Subgroup of motions without a translation?

Question

Let $G \subset M$ be the smallest subgroup of rigid transformations in $\mathbb R^2$ containing a rotation of $1$ radian about $(0,0)$ (call this element $m_1$) and a rotation of $\frac{\pi}{4}$ radians fixing $(1,0)$ (call this $m_2$).

I suspect that $G$ contains no translation, since I don't see how we can apply $m_1$ and $m_2$ finitely many times and somehow remove the $1$ radian.

Yet Artin's Algebra seems to insist otherwise. An exercise in the chapter about symmetry asks us to prove that a subgroup of rigid motions in $\mathbb R^2$ containing rotations that fix $2$ separate points contains a translation.

So who is right?

Answer 1

The angles of rotation add: that is, if you do $m_1$ first and then $m_2$, you will have a total rotation about some point of $1+\pi/4$. From an advanced standpoint, there’s a map from the group of rigid motions of the plane to the circle group that’s a homomorphism. Now all you need to do is get the total rotation equal to zero, say by doing $m_1\circ m_2\circ m_1^{-1}\circ m_2^{-1}$. This is a rigid motion, certainly not identity, with rotation zero. So, a translation.

Answer 2

Call the two transformations $a$ and $b$. Then conjugating, $b^{-1}ab$ is a rotation of 1 radian about a new point $(1-1/\sqrt{2},-1/\sqrt{2})$ distinct from the first two. Call this conjugate $c$.

Then $bc^{-1}$ is a translation along the line perpendicular to their two fixed points, like waddling along with a board glued on your feet.