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Let $r_1,r_2$ be two plane rotations of finite order. That's equivalent to rotations by a rational multiple of $\pi$.

Assume they're not commuting (which is equivalent to the rotations being around different centers).

If the rotations are not both rotations by $\pi$, show that the group generated by $r_1,r_2$ contain two translations, along different lines.

partial answer: the commutator $[r_1,r_2]=r_1 r_2 r_1^{-1} r_2^{-1}$ is a translation, and not a trivial one (i.e. the identity) since the rotations do not commute.

Another commutator $[r_1^{-1},r_2^{-1}]=r_1^{-1} r_2^{-1} r_1 r_2$ is also a translation, and again not a trivial one. It is a translation along a different line, as long as the rotation angles of $r_1$ and $r_2$ do not add up to a whole multiple of $\pi$. What about the case where they do add up to a whole multiple of $\pi$ ?

Edit: Using Daniel's hint below, here's what we have. Denote $$r_j(z) = e^{i \theta_j \pi}z + b_j.$$ We have $\theta_j \in \mathbb{Q}$, (since the rotations are of finite order), and $b_j\in \mathbb{C}$. Computing the commutators, we get: $$ [r_1,r_2](z) = z + b_1(1-e^{i \theta_2 \pi}) - b_2(1-e^{i \theta_1 \pi}),$$ $$ r_1^2(z) = e^{2i \theta_1 \pi} + b_1 (1+e^{i \theta_1 \pi})$$ and $$ [r_1^2,r_2](z) = z + b_1(1-e^{i \theta_2 \pi})(1+e^{i \theta_1 \pi}) - b_2(1-e^{2i \theta_1 \pi}).$$ How can we show that these two translations are along different lines ? put another way, how do we show $$ v_1 := b_1(1-e^{i \theta_2 \pi}) - b_2(1-e^{i \theta_1 \pi})$$ and $$ v_2 := b_1(1-e^{i \theta_2 \pi})(1+e^{i \theta_1 \pi}) - b_2(1-e^{2i \theta_1 \pi}) =\\ =b_1(1-e^{i \theta_2 \pi})(1+e^{i \theta_1 \pi}) - b_2(1-e^{i \theta_1 \pi})(1+e^{i \theta_1 \pi})=\\ = (1+e^{i \theta_1 \pi}) v_1$$ are linearly independent as vectors in $\mathbb{C}$ over $\mathbb{R}$ ? But this is clear, assuming (as we did) that $r_1$ is an actual rotation (hence $e^{i \theta_1 \pi} \neq 1 $), and $r_1$ is not a $\pi$-rotation (hence $e^{i \theta_1 \pi} \neq -1 $).

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    $\begingroup$ Consider $[r_1^2, r_2]$ if $r_1$ is not a rotation by $\pi$. $\endgroup$ Aug 1 '13 at 13:00
  • $\begingroup$ @DanielFischer: Thanks. It certainly is a translation. How do we know it's along a different line (w.r.t. $[r_1,r_2]$) $\endgroup$
    – Teddy
    Aug 2 '13 at 6:56
  • $\begingroup$ I get it. $[r_1^2,r_2]=r_1[r_1,r_2]r_1^{-1}$ is the conjugated translation, which is a translation along a different line since $r_1$ is not a rotation by $\pi$. $\endgroup$
    – Teddy
    Aug 2 '13 at 8:05
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As suggested above, the group $G=<r_1, r_2>$ contains a nontrivial translation $\tau$ along a line $L$. Then for every $g\in G$, the conjugate $g\tau g^{-1}$ is a translation along the line $g(L)$. Now, take $g=r_i$ which is not an order 2 rotation. Then $g(L)$ is not parallel to $L$ and, hence, you are done.

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