Group Q: there exists x so that xax=b <=> c^2=ab for s

[daon]

I need to prove the following:

There exists an x in G s.t. xax=b <=> c²=ab for some c.

I have shown the \(\displaystyle =>\) and have found that c=x^-1b.

xax=b <=>
xa=bx^-1 <=>
xab=bx^-1b <=>
ab=x^-1bx^-1b = (x^-1b)²


Is that right?

But I am having trouble showing the <= direction.

 

[pka]

What you have does indeed work. Here is the other way.

\(\displaystyle \L
\begin{array}{rcl}
c^2 & = & ab \\
e & = & c^{ - 1} abc^{ - 1} \\
b & = & bc^{ - 1} abc^{ - 1} \\
b & = & \left( {bc^{ - 1} } \right)a\left( {bc^{ - 1} } \right)\, \Rightarrow \;x = bc^{ - 1} \; \\
\end{array}\)

 

[daon]

Aggh! Thanks.