I know that normal subgroups are the union of some conjugacy classes

Conjugacy classes are represented by the the columns in a matrix

How could we use character values in the table to determine normal subgroups?


1 Answer 1


This is quite well-known and can be found in books on representation theory. Here is an explanation, which is far from being original.

First fact : $N$ is a normal subgroup of a finite group $G$ if and only if there exists a character $\chi$ of $G$ such that $N = \ker \chi := \{g \in G | \chi(g)=\chi(1)\}$. Indeed, if $N$ is normal then $G$ acts on the complex algebra $\mathbf{C}[G/N] = \displaystyle \bigoplus_{gN \in G/N} \mathbf{C} e_{gN}$ by $h \cdot e_{gN}=e_{hgN}$. This is a linear representation of $G$ (coming from the regular representation of $G/N$). Let $\chi$ be its character. It is easy to check that $\chi(h) = 0$ if $h \notin N$ and $\chi(h) = \mathrm{Card}(G/N) = \chi(1)$ if $h \in N$. So $N = \ker \chi$. Conversely, using the fact that a character is constant on every conjugacy class, any subgroup of the form $\ker \chi$ is normal.

Second fact : if $\rho : G \to \mathrm{GL}(V)$ is the representation associated to the character $\chi$ then $\ker \rho = \ker \chi$. The inclusion $\subseteq$ is trivial. Conversely, assume $\chi(g) = \chi(1) = \dim V$. Since the eigenvalues of $\rho(g)$ are roots of $1$ and $\chi(g)$ is the sum of the eigenvalues (with multiplicities), these eigenvalues are forced to be all equal to $1$. So $\rho(g) = \mathrm{id}_V$, that is to say $g \in \ker \rho$.

Third fact : if $\chi = \displaystyle \sum_{i=1}^r n_i \chi_i$ (where the $\chi_i$ are pairwise distinct irreducible characters and $n_i \geq 1$) then $\ker \chi = \displaystyle \bigcap_{i=1}^r \ker \chi_i$. Writing $\rho, \rho_1,\ldots,\rho_r$ for the corresponding representations, $\rho$ is the direct sum of copies of $\rho_1,\ldots,\rho_r$ so $\ker \rho = \displaystyle \bigcap_{i=1}^r \ker \rho_i$. Then apply the second fact.

Conclusion : with your character table, you can read the subgroups $N_i:=\ker \chi_i$. Then the normal subgroups of $G$ are exactly the intersections of some $N_i$.

  • 1
    $\begingroup$ Good summary! I've always had trouble retaining this, even after a couple readings of texts. This makes it easier. $\endgroup$
    – rschwieb
    May 17, 2016 at 16:29

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