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Let $G$ be a group of order 10 having 4 conjugacy classes and the above character table. Complete the table.

It's easy to get the degrees of the remaining 2 irreducible characters to be 2 and 1. But I am having trouble finding out the rest 6 values. I am using the orthogonality relations but the equations are messy.

Please help. I know there must be an easier way.

saubhik
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    Irreducible characters are closed under Galois conjugation (exercise; see http://math.stackexchange.com/questions/42879/shortest-way-of-proving-that-the-galois-conjugate-of-a-character-is-still-a-char for some discussion), so the Galois conjugate of $\chi_2$ is another irreducible character, and then finding the final character is straightforward. – Qiaochu Yuan Apr 23 '15 at 08:56
  • An alternate approach: if there are two characters of degree $1$, then the abelianization of $G$ must be $\mathbb{Z}_2$, with the two characters corresponding to the trivial and sign character of $\mathbb{Z}_2$. The kernel of the sign character must be a union of conjugacy classes of order $5$ containing the identity, hence must be the first three conjugacy classes, and so the sign character must be $1, 1, 1, -1$. – Qiaochu Yuan Apr 23 '15 at 09:01
  • A third approach, perhaps the least satisfying, to just determine $G$. We know that $G$ has abelianization $\mathbb{Z}_2$. Since it has an element of order $2$ by Cauchy's theorem, it must be a semidirect product $\mathbb{Z}_5 \rtimes \mathbb{Z}_2$. There are exactly two such semidirect products (in fact there are exactly two groups of order $10$!), and exactly one of them is nonabelian, namely the one where $\mathbb{Z}_2$ acts by $-1$. This is the dihedral group $D_5$, and you can in fact compute the character tables of all of the dihedral groups $D_n$ by inducing from $C_n$. – Qiaochu Yuan Apr 23 '15 at 09:10
  • @QiaochuYuan thanks a lot! – saubhik Apr 23 '15 at 09:15

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