Why is the kernel of a character equal to the kernel of the representation that affords the character?

closed as too localized by S. Carnahan♦ May 16 '11 at 6:56
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This is too elementary a question for MO. In the case of complex characters, the kernel of a character $\chi$ of a finite group $G$ is $\{g \in G: \chi(g) = \chi(1) \}.$ On the other hand, if the character $\chi$ is afforded by a representation $\sigma$ (that is, $\chi(g) = {\rm trace}(g\sigma)$ for all $g \in G$), then for each $g \in G$, the eigenvalues of $g \sigma$ are all $o(g)$th roots of unity, where $o(g)$ is the order of $g$. Hence the triangle inequality yields $\chi(g) \leq \chi(1)$, and the only way we can have $\chi(g) = \chi(1)$ is if all eigenvalues of $g\sigma$ are equal to 1. Since $g\sigma$ has finite order, in that case, $g \sigma$ must be the identity matrix. Thus ${\rm ker}\chi$ is precisely equal to ${\rm ker} \sigma$, as the inclusion ${\rm ker}\sigma \leq {\rm ker}\chi$ is clear. Added later, in light of the discussion below: ``kernel" implicitly refers to the kernel of a group homomorphism here (actually, several group homomorphisms, as we will see). The (complex) character $\chi$ may be afforded by several different homomorphisms $\sigma: G \to {\rm GL}(n,\mathbb{C})$, where $n = \chi(1)$. But all such representations are equivalent( that is. detemined up to conjugation by a matrix in ${\rm GL}(n,\mathbb{C})).$ Hence all such representations have the same kernel in the grouptheoretic sense. Furthermore, the argument above shows that the kernel can be seen directly from the character values. Hence the kernel of a character is really the kernel of an equivalence class of representations. One of the (many) advantages of working with (complex) characters is that all normal subgroups of a finite group $G$ can be determined by inspection of the character table of $G$. It would also be possible to speak of the kernel of an algebra homomorphim, but that would be the set of all elements of the group algebra $\mathbb{C}G$ sent to the zero matrix by a chosen representation $\sigma$, and we would need to consider the algebra homomorphism as mapping into the full matrix ring $M_{n}(\mathbb{C})$. However, this kernel is not so easy to read from the character values on group elements. 


My question is why the kernel of a character $\chi$ of a finite group $G$ is $ \{g \in G: \chi(g) = \chi(1) \} $, instead of $ \{ g\in G: \chi(g) = 0 \} $? 

