abstract-algebra
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.
The definition of $K$-conjugacy is given on the first page of the paper. If $K$ is not algebraically closed, it might be missing some roots of unity. For instance, if $q=13$ and $p=5$, then the elements of $G$ have eigenvalues that are $5$th roots of unity, but the field $\mathbb{Z}/13\mathbb{Z}$ has no $5$th roots of unity. If we try to adjoin a 5th root of unity $\zeta_5$ to $\mathbb{Z}/13\mathbb{Z}$, then we end up adjoin all of its Frobenius conjugates too: $\zeta_5^{13}, \zeta_5^{13^2}, \zeta_5^{13^3}, \zeta_5^{13^4} = \zeta_5$. This last equality follows since $13^4 \equiv 1 \mod 5$.
Since we can't tell the eigenvalue $\zeta_5$ from $\zeta_5^{13}$ using the field $\mathbb{Z}/13\mathbb{Z}$, we also cannot tell the difference between $g$ and $g^{13}$ using $\mathbb{Z}/13\mathbb{Z}$-representations. In particular, if $g$ is a rotation of order 5, then it is $K$-conjugate to each of $g^{13}, g^{13^2}, g^{13^3}, g^{13^4}=g$.
You can imagine the field automorphisms of $K$ act not only on character values, but also on group elements.
Of course, this only works on $K$-regular classes, so don't take $q \in \{2,p\}$ or don't count the reflections (when $q=2$) or the rotations (when $q=p$) at all.
As suggested in the Math Review, see Curtis–Reiner's Representation Theory (not Methods) page 306, §42.8.
The standard realization of a dihedral group as a matrix group views it as a group of symmetries of a convex polygon centered at the origin.
$$\left\langle \begin{bmatrix} -1 & . \\ . & 1 \end{bmatrix}, \begin{bmatrix} \cos(2\pi/p) & -\sin(2\pi/p) \\ \sin(2\pi/p) & \cos(2\pi/p) \end{bmatrix}\right\rangle$$
The matrix corresponding to a rotation has eigenvalues $\zeta_p$ and $\bar \zeta_p = 1/\zeta_p$. The trace of the matrix is just $\zeta_p + \zeta_p^{-1} = 2\cos(2\pi/p)$. It turns out to be easier to change fields if we diagonalize the rotation matrix over $\mathbb{C}$ first.
$$\left\langle \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} \zeta_p & 0 \\ 0 & \bar \zeta_p \end{bmatrix} \right\rangle$$
Notice the rotation matrix still has trace $2\cos(2\pi/2) = \zeta_p + \zeta_p^{-1} \in \mathbb{R}$. This is another example of $K$-conjugacy, where a real matrix cannot have just one complex eigenvalue. The complex eigenvalue come in pairs, related by the Galois group. Notice also that on roots of unity, the action of the Galois groups is $z \mapsto z^{-1}$. In other words, here $q=-1$. The trace of the reflection is always going to be 0, which also is a real number.
Before trying to write our group as matrices over $\mathbb{Z}/q\mathbb{Z}$, let's just try to get our matrices to have trace inside $\mathbb{Z}/q\mathbb{Z}$. Inside the algebraic closure of $\mathbb{Z}/q\mathbb{Z}$ there is a $\zeta_p$, a primitive $p$th root of unity (assuming $p\neq q$). Any polynomial in $\zeta_p$ is also in the algebraic closure. When is one of those polynomials in $\mathbb{Z}/q\mathbb{Z}$ itself? Exactly when raising it to the $q$th power does not change it. In particular, when we include $\zeta_p$, we have to include $\zeta_p^{q}, \zeta_p^{q^2}, \ldots$.
If we do this for $p=5$, $q=13$, we get
$$\left\langle \left[\begin{array}{rr|rr} . & . & 1 & . \\ . & . & . & 1 \\ \hline 1 & . & . & . \\ . & 1 & . & . \end{array}\right], \left[\begin{smallmatrix} \zeta_p & . & . & . \\ . & \zeta_p^q & . & . \\ . & . & \zeta_p^{q^2} & . \\ . & . & . & \zeta_p^{q^3} \end{smallmatrix}\right] \right\rangle$$
Notice how conjugating by the reflection swaps the top 2 $\zeta_p^{q^i}$ with the bottom two. Since this is the same as applying $x \mapsto x^{q^2}$ it sure is nice that $-1 \equiv q^2 \mod p$. Hence we still get a dihedral group over an algebraic closure of $K$, and all of the traces of the matrices lie in $K$ itself. Of course we need to write this over $K$ itself, but replacing the rotation matrix with the companion matrix of its minimal polynomial and the reflection with the permutation matrix that swaps the right basis vectors, we get:
$$\left\langle \begin{bmatrix} . & . & . & 1 \\ . & . & 1 & . \\ . & 1 & . & . \\ 1 & . & . & . \end{bmatrix}, \begin{bmatrix} . & 1 & . & . \\ . & . & 1 & . \\ . & . & . & 1 \\ -1 & -1 & -1 & -1 \end{bmatrix} \right\rangle$$
We've used up all the $\zeta_p^i$ there are, and so this is the only faithful $\mathbb{Z}/q\mathbb{Z}$ representation.
Things work almost identically for any $q$ such that $q$ has multiplicative order $p-1$ mod $p$. The left matrix is always the "reversal" permutation, and the right matrix is always the "1s on the super, -1s on the bottom row" except of course we get $(p-1) \times (p-1)$ matrices in order make sure $\zeta_p$ being used implies $\zeta_p^q$ is used. Since the order of $q$ mod $p$ is $p-1$, that uses up all of the $\zeta_p$.
If $q$ has smaller order mod $p$, say $(p-1)/k$, then you only need to include the $\zeta_p^{q^i}$, and that doesn't use up all of the $\zeta_p^i$. You'll get $k$ bunches of $(p-1)/k$ roots of unity. Each of the $k$ bunches produces its own irreducible $K$-representation. I believe the formulas for the matrix entries over $K$ are a little more complicated, but the "diagonal" version over the algebraic closure (with trace still in $K$) is unchanged and very easy.
If you only care about traces, then more or less everything I said holds and is very practical for every finite group of order coprime to $q$. When the order is not coprime to $q$, you need to worry about $K$-regular classes.
There is a computationally easier way to do this, and you do not even need to revisit the geometric intepretation of $\operatorname{Dih}(6)$. Let $\phi:\operatorname{Dih}(6)\times\{1,2,3,4,5,6\}\rightarrow\{1,2,3,4,5,6\}$ be the group action of the dihedral group of a regular hexagon on the six vertices. Let $\varphi:\operatorname{Dih}(6)\rightarrow\operatorname{Sym}(6)$ be the permutation representation afforded by the action. Then write $\varphi(r)=(123456)$ and $\varphi(s)=(13)(46)$. Then since $\varphi$ is a homomorphism, and $\operatorname{Dih}(6)=\{1,r,r^2,\dots,s,sr,sr^2\dots\}$ is generated by $r$ and $s$, you can compute the cycle representation of each element of $\operatorname{Dih}(6)$, e.g. $\varphi(sr^2)=\varphi(s)\varphi(r)^2=(13)(46)(123456)^2=\dots$ (check it yourself :D).
P.S. Hmm... it seems that your answer is correct, but you should check it again using the group action approach anyway.
Best Answer
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 $3$.