Show identity of subgroup is same as identity of group

Solution 1:

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.

Solution 2:

Hint:

Start with $1_{H}^2 = 1_{H}$.

Solution 3:

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 $H$.)

Solution 4:

The identity of $H$ is an element of $G$ satisfying $x^2 = x$, and the only such element can be $1_G$.

Solution 5:

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, $1_H = 1_G$.