Let G be a group and H be a subgroup of G. Define N(H)={x is a member of G|xHx^-1=H} Prove that N(H) (called normalizer of H) is a subgroup of G.
This is a straightforward problem. Just use what it means is a definition of a subgroup. If $\displaystyle x,y\in N(H)$ can you prove $\displaystyle xy\in H$? Once you prove closure the others ones should be even simpler.
x,y is a member of N(H)
x^2=e and y^2=e
Prove (xy^-1)^2=e
(ab^-1)^2= ab^-1 * ab^-1= a^2(b^-1)^2=a^2(b^2)^-1=ee^-1=e
So xy^-1 belongs to N(H)
So N(H) is a subgroup of G