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We know that any character on a finite group is a class function i.e. they each take a constant value on a given conjugacy class . Is the converse true ? that is let $G$ be a finite group , $g_1,g_2 \in G$ such that for any representation $\rho :G\to GL(V)$ ( where $V$ is a finite dimensional vector space over $\mathbb C$ ) , $\chi _\rho (g_1)=\chi_\rho (g_2)$ , (where $\chi (g):=trace(\rho(g))$) then is it true that $g_1,g_2$ belongs to the same conjugacy class ? Please help . Thanks in advance

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  • $\begingroup$ Yes, this is true. $\endgroup$ – Qiaochu Yuan Jan 8 '16 at 15:29
  • $\begingroup$ @QiaochuYuan: Could you please elaborate on the derivation ? $\endgroup$ – user228169 Jan 8 '16 at 15:33
  • $\begingroup$ This follows once you show that the space of characters has dimension the number of conjugacy classes of $G$, or equivalently that the number of irreducible characters is equal to the number of conjugacy classes of $G$ ("the character table is square"). There are various ways of proving this; one is to show that both of these numbers is the dimension of the center $Z(\mathbb{C}[G])$ of the group algebra. $\endgroup$ – Qiaochu Yuan Jan 8 '16 at 15:41
  • $\begingroup$ Duplicate of math.stackexchange.com/questions/189900 $\endgroup$ – David E Speyer May 21 at 15:29
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The Second Orthogonality relation (column orthogonality of character tables) implies that if $g_1,g_2 \in G$, then $$\sum_{\chi \in Irr(G)}\chi(g_1)\overline{\chi(g_2)}=0 \text{ iff } g_1 \nsim g_2$$ where $\nsim$ indicates not being conjugate to each other. So if all characters $\chi$ take the same values on $g_1$ and $g_2$, then in particular this holds true for irreducible characters. But the above sum then becomes $\sum_{\chi \in Irr(G)}|\chi(g_1|^2$, and this sum is $0$ if and only if $\chi(g_1)=0$ for all $\chi \in Irr(G)$. But this leads to a contradiction for any linear character.

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