# Special Linear Group is a Normal Subgroup of General Linear Group ## Problem 332

Let $G=\GL(n, \R)$ be the general linear group of degree $n$, that is, the group of all $n\times n$ invertible matrices.
Consider the subset of $G$ defined by
$\SL(n, \R)=\{X\in \GL(n,\R) \mid \det(X)=1\}.$ Prove that $\SL(n, \R)$ is a subgroup of $G$. Furthermore, prove that $\SL(n,\R)$ is a normal subgroup of $G$.
The subgroup $\SL(n,\R)$ is called special linear group Add to solve later

## Hint.

We are going to use the following facts from linear algebra about the determinant of a matrix.
For any $n\times n$ matrices $A, B$, we have
\begin{align*}
\det(AB)&=\det(A)\det(B) \text{ and }\\
\det(A^{-1})&=\det(A)^{-1}
\end{align*}
if $A$ is invertible.

We give two proofs.
The first one proves that $\SL(n,\R)$ is a normal subgroup of $\GL(n,\R)$ by directly verifying the defining property.

The second proof uses a fact about group homomorphism. If you are familiar with group homomorphism, the second proof is concise and nice.

## Proof 1.

### The special linear group $\SL(n,\R)$ is a subgroup.

Let $X, Y\in \SL(n,\R)$ be arbitrary elements. We have
$\det(X)=\det(Y)=1$ by definition of $\SL(n,\R)$.

Then we obtain
\begin{align*}
\det(XY)=\det(X)\det(Y)=1,
\end{align*}
and hence $XY$ is in $\SL(n,\R)$.

Also, we have
$\det(X^{-1})=\det(X)^{-1}=1,$ and it follows that $X^{-1}$ is in $\SL(n,\R)$.
Thus, $\SL(n,\R)$ is a subgroup of $G$.

### The special linear group $\SL(n,\R)$ is normal.

To prove that $\SL(n,\R)$ is a normal subgroup of $G$, let $X\in \SL(n,\R)$ and let $P\in G$.
Then we have
\begin{align*}
\det(PXP^{-1})=\det(P)\det(X)\det(P)^{-1}=\det(X)=1,
\end{align*}
and hence the conjugate $PXP^{-1}$ is in $\SL(n,\R)$.
Therefore, $\SL(n,\R)$ is a normal subgroup of $G$.

## Proof 2.

Let $\R^*$ be the multiplicative group of nonzero real numbers.

Let $\phi:\GL(n,\R) \to \R^*$ be the map given by
$\phi(X)=\det(X),$ for each $X\in \GL(n,\R)$.

Note that this is well-defined since $\det(X)\neq 0$ for $X\in \GL(n,\R)$.

By the property of the determinant, we know that
$\det(XY)=\det(X)\det(Y)$ for any $X, Y \in \GL(n,\R)$.

This implies that the map $\phi$ is a group homomorphism.

Then the kernel of $\phi$ is given by
\begin{align*}
\ker(\phi)=\{X \in \GL(n,\R) \mid \phi(X)=\det(X)=1\}=\SL(n, \R).
\end{align*}

As the kernel of the group homomorphism $\phi:\GL(n,\R) \to \R^*$ is alway a normal subgroup of $\GL(n,\R)$, we conclude that $\SL(n, \R)$ is a normal subgroup of $\GL(n,\R)$. Add to solve later

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