Prove $f(x)=a*x*a^{-1}$ Group Homomorphism

Question

If $f:(G,*) \to (G,*)$ , $a$ in $G$

Prove that $f$ is a group homomorphism , where

$$f(x)=a*x*a^{-1},\ \ \ \ x\in G.$$

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My answer:

We should to prove this :

$$f(x*y)=f(x)*f(y).$$

And from definition in the question:

$$\tag{1}f(x*y)=a*x*y*a^{-1}$$ and $$\tag{2}f(x)=a*x*a^{-1}$$ and $$\tag{3}f(y)=a*y*a^{-1}.$$

But How can I prove $(1)=(2)*(3)$?

Answer 1

Sometimes, when you can't complete one side of an equality, you should try with the other one: $$ f(x)*f(y)=(a*x*a^{-1})(a* y* a^{-1})=a*x*(a^{-1}*a)*y*a^{-1}\\ =a*x*e*y*a^{-1} =a*x*y*a^{-1} =f(x*y). $$