Let G be a group, and let $\displaystyle H \leq G$

(That means H is the normal subgroup of G, right?)

a)Show that $\displaystyle aHa^{-1}$ is also a subgroup of G.

My proof:

Suppose that x,y is in G, so that $\displaystyle xHx^{-1}, yHy^{-1} \in aHa^{-1}$

Since H is normal in G, pick $\displaystyle Hx^{-1}=x_{1}H, yH=Hy_{1},x_{1},y_{1} \in G$

$\displaystyle (xHx^{-1})(yHy^{-1}) = x(x_{1}H)(Hy_{1}y^{-1}) = xx_{1}Hy_{1}y^{-1}$

Now, $\displaystyle (xHx^{-1})^{-1}=x^{-1}Hx \in aHa^{-1}$

Is this right?

b) If H is finite of order k, then $\displaystyle aHa^{-1}$ is also finite of order k.