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I am trying to prove that given two isometry on the plane that consist of rotations of the plane around distinct points contain a translation in their group (please see Groups containing two rotations.)

I have showed that they need to be in the kernel and everything but cannot show that $fg$ is not equal to $gf$.

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I know that $a$ and $p(a)$ have the same length as orthogonal operators preserve lengths. It makes sense that they don’t equal but how do we know for sure? So basically is: $a+b’=b+a’$ where $a$ and $a’$ have the same length same goes for $b$. How to show this is not possible for non trivial vectors.

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It's important to also assume that $f$ and $g$ are nontrivial rotations, or else it is indeed true that $fg = gf$.

So, given that $f$ is a nontrivial rotation about the point $p$ and $g$ is a nontrivial rotation about the point $q$, and $p \neq q$, why don't we check what $fg$ and $gf$ do to the points $p$ and $q$ themselves?

Since $g(q) = q$, we have $$ fg(q) = f(g(q)) = f(q) $$

But since $f$ is not centered at $q$, we have that $f(q) \neq q$. Therefore $$ gf(q) = g(f(q)) \neq f(q) $$ using the fact that $g$ is a nontrivial rotation, and hence $g(x) \neq x$ for all $x \neq q$. Thus, we conclude that $$ fg(q) = f(q) \neq gf(q) $$ $$ \implies fg \neq gf $$

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  • $\begingroup$ Thank you, this was helpful $\endgroup$
    – Dhdh
    Dec 4 '19 at 20:54

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