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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 saidenter image description here

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.2012-10-15

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"Subgroup of $G$" means "subgroup under the operation inherited from $G$". The operation on the big group is multiplication, not addition.

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    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 $gL(n,{\bf R})$.2012-10-15
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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.