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

daon

Senior Member
Joined
Jan 27, 2006
Messages
1,284
I need to prove the following:

There exists an x in G s.t. xax=b <=> c<sup>2</sup>=ab for some c.

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

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


Is that right?

But I am having trouble showing the <= direction.
 
What you have does indeed work. Here is the other way.

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