SOLUTION: This is my question: Suppose that G is a group and g,h are elements of G. There exists a k in G such that kgk=h if and only if gh=m^2 for some m in G. I have already have: L

Algebra ->  Proofs -> SOLUTION: This is my question: Suppose that G is a group and g,h are elements of G. There exists a k in G such that kgk=h if and only if gh=m^2 for some m in G. I have already have: L      Log On


   

Question 403583: This is my question:
Suppose that G is a group and g,h are elements of G. There exists a k in G such that kgk=h if and only if gh=m^2 for some m in G.
I have already have:
Let k be in G and kgk=h. We can perform the operation on each side of the equation to get gkgk=gh=(gk)^2. m=gk and by the closure component of a group we know that gk is in G so we know that m is in G also.

I do not get how to prove it the other way. Please can you help.

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
You have proved the sufficiency part, or the (==>) part.
To prove the necessity part, or the (<==), we proceed as follows:
Assume gh++=m%5E2 for some m in G. We have to find to find k such that k*g*k = h.
Now gh++=m%5E2 ==> m%5E%28-1%29%28gh%29m%5E%28-1%29+=+1%5BG%5D. (We can do this by the existence of the inverse element in G, as well the existence of identity element 1%5BG%5D).
Left multiply both sides by h, to get
hm%5E%28-1%29%28gh%29m%5E%28-1%29+=+h
By associativity,
%28hm%5E%28-1%29%29g%28hm%5E%28-1%29%29+=+h.
Therefore k exists, and is equal to hm%5E%28-1%29.