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

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) (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 for some m in G. We have to find to find k such that k*g*k = h.
Now ==> . (We can do this by the existence of the inverse element in G, as well the existence of identity element ).
Left multiply both sides by h, to get

By associativity,
.
Therefore k exists, and is equal to .
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