Why isn't $GL_n(\mathbb{R})$ a ring? Definition :  A ring R is a set R equipped with two binary operations $+$ and $\cdot$ on R satisfying the following conditions:
(1)  R is an abelian group under $+$.
(2)  $\cdot$ is associative and has an identity element.
(3a) $\cdot$ is left-distributive over $+$, that is, $a \cdot (b+c) = (a \cdot b) + (a \cdot c)$ for all a,b,c in R.
(3b) $\cdot$ is right-distributive over $+$, that is $(b+c) \cdot a = (b \cdot a) + (c \cdot a)$ for all a,b,c in R.
$GL_n(\mathbb{R}) = \{A \in M_n(\mathbb{R}) : det(A) \neq 0 \}$
I feel like property 1 fails, but I am unsure why.
Thank you for your time and help.

Yes, property (1) fails, because if $+$ denotes the usual addition of matrices, $GL_n(\mathbb{R})$ isn't even closed under $+$.  For instance, if $I$ is the identity matrix, then $I\in GL_n(\mathbb{R})$ and $-I\in GL_n(\mathbb{R})$, but $I+(-I)=0\not\in GL_n(\mathbb{R})$.
