# Help with Normalizers

1. Feb 15, 2008

### wegmanstuna

For a subgroup H of G and a fixed element a ∈ G,
let H^a = {x∈ G / axa^-1 ∈ H}, it's normalizer N(H) = {a∈G / H^a=H}

Show that for any subgroup H of G, N(H) is a subgroup of G.

I know that for the first one I need to show that closure holds, an identity exists, and inverses exist. But I don't even know where to start with closure!!!
Help!!!

2. Feb 16, 2008

### morphism

If a and b are in N(H), write down what H^(ab) is. Now maybe you can try showing that $H^{ab} \subset H$ and $H \subset H^{ab}$.