Consider the so called general linear group
GL(n, R) = {A ∈ Rn×n: det A is not equal 0} with the matrix multiplication as the group operation. Prove that the subset
SL(n, R) = {A ∈ Rn×n: det A = 1} is a normal subgroup of GL(n, R).

Question

Consider the so called general linear group
GL(n, R) = {A ∈ Rn×n: det A is not equal 0} with the matrix multiplication as the group operation. Prove that the subset
SL(n, R) = {A ∈ Rn×n: det A = 1} is a normal subgroup of GL(n, R).

anonymous · Answer

We at least know it is a subset, given that $\det A = 1 \implies \det A \neq 0$.

anonymous · Answer

Given that $\det(AB) = \det(A)\det(B)$, we also know it'll be closed.

anonymous · Answer

I don't remember all the properties of a normal subgroup.

anonymous · Answer

H is normal of G if xHx^-1=H

rsadhvika · Answer

You just need to show $\det(PAP^{-1} )= 1$

rsadhvika · Answer

where $P \in GL(n,R)$ and  $A \in SL(n,R)$

anonymous · Answer

how can we do that? can you help me please?

rsadhvika · Answer

use below : 

$\det(P^{-1}) = \dfrac{1}{\det(P)}$

anonymous · Answer

is it the same as DET B.DET A. DET P-1=1

rsadhvika · Answer

Yes

$$\det(PAP^{-1} )\~\= \det(P)\det(A)\det(P^{-1})\~\ =  \det(P)\det(A)\dfrac{1}{\det(P)} \~\= \det(A) \~\=1$$