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Below is the cycle graph of $\operatorname{Dih}_4$. What I don't understand is that, since $(ba)^2=a^2$, why there isn't a link between $ba$ and $a^2$, and hence of course also $a^2$ and $ba^3$? I can see that "$e$ - $ba$" is not a cycle at all, because $(ba)^2\ne e$.

   / \
  a   a^3
   \ /
 / / \    \
b ba ba^2 ba^3
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But $(ba)^2=baba=1$ surely...(as $bab=a^{-1}$) – user1729 Nov 8 '11 at 10:40
@user1729: No, $(ba)^2=b^2 a^2=a^2$. ($bab=a$, $a^3=a^{-1}$) – Voldemort Nov 8 '11 at 10:44
@ Voldemort: No, $(ba)^2=a^{-1}a=1$. (bab=a^{-1}, a^3=a^{-1}) – user1729 Nov 8 '11 at 10:48
How is it that you are defining the Dihedral group? If you are given it in terms of a presentation (see "Equivalent Definitions" in the wikipedia article) then seeing that $(ba)^2=1$ is easy. Otherwise, I would recommend playing around with squares some more, or working out a permutation representation and playing with that (to work out a permutation representation, label the corners of your square $1$, $2$, $3$ and $4$ then work out what a rotation does to them and what a flip does to them. A flip sends $1$ to what? $2$ to what? etc.). – user1729 Nov 8 '11 at 10:50
up vote 2 down vote accepted

...but $(ba)^2=baba=1$ $bab=a^{-1}$. So, $a^2\neq (ba)^2$, so they should not appear in the same cycle.

Indeed, taking an arbitrary element of $D_4$ which is not a power of $a$, $a^ib$ say, then $(a^ib)^2=a^iba^ib=a^ia^{-i}=1$ as $ba^ib=a^{-i}$. So, basically, every element has order two apart from $a$, which has order $4$, and $e$, which is trivial.

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I'm sorry. It turns out I didn't quite think this through, the Dihedral Group is not commutative. Thank you. – Voldemort Nov 8 '11 at 10:54
That's all right - for my part I should have spotted that that was your problem earlier! – user1729 Nov 8 '11 at 11:03

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