# Character Table from Generators of a Group

Let $$G=\langle x,y|x^5=y^4=yxy^{-1}x^{-2}=1\rangle$$ be a group. How would I construct the full character table of this group with no other given information?

Here is what I know regarding characters:

-If $$\chi$$ is an irreducible character of dimension $$\chi(1)=n$$, then $$n$$ divides the order of $$G$$.

-If $$G$$ has $$r$$ conjugacy classes, then the number of irreducible characters is equal to $$r$$.

-A character has dimension $$1$$ if it is irreducible. Otherwise it is the sum of irreducible characters.

-Characters are constant on conjugacy classes of $$G$$.

• Depends on how much character theory you know. Do you know that the degrees of irreducible characters divide the order of the group? – Tobias Kildetoft Dec 4 '18 at 7:17
• @TobiasKildetoft Yes. Don't know anything about modules though, just as a note. – JB071098 Dec 4 '18 at 7:25
• So do you see how to find the degree 1 characters? And do you see what degree(s) to look for for the rest? – Tobias Kildetoft Dec 4 '18 at 7:31
• @TobiasKildetoft I will edit the post to contain everything I know. – JB071098 Dec 4 '18 at 7:34

Here are some thoughts to get you started. The group is a semidirect product of cyclic groups $$A$$ and $$B$$ of order $$5$$ and $$4$$, where no element of $$B$$ centralizes any element of $$A$$, and the nontrivial elements of $$A$$ are all conjugate. So is has no elements of order $$10$$ or $$20$$, all elements lie either in $$A$$ or a conjugate of $$B$$. So there are a total of five conjugacy classes, two containing elements of order $$4$$ and one each containing elements of orders $$1,2,5$$.
Hence there are a total of $$5$$ irreducible characters. Since $$G/A = AB/A \cong B$$ is cyclic of order $$4$$, there are $$4$$ characters of degree $$1$$ with $$A$$ in their kernel, and you should be able to write those down. So there is just one more, and since the sum of the squares of the degrees of the irreducible characters is $$|G|=20$$, the other must have degree $$4$$, and you can work out its values by using the orthogonality relations.
• @JB071098 It is also possible to explicitly obtain the irreducible $4$-dimensional representation by linearizing the action on the cosets of the subgroup generated by $y$ and taking the complement of the trivial representation in that (but using orthogonality is probably easier, especially if you combine it with using that the character will be $0$ on any element which is not in the kernel of all the $1$-dimensional ones). – Tobias Kildetoft Dec 4 '18 at 8:57