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7 minor typos

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 homomomorphisms 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 group-theoretic 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 fom from the character values on group elements.

6 Minor textual changes.

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 homomomorphisms $\sigma: G \to {\rm GL}(n,\mathbb{C})$, 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 group-theoretic 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 a ring 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 ring algebra homomorphism as mapping into the full matrix ring $M_{n}(\mathbb{C})$. However, this kernel is not so easy to read fom the character values on group elements.

5 minor typos

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 homomomorphisms $\sigma: G \to {\rm GL}(n,\mathbb{C})$, where $n = \chi(1)$. But all such representations are equivalent( that is. detemined up to conjugation by a matrix in ${\rm GL}(n,\mathbb{C}).$ GL}(n,\mathbb{C})).$Hence all such representations have the same kernel in the group-theoretic sense. Furthermore, the argument above shows that the kernel can be seen directly from the character values. Hence the kernel of a character is really the kernel of an equivalence class of representations. One of the (many) advantages of working with (complex) characters is that all normal subgroups of a finite group$G$can be determined by inspection of the character table of$G$. It would also be possible to speak of the kernel of a ring homomorphim, but that would be the set of all elements of the group algebra$\mathbb{C}G$sent to the zero matrix by a chosen representation$\sigma$, and we would need to consider the ring homomorphism as mapping into the full matrix ring$M_{n}(\mathbb{C})\$. However, this kernel is not so easy to read fom the character values on group elements.

4 Minor textual changes.
3 Added notes in light of discussion below
2 minor typos
1