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I have given a group and I have to prepare the character table of this given group. I know that firstly I have to find the conjugacy classes of the given group. The group is below:

$T_{4n}=\{a,b:a^{2n}=1, a^n=b^2=1, b^{-1}ab=a^{-1}\}$

and I need to find for $n=5$ i.e. $T_{20}$. I think there is a case about $n$, if $n$ is odd or $n$ is even. I hope that I can prepare the character table after finding the conjugacy classes.

How can I find conjugacy classes of this group?

Thanks for any help...

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Are you sure that is the presentation you have been given? Usually, people don't include redundant relations in their presentations. – Tobias Kildetoft Dec 6 '13 at 8:26
yes,sure that is the presentation. – juliet Dec 6 '13 at 8:29
The thing is, that the given presentation is the dihedral group of order $2n$, and it is called $T_{4n}$. So I am still convinced there is something wrong. – Tobias Kildetoft Dec 6 '13 at 8:31
It might also be a dicyclic group of order 20, using $a^{2n}=1, a^n=b^2, b^{-1}ab=a^{-1}$. – D. N. Dec 6 '13 at 8:34
@DeiborlangNongsiang That is assuming it was written incorrectly. In the current form, it is dihedral. – Tobias Kildetoft Dec 6 '13 at 8:47

Assuming that your given presentation is $T_{4n}=\langle a,b\mid a^{2n}, a^nb^2,abab^{-1}\rangle$,

let $c(x)$ denote the set of all conjugates of $x$ in $T_{4n}$ then

$c(1)=\{1\}$, $c(a^n)=\{a^n\}$, $c(a^i)=\{a^{\pm i}\}$ for $0<i<2n$ and $i\ne n$, $c(a^{2i}b)=\{a^{2j}b\mid 0\leqslant j< n\}$, $c(a^{2i+1}b)=\{a^{2j+1}b\mid 0\leqslant j< n\}$ for $0\leqslant i<2n$.

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