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.

Printable View

- October 13th 2008, 07:12 AMmandy123prove it is a subgroup
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.

- October 13th 2008, 08:20 AMThePerfectHacker
- October 13th 2008, 11:41 AMmandy123
so would this work

a^2=e and e^2=e so H is nonempty

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