# 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

To prevent this from being left unanswered:

1. You are incorrect: multiplication of matrices is associative (it corresponds to composition of linear transformations, and composition of functions is associative).

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 $A(BC) = I(II) = II = I$.

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