Find the inverse element of a group $(G,\circ),$ $x\circ y=x*a*y$

Question

Let $(G,*,{}^{-1},e)$ be a group and $a\in G$ fixed element. We define a new operation $\circ$ on $G:$ $$x\circ y=x*a*y.$$ I have to show that $(G,\circ)$ is, also, a group. I don't know how to find inverse element of a element $x.$ I get that neutral in this structure is fixed element $a.$ Then $$x\circ x^{-1}=a$$ and $x*a*x^{-1}=a.$ What to do next?

Answer 1

Firstly, $a$ is not the neutral element, as for example you have $a \circ x = a \ast a \ast x \neq x$. The element you are looking for is $a^{-1}$ since for any $x \in G$ you have $a^{-1} \circ x = a^{-1} \ast a \ast x = x = x \ast a \ast a^{-1} = x \circ a^{-1}$.

Secondly, let $x \in G$. Then I claim that $a^{-1} \ast x^{-1} \ast a^{-1}$ is the inverse element of $x$. Try showing this.

Answer 2

Hint: since G is a group there exists a unique element, $y$, such that $a*y = x^{-1}$