Exercise 1711Algebra › GroupsL1Added 3 Nov 2026

The special linear group $SL_n(\mathbb{R})$

#special linear group#determinant#normal subgroup

Statement

Show that $SL_n(\mathbb{R})=\{A\in GL_n(\mathbb{R}):\det A=1\}$ is a normal subgroup of $GL_n(\mathbb{R})$ .

Hint

$SL_n$ is the kernel of the determinant homomorphism.

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Solution

Homomorphism. $\det:GL_n(\mathbb{R})\to(\mathbb{R}^*,\times)$ satisfies $\det(AB)=\det A\det B$ .

Kernel. $\ker\det=\{A:\det A=1\}=SL_n(\mathbb{R})$ .

Kernel is normal. The kernel of any homomorphism is a normal subgroup: for $A\in SL_n$ and $B\in GL_n$ , $\det(BAB^{-1})=\det B\cdot1\cdot(\det B)^{-1}=1$ .

Conclusion. $SL_n(\mathbb{R})$ is a normal subgroup of $GL_n(\mathbb{R})$ ; the quotient is $GL_n/SL_n\cong\mathbb{R}^*$ .

The special linear group $SL_n(\mathbb{R})$

Statement

Solution

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