This is another homework question I can't figure out.
For $|G|$ even, $\forall x\in G\exists b\in G\setminus{\{x^{-1}\}}$ such that $bxb = x^{-1}$
I tried to toy with associativity but to no avail. Also, I can't see the relevance of $|G|$ being even. Any hint (I'm not here for the solution) is appreciated.
Thanks for your attention.
Update: I thought I would show you the proof I've written since that's the least I can trade for the effort you took to help me. I'm eager for any kind of feedback so, if you're about to comment on it, show no mercy. I'm trying to improve.
Let $x \in G$. There's left to prove that there exists $b \in G\setminus{\{x^{-1}\}}$ such that:
$bxb = x^{-1}$
$\Longleftrightarrow (bxb)x = x(bxb) = e$
$\Longleftrightarrow (bx)bx = xb(xb) = e$
$\Longleftrightarrow (bx)^{2} = (xb)^{2} = e$
$\Longleftrightarrow bx = (bx)^{-1}$ and $xb = (xb)^{-1}$
$\Longleftrightarrow bx = x^{-1}b^{-1}$ and $xb = b^{-1}x^{-1}$Let $a \in G\setminus{\{e\}}$ such that $a = a^{-1}$. (We know such $a$ exists by a previous result). Let $b = ax^{-1}$. Then:
$bx = x^{-1}b^{-1}$ and $xb = b^{-1}x^{-1}$
$\Longleftrightarrow ax^{-1}x = x^{-1}(ax^{-1})^{-1}$ and $xax^{-1} = (ax^{-1})^{-1}x^{-1}$
$\Longleftrightarrow a = x^{-1}xa^{-1}$ and $xax^{-1} = xa^{-1}x^{-1}$
$\Longleftrightarrow a = a^{-1}$ and $xax^{-1} = xax^{-1}$
What do you think of it?