Does the identity element of a group have an inverse? I can't seem to find anything on the topic.
 A: Let $1$ be the identity element. Then its inverse $a$ is defined by
$$
1\cdot a = 1
$$
But $1 \cdot x = x$ so in particular $1 \cdot a = a$. Therefore, $a = 1$ and $1$ is its own inverse.
A: If $e$ is the neutral element of $G$ then $eg=ge=g$ for all $g\in G$, in particular for $g=e$ then $ee=e$, ie $e=e^{-1}$.
A: All elements of a group have an inverse. This is a requirement in the definition of a group.
For an element $g$ in a group $G$, an inverse of $g$ is an element $b$ such that $gb = e$ where $e$ is the identity in the group. (Since the inverse of an element is unique, we usually denoted the inverse of $g$ $g^{-1}$ or $-g$.)
Note now that $ee =e$, so by definition $e$ is an inverse of itself.
You might be wondering if other elements might be their own inverses. The answer to this is yes. For example in the group $\mathbb{Z}_2$ of order $2$, both elements are their own inverse.
A: Let $g = e^{-1}$.
Then $g*e = e*g = e$.  Solve for $g$.
Then $g*e = e \implies g*e*e^{-1} = e*e^{-1} => g = e$.
This gets kind of silly the more you think about it.
