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I am given the dihedral group of order 12: $D_{12}=<a,b:a^6=b^2=3, ba=a^5b>$, where $a$ is a rotation of a hexagon by 60 degrees, and $b$ is a reflection across a diagonal of two vertices.

I am looking to find the conjugacy class $cl_{D_{12}}(b)$. I have, $cl_{D_{12}}(b)$ = $\{b, aba^{-1}, a^2ba^{-2}, a^3ba^{-3}\}$.

Have I found this conjugacy class correctly?

Thanks.

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  • $\begingroup$ I think you mean $a^6 = b^2 = 1$, not $3$. $\endgroup$ – rogerl Nov 25 '18 at 1:07
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Note that $a^3 = a^{-3}$, so that $$a^3ba^{-3} = a^3ba^3 = a^3a^{15}b = a^{18}b = b.$$

It might be easier if you think of the elements of $D_{12}$ as $$1, a, a^2, a^3, a^4, a^5, b, ba, ba^2, ba^3, ba^4, ba^5,$$ and remember that always $a^kb = ba^{6-k}$.

As a hint, $D_{12}$ has two conjugacy classes of size $1$, two of size $2$, and $2$ of size $3$.

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By hand, we have $b,aba^{-1},a^2ba-{2},a^3ba^{-3},a^4ba{-4},a^5ba^{-5},bab(ba)^{-1},ba^2b(ba^2)^{-1},ba^3b(ba^3)^{-1},ba^4b(ba^4)^{-1},ba^5b(ba^5)^{-1}$.

But these simplify to $\{b,ba^4,ba^2\}$, using the relations.

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