# show that $((GL(2,\mathbb{R}),\bullet)$ is a group

Let $G(2,\mathbb R)=\{\text{All invertible }2 \times 2\text{ matrices over }\mathbb{R}\}$. Then i want to show that $((G(2,\mathbb{R}),\bullet)$ is a group, where $\bullet$ is multiplication of matrices.

I think is not a group because $\bullet$ is not associative, for example for all $A,B,C$ in the set then $(AB)C$ is not equal to $A(BC)$

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Matrix multiplication is indeed associative. Try writing down (AB)C and A(BC) for a few examples of $2\times 2$ matrices A,B, and C. – Brad Dec 5 '11 at 2:56
What is $\bullet$? Multiplication of matrices? Then multiplication of matrices is associative. Matrix multiplication is essentially composition of functions, and composition of functions is associative. – Arturo Magidin Dec 5 '11 at 2:56
Did you know that "GL" in $GL(2,\mathbb R)$ stands for "general linear group"? – matt Dec 5 '11 at 3:01
yes ● is a multiplication of matrices – neema Dec 5 '11 at 3:14
@neema You can verify associativity, then. Write down three completely general matrices $A = \begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix}$, $B = (b_{ij})$, and $C = (c_{ij})$. Compute $A(BC)$ and $(AB)C$. You won't even have to use the fact that these matrices are invertible, although that is of course crucial elsewhere. – Dylan Moreland Dec 5 '11 at 3:19

2. Even if you were right that multiplication of matrices is not associative, it would still be incorrect to say that "for every $A$, $B$, $C$ in the set, $A(BC)$ is not equal to $(AB)C$." The negation of "For every $A$, $B$, and $C$ in the set, $A(BC)$ is equal to $(AB)C$" is "there exist matrices $A$, $B$, and $C$ in the set such that $A(BC)$ is not equal to $(AB)C$". In fact, it is easy to see that your asserted claim cannot hold even without knowing if matrix multiplication is associative, since letting $A=B=C$ be the identity matrix, we would have $(AB)C = (II)I = II = I$ and $A(BC) = I(II) = II = I$.