Question

Prove that $SL\left(2,R\right)$ is a normal subgroup of$GL\left(2,R\right)$ . [Hint: $SL\left(2,R\right)$ is defined in Exercise 23 of section 7.1.Use Exercise 17 above and Exercise 32 of section 7.4]

Short answer

It is proved that $SL\left(2,R\right)$is a normal subgroup of $GL\left(2,R\right).$

Step-by-step solution

Required Theorem

Theorem 8.11: The following conditions on a subgroup N of a group G are equivalent:

1. Nis a normal subgroup of G.
2. $a^{-1}Na\subseteq N$for every $a\in G$, Where$a^{-1}Na\subseteq N$$\left\{a^{-1}Na| n\in N\right\}$
3. $aN{a}^{-1}\subseteq N$for every $a\in G$ , Where $aN{a}^{-1}\subseteq N$$\left\{aN{a}^{-1}| n\in N\right\}$
4. $a^{-1}Na\subseteq N$for every $a\in G$ .
5. $aN{a}^{-1}\subseteq N$ for every $a\in G$ .

Proving that SL2,R  is a normal subgroup of  GL2,R

Consider two arbitrary elements ,

Now, in order to prove that$SL\left(2,R\right)$is a normal subgroup of $GL\left(2,R\right)$, we must show that:

$BA{B}^{-1}\subseteq SL\left(2,R\right),$for all $B\in GL\left(2,R\right).$

Taking a determinant, we get:

$\begin{array}{c}\det \left(BA{B}^{-1}\right)=\det B \det A\det B^{-1}\\ =\det A\det B \det B^{-1}\\ =\det A\\ =1\end{array}$

From the above result and from the definition of $SL\left(2,R\right)$, we know$1\in SL\left(2,R\right)$

Therefore, $BA{B}^{-1}\in SL\left(2,R\right),$for $B\in GL\left(2,R\right).$

This implies $BA{B}^{-1}\subseteq SL\left(2,R\right),$for all $B\in GL\left(2,R\right).$

Therefore, from the above result and theorem 8.11, we can conclude that $SL\left(2,R\right)$is a normal subgroup of $GL\left(2,R\right).$