Prove that if $H$ is a subgroup of $G$, then the identity $e$ of $G$ is in $H$.

Question

I am a little stuck in understanding the logic to this proof I found.

Prove that if $H$ is a subgroup of $G$, then the identity $e$ of $G$ is in $H$.

Since $H$ is a subgroup, it must have an identity $e_{H}$. Note that $e_{H}\circ e_{H}=e_{H}$ and $e\circ e_{H}=e_{H}\circ e=e_{H}$. Then $e_{H}\circ e_{H}=e\circ e_{H}$. Then $e_{H}\circ e_{H}\circ e=e\circ e_{H}\circ e$. Then $e_{H}=e$.

The last part I do not understand. I understand the first side of the equation becomes $e_{H}$, but why does $e\circ e_{H}\circ e=e$?

Answer 1

I don't understand the logic in the last step. But, in any case, it looks overly complicated.

In any group, if $b^2=b$, then $b=e$. This follows simply by applying $b^{-1}$ to the equality: you get $b^{-1}b^2=b^{-1}b$, that is $b=e$.