Question

Let $N=\left\{A\in GL\left(2,R\right)| det A\in Q\right\}$. Prove that N is a normal subgroup of$GL\left(2,R\right)$ . [Hint:Exercise 32 of section 7.4 ]

Short answer

It is proved that $N=\left\{A\in GL\left(2,R\right)| \det A\in \right\}$is a normal subgroup of $GL\left(2,R\right)$$GL\left(2,R\right)$.

Step-by-step solution

Important 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 N=A∈ GL2,R|   detA∈Q  is a normal subgroup of GL2,R

Let’s consider an arbitrary element $a\in GL\left(2,R\right)$,$A\in N$.

If N is a normal subgroup of $GL\left(2,R\right)$, then for $a\in GL\left(2,R\right)$

$aA{a}^{-1}\in N$

Using homomorphism

$\begin{array}{c}\det \left(aA{a}^{-1}\right)= \det a . \det A .\det a^{-1}\\ =\det A\end{array}$

Considering $\det A\in$, and the definition of N given above, we can write$aA{a}^{-1}\in N$

This implies $aN{a}^{-1}\in N\Rightarrow aN{a}^{-1}\subseteq N$for $a\in GL\left(2,R\right)$.

From the above result and theorem 8.11, it is proved that $N=\left\{A\in GL\left(2,R\right)| \det A\in \right\}$ is a normal subgroup of $GL\left(2,R\right)$.