# What is the relationship between SL(n) and SO(n)?

I'm in Linear Algebra right now and we're mostly just working with vector spaces, but they're introducing us to the basic concepts of fields and groups in preparation taking for Abstract Algebra later on.

Let us assume the underlying field is $\mathbb{R}$. I know that $SL(n)$ is the group of all $n\times n$ matrices that are both invertible and have a determinant of $1$. I also understand that $SO(n)$ is the group of all $n\times n$ orthogonal matrices that have a determinant of $1$. But what is the relationship between the two groups? I think that $SO(n) \subset SL(n)$ but also $SO(n) \neq SL(n)$, since there are probably matrices in $SL(n)$ that are not orthogonal?

• That's right. $SO(n)$ is a subgroup of $SL(n)$, and for $n > 1$, it is a proper subgroup. – Daniel Fischer Dec 4 '13 at 20:53

Indeed, because every matrix in $SO(n)$ has determinant one, $SO(n)\subset SL(n)$. There are also matrices in $SL(n)$ which are not orthogonal; for example, upper triangular matrices whose diagonals multiply to $1$ have unit determinant but are not symmetric.
There are deeper relationships, though. The group $SL(n)$ has an Iwasawa decomposition, a way of writing any matrix in $SL(n)$ as a product of three simpler matrices: $$\bigg( \mbox{orthogonal}\bigg) \bigg( \mbox{diagonal}\bigg) \bigg( \mbox{upper triangular}\bigg).$$ One consequence is that, as a space rather than as a group, $SL(n)$ is the product $SO(n)\times \mathbb{R}^k$ (for some dimension $k$). The group $SO(n)$ is called a "maximal compact subgroup."
You are correct. For example, $\begin{pmatrix}1&2\\0&1\end{pmatrix}\in SL(2)$ but not in $SO(2)$.