Proving Group Homomorphism B: G-->Aut(K) for Normal Subgroup K of Group G

[chibulls59]

Homework Statement

Let G be a group and K be a normal subgroup of G.
Let x be an element of G
Define A_x: K-->K given by A_x(k)=xkx^-1
Prove that B: G-->Aut(K) given by B(x)=A_x is a well defined group homomorphism

Homework Equations

The Attempt at a Solution

I found B(xy)=xyky^-1x^-1 and B(x)B(y)=xkx^-1yky^-1
I can't get B(xy)=B(x)B(y). It has been a long time since we have covered group theory as we have moved on to ring theory so there may be something I am missing in regards to normal subgroups.

 

[Stephen Tashi]

The product you called $$B(x)B(y)$$ isn't relevant to showing that B is a homomorphism. What you must show (in slightly improved notation) is that
$$B_{xy}(k) = B_x( B_y(k))$$ or $$B_{xy}(k) = B_y( B_x(k))$$, depending on how your materials interpret a product of mappings as their composition.

$$A_x(k)$$ wouldn't necessarily be an automorphism of $$K$$ onto $$K$$ were $$K$$ not a normal subgroup.