why square matrix is group for multipy

Question

There are 4 requirements in order to be group for $$(R^{n*n},*), n\in{N}$$

1. Closer $ \forall{A,B}\in{R^{n*n}}:A*B\in{R^{n*n}}$
2. Associativity $\forall{A,B,C}\in{R^{n*n}}:A*(B*C)=(A*B)*C$
3. Neutral element $\exists{I}\in{R^{n*n}},\forall{A}\in{R^{n*n}}:I*A=A*I=A$
4. Inverse element $\exists{X}\in{R^{n*n}},\forall{A}\in{R^{n*n}}:X*A=A*X=I$

so the question is, i dont understand why square matrix satisfy the inverse element requirements.

in my mathematic material, it say square matrix is group but not every square matrixs are invertible, and in inverse element requirement, it use $\forall$ notation.

for my understanding, the requirement should change like this($\forall$->$\exists$) $$\exists{X,A}\in{R^{n*n}}:X*A=A*X=I$$

when we say $(R^{n*n},*), n\in{N}$, does it imply that it includes only matrix which is invertible? or maybe i am missing some simple thing hopefully.

Answer 1

You are correct that not all square matrices are invertible. Only the invertible ones are in the group. By the way, the group is called the general linear group.

Also, as pointed out in a comment by AHusain, the fourth requirement should be written $\forall{A}\in{R^{n*n}, \exists{X}\in{R^{n*n}}}:X*A=A*X=I$, so $X$ can depend on $A$.