# Normal subgroups

• Dec 13th 2008, 05:32 PM
charleschafsky
Normal subgroups
This will be my second post of today. Thanks to all who helped before :)

Let G = GLn(R) be the group of all invertible n x n matrices with real entries, and let N = {A exists in G | det(A) > 0}
Show that N is a normal subgroup of G.

This is the question. I know a few theorems about normal subgroups, and I think I've made some headway. However, I can't seem to piece it all together.

Obviously, N is the group of invertible, square matrices with positive determinants. This theorem appears useful to me...

(a^-1)Na is a subset of N, for all a that exist in G.

By the invertible matrix theorem, I know that the determinant of a (and a^-1) exist. I also know that the determinant of these two values is not 0. I also know that I'm incredibly tempted to just assume commutativity and swap the equation to a(a^-1)N is a subgroup of N, but that's absolutely useless and you can't assume commutativity. Thus, I am stuck here. This seems like the correct theorem to be using, but I can't seem to put it to use.

Afterwards, I am to find the quotient group G/N, but I can see this being much more clear once I've solved this issue.
• Dec 13th 2008, 05:36 PM
ThePerfectHacker
Quote:

Originally Posted by charleschafsky

Let G = GLn(R) be the group of all invertible n x n matrices with real entries, and let N = {A exists in G | det(A) > 0}
Show that N is a normal subgroup of G.

If $\displaystyle n\in N$ and $\displaystyle g\in G$ then $\displaystyle \det (gng^{-1}) = \det (g) \det (n) \det (g)^{-1} = \det (n) > 0$ and therefore $\displaystyle gng^{-1} \in N$.
This means that $\displaystyle N\triangleleft G$ (you need to also show that $\displaystyle N$ is a subgroup first !)
• Dec 13th 2008, 05:39 PM
charleschafsky
Determinants, the solution to everything! Thanks!