- #1
wegmanstuna
- 3
- 0
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!
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!