# Subgroups of matrices. Why multiplication?

Determine whether the given set of invertible n x n matrices with real number entries is a subgroup of $GL(n,\mathbb{R})$

The n x n matrices with determinant 2

The key said

I understand that's how you compute determinants, but why did they pick out multiplication? Why couldn't they say for instance

"If detA = detB = 2, then we also see that detA + detB = 4 and it is not closed under addition".

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Note that it is not true that $\det(AB)=\det A+\det B$, nor is it true that $\det(A+B)=\det A+\det B$, so the fact $\det A+\det B=4$ would be worthless, whereas $\det(A)\det(B)=4$ tells you $AB$ has determinant $4$ and hence $AB$ is not in the set. –  anon Oct 15 '12 at 5:28

"Subgroup of $G$" means "subgroup under the operation inherited from $G$". The operation on the big group is multiplication, not addition.
So $gL(n,\mathbb{R})$ means addition then? –  Hawk Oct 15 '12 at 5:02
@jak: $GL(n,F)$ (capital letters) stands for the general linear group of $F^n$ ($F$ is the base field), which consists of invertible $n\times n$ matrices with the group operation being multiplication. In the context of Lie theory, $\mathfrak{gl}(V)$ (lowercase fraktur font) is the general linear algebra: it contains all endomorphisms of $V$, and is an algebra $-$ it has both addition and multiplication as two different binary operations. –  anon Oct 15 '12 at 5:17
In the context of matrices, "invertible" means "having a multiplicative inverse". So you've got the matrices that have a multiplicative inverse. That set is not closed under addition, so when we say it's a group, we must be talking about multiplication. I don't know if your lower-case $g$ is a typo or a genuine distinction; I've never seen any $gL(n,{\bf R})$. –  Gerry Myerson Oct 15 '12 at 6:35
It's because the group operation in $GL(n,\mathbb{R})$ is taken to be matrix multiplication so we want to see if the product of two matrices with determinant 2 still has determinant 2 when we are trying to determine whether the set of such matrices is a subgroup.