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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.

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

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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 $\operatorname{det} (BAB^{-1})=1.$

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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.

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