit group.
Then there are subsemigroups of $M$ that don't share $M$'s identity, but are groups with the same law as in $M$:
In $\Bbb{Z}/15$, $3$ is not a unit, however it's powers $3^k, k \geq 1$ form a group:
$$
3,3^2,3^3=12, 3^4 = 6, 3^5=3, \text{(repeat)}
$$
The group's identity is $6$!  We have $9\cdot 6 = 9$, etc.
