Lemma right inverse is also left inverse

Question

I read about a Lemma that says, for any Group $\langle G, \circ\rangle$ with $\forall a \in G $ the right inverse is also the left inverse. Without explicitly being said that the group is an abelian group the lemma shows:

$a^{-1} \circ a = a^{-1} \circ a \mathbb{1} = a^{-1} \circ a \circ a^{-1} \circ (a^{-1})^{-1} = \mathbb{1}$

I don't fully understand the logic behind that. In the calculation it's already assumed that $a^{-1} \circ a = \mathbb{1}$ and therefor the right inverse is also the left inverse. I do trust my book but I would like to be able to understand how the calculation can be understood.

Answer 1

They are not assuming that $a^{-1}\circ a=1$, they are starting with the LHS, eventually solving it and proving to be equal to identity. First they postmultiply by identity, then write $1=a^{-1}\circ (a^{-1})^{-1}$, where $(a^{-1})^{-1}$ is right inverse of $a^{-1}$. Then they use associativity to multiply the inner ones to get $a\circ a^{-1}=1$, since $a^{-1}$ is right inverse. Then they multiply the outer elements in the same way.

Answer 2

Maybe it's clearer if we use other names instead of $a^{-1}$.

Assume that for $a$ we have $b$ and $c$ such that $b\circ a=1$ and $a\circ c=1$. Then $b=b\circ1= b\circ (a\circ c)=(b\circ a)\circ c=1\circ c=c$, so $b=c$ and we can call them $a^{-1}$