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I've just started studying representation theory of finite groups and I'm having trouble finding the character table of the group $G:=\left\langle x,y | x^7=e=y^3, y^{-1}xy=x^2\right\rangle$.

This is what I've got so far (not sure if it's right):

$\begin{array}{c|c c c c c} \text{Class length} & 1 & 3 & 3 & 7 & 7 \\ \hline \text{Class rep.} & e & x & x^3 & y & y^2\\ \hline\hline 1&1&1&1&1&1\\ \tau_1 & 1 & 1 & 1 & \dfrac{1}{2}(-1+\sqrt{3} i) & \dfrac{1}{2}(-1-\sqrt{3} i)\\ \tau_2 & 1 & 1 & 1 & \dfrac{1}{2}(-1-\sqrt{3} i) & \dfrac{1}{2}(-1+\sqrt{3} i)\\ \tau_3 & 3\\ \tau_4 & 3\\ \end{array}$

But I don't know how to find the 3 dimensional ones. Could anyone offer help?

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There are a lot of ways to proceed depending on what tools you have at your disposal. You can aggressively exploit orthogonality and the fact that you know the character of the regular representation. You can try to write down induced representations. You can try to write down permutation representations... – Qiaochu Yuan Mar 9 '12 at 19:25
It seems reasonable to try those induced characters of linear ones of a normal subgroup of index 3, which this group must possess. I have not yet worked out the details, but if they turn out to be the remaining irreducible characters, then they are all monomial and hence G is a M-group. Well, all speculations. – awllower May 23 '12 at 2:21

Consider the representation $\tau_3$ (or $\tau_4$) and an eigenvalue $w$ of the the of $x$. Then $w$ is a $7^{th}$ root of unity. Since $x^2$ and $x^4$ are conjugate to $x$, $w^2$ and $w^4$ are also eigenvalues of $x$. All eigenvalues of $x$ cannot be $1$ by column orthonormality. So, $w$, $w^2$ and $w^4$ are distinct. Filling the rest of the table should not be hard now.

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These representations also emerge by inducing a non-trivial character of the cyclic 7-subgroup $H$. Say, if $\chi(x)=w$, then $\mathrm{Ind}_H^G(\chi)(x)=w+w^2+w^4.$ – Jyrki Lahtonen Jun 9 '12 at 15:35

By column orthogonality you can immediately work out the entries in the last two columns. For the first two missing columns, the key observation is that x and $x^{-1}$ are not conjugate. This means that the first two missing columns are complex conjugates of each other. You then have to split into cases based on whether the entries are real or not, but column orthogonality (both the column with itself and with the identity column) gets you down to only one possibility.

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