# $\{A \in GL_2(\mathbb R):|\det(A)|=1\}$ is a normal subgroup of $GL_2(\mathbb R)$

I have to prove that $$O_2(\mathbb R)=\{A \in GL_2(\mathbb R):|\det(A)|=1\}$$ is a normal subgroup of $$GL_2(\mathbb R)$$.

I tried to go on with the definition of normal subgroup but I don't really know how to use it in this case. I've also tried with some equivalent properties like that the left co-class has to be the same as the right co-class but I don't reach any solid conclusion.

• The equality is incorrect. The set you wrote is not $O_2$. (It is a normal subgroup though, unlike $O_2$.) Apr 13, 2020 at 11:25
• Otherwise, hint: the inverse image of a normal subgroup by a group morphism is a normal subgroup. Apr 13, 2020 at 11:29

You can do it using the definition. Just take $$A \in \text{SL}_2(\mathbb R)$$ (that is, a $$2 \times 2$$ real matrix $$A$$ such that $$\operatorname{det} A = 1$$) and $$B \in \text{GL}_2(\mathbb R),$$ and show that $$BAB^{-1} \in \text{SL}_2(\mathbb R).$$ That is, show that $$\operatorname{det} (BAB^{-1})=1.$$
If you can use homomorphisms, then $$A \mapsto |\det(A)|$$ is a group homomorphism $$GL_2(\mathbb R) \to \mathbb R^\times_+$$ whose kernel is the set in question.