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Given a group generated by two rotations of the plane about distinct points, how to show that it contains a translation?

Attempt: Let the two rotations be $f$, $g$, then consider the motion $f \circ g \circ f^{-1} \circ g^{-1}$. This would give a rotation of 0 degree, so it is either the identity or a translation. If $f$ and $g$ does not commute, then it is a translation.

Question:

1) I suspect the non-commutativity of $f$ and $g$ comes from the fact that they are rotations about two different points, but I can't seem to find a way to prove this. Do you think this approach is right and if so, do you have any suggestion on continuing?

2) If the above approach is not right, can you suggest a way for me to work on this problem?

Thank you.

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    $\begingroup$ Your idea is good, but it's not clear how much you are expected to do to carry this out. Are you allowed to pick your own two rotations about distinct points (and compute with those specific rotations), or have you been asked to demonstrate this in full generality? $\endgroup$
    – hardmath
    Oct 19 '15 at 2:50
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Your idea is on the mark, so that all you need is the tools to put your program into effect.

Here’s your approach. You represent a point $(x,y)$ in the plane by a column vector with entries $x,y,1$, and you represent a rigid orientation-preserving motion of the Euclidean plane by a $3\times3$ matrix of the form $$ \begin{pmatrix} \cos\theta&-\sin\theta&a\\ \sin\theta&\cos\theta&b\\ 0&0&1 \end{pmatrix}\,. $$ This is the motion that amounts to rotation about the origin by $\theta$, followed by a translation of $a$ units to the right, $b$ units upwards.

I think you can work out, if $\theta\ne0$, what the fixed point of this motion is, and how to represent a rotation about another point than the origin. Then you can prove your claim.

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