Efficient computation of conjugacy classes of a small group.

I'm trying to construct a character table for a group of order 54 given by:

$$\langle a,b : a^9 = b^6 = 1, b^{-1} a b = a^2\rangle$$

To do this first I need to compute conjugacy classes. This feels like a tedious task and I'm looking for some guidance on how to compute those efficiently.

$$G = \{a^mb^n:0\le m\le 8,0\le n\le 5\}$$. For an element $$a^mb^n\in G$$, $$b^{-1}a^mb^nb = a^{2m}b^n$$ and thus for a fixed $$n$$, $$a^mb^n$$ (with $$m = 1,2,4,5,7,8$$ or with $$m = 3,6$$) are in a same conjugacy class. Moreover, $$a^{-1}a^mb^na = a^{m-1}b^nab^{-n}b^n = a^{m-1}a^{5^n}b^n$$since $$ba^2b^{-1} = a = (a^2)^5$$ and $$a^2$$ is also a generator of $$\langle a\rangle$$. Note that a conjugate by $$a$$ or $$b$$ does not change $$n$$. Therefore, we only need to discuss $$n$$.

(1) $$n=0$$, then $$a^{-1}a^mb^na = a^mb^n$$, so the conjugate by $$a$$ is fixed. So $$\{1\},\{a,a^2,a^4,a^5,a^7,a^8\},\{a^3,a^6\}$$ are conjugacy classes.

(2) $$n=1$$, then $$a^{-1}a^mb^na = a^{m+4}b^n$$. So $$\{b,ab,\dots, a^8b\}$$ is a conjugacy class, since $$m+4$$ runs all the $$m$$'s.

(3) $$n=2$$, then $$a^{-1}a^mb^na = a^{m+24}b^n = a^{m+6}b^n$$. Now $$a^3b^2$$, $$a^6b^2$$ and $$b^2$$ are in the same conjugacy class, while other $$m$$ forms another conjugacy class.

(4) $$n=3$$, then $$a^{-1}a^mb^na = a^{m+124}b^n = a^{m+7}b^n$$. The same as (2) since it runs all the $$m$$'s.

(5) $$n=4$$, then $$a^{-1}a^mb^na = a^{m+3}b^n$$. The same as (3).

(6) $$n=5$$, then $$a^{-1}a^mb^na = a^{m+1}b^n$$. The same as (2).

• Maybe there are some mistakes in calculation, but the key is that the conjugate does not change $n$. – Hongyi Huang May 24 at 11:14

From the defining relations try to get a description of all the elements.

Any element is a "word" involving $$a$$ and $$b$$.

Last condition $$b^{-1}ab = a^2$$ can be rewritten as $$ab= ba^2$$. This modified version says whenever a comes ahead of b it can be pushed to come behind b, but as $$a^2$$. So all words in a and b can be reformulated to have b's in the front and a's at end. As $$a^9=b^6=1$$, any element can be written as $$b^m a^n$$ with $$m=0,1,2,\ldots, 8, \ n=0,1,2,\ldots, 5$$, giving 54 possibilities.

Now try to describe conjuates of elements of type $$a^k$$, then $$b^k$$ and then $$a^k b^l$$ by any $$g$$ ($$g$$ has to be of the form $$b^m a^n$$ as above). It is doable.