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I'm having trouble with the associativity part of showing that $\text{Aff}(\mathbb{R}^2)$ is a group.

Recall that if $\left[ A, \overline{r} \right]$ denotes an affine transformation (where $A$ is the matrix representation of the linear transformation and $\overline{r}$ is the translation vector), then multiplication in $\text{Aff}(\mathbb{R}^2)$ is defined by $$\left[ A, \overline{r} \right] \cdot \left[ B, \overline{s} \right] := [AB,\; B^{-1}\overline{r} + \overline{s} ].$$

We must show that $$\left(\left[ A, \overline{r} \right] \cdot \left[ B, \overline{s}\right] \right) \cdot [ C, \overline{t} ] = \left[ A, \overline{r} \right] \cdot \left([ B, \overline{s}] \cdot [ C, \overline{t}] \right)$$

The expansion of the left side is $[ABC,\; C^{-1}(B^{-1}\overline{r} + \overline{s}) + \overline{t}]$. I've tried working from the other side and making them meet in the middle, but I can't seem to massage them the right way so that they're equal. What's the trick here?

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Just basic properties of matrices: $$\begin{align*} \Bigl( [A,\overline{r}][B,\overline{s}]\Bigr)[C,\overline{t}] &= [AB,B^{-1}\overline{r}+\overline{s}][C,\overline{t}]\\ &= [(AB)C,C^{-1}(B^{-1}\overline{r}+\overline{s})+\overline{t}]\\ &= [ABC, C^{-1}B^{-1}\overline{r} + C^{-1}\overline{s} + \overline{t}].\\ \ [A,\overline{r}]\Bigl([B,\overline{s}][C,\overline{t}]\Bigr) &= [A,\overline{r}][BC,C^{-1}\overline{s}+\overline{t}]\\ &= [A(BC), (BC)^{-1}\overline{r} + C^{-1}\overline{s} + \overline{t}]. \end{align*}$$ Now, from linear algebra, how can you express $(BC)^{-1}$ in terms of $B^{-1}$ and $C^{-1}$?

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It seems I made a stupid mistake that may indicate I have dyslexia, or not enough sleep. Thanks for the answer! Also, $(BC)^{-1} = C^{-1}B^{-1}$. Great! –  Will Feb 1 '12 at 5:54
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