Let us show that we can identify a couple of important normal subgroups of $G$ just be identified just by looking at the character table. These are the derived group $G'$ (see Proposition 2.7.1) and the center $Z (G)$.

**Proposition 3.2.1:***
Let $G$ be a finite group and let $g \in G$. Then $g \in G'$ if and only if
$\chi (g) = 1$ for all linear characters $\chi$ of $G$.
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**Proof. **(Click to Expand/Collapse)

By Proposition 2.7.1 the linear characters of $G$ are the
same as the linear characters of the abelian group $G^{\operatorname{ab}} = G / G'$.
Thus if $\bar{g}$ is the image of $g$ in $G / G'$, then $\bar{g} = 1$ if and
only if $\chi ( \bar{g}) = 1$ for all linear characters of $G^{\operatorname{ab}}$
(see Proposition 2.2.2). Thus $g \in G'$ if and only if $\chi
(g) = 1$ for all linear characters of $G$.

The next result is closely related to Proposition 2.7.4,
establishing the existence of a central character of any irreducible
representation $(\pi, V)$. Recall that this is a *linear* character
$\omega : Z (G) \longrightarrow \mathbb{C}^{\times}$ such that $\pi (z)$ acts
by the scalar $\omega (z)$ when $z \in Z (G)$. The next result is a partial
converse to this fact.

**Proposition 3.2.2:***
Let $g \in G$. Then $g \in Z (G)$ if and only if $| \chi_{\pi} (g) | = |
\chi_{\pi} (1) |$ for every irreducible character.
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**Proof. **(Click to Expand/Collapse)

If $\omega$ is the central character of $\pi$, then by
Proposition 2.7.4 we have $\chi_{\pi} (z) = d \omega (z)$
when $z \in Z (G)$, where $d = \chi (1) = \dim (V)$ is the degree of $\chi$.
Since $\omega (z)$ is a root of unity, this means that $| \chi_{\pi} (g) | =
d = \chi_{\pi} (1)$. This proves one direction.

Conversely, suppose that $| \chi_{\pi} (g) | = \chi_{\pi} (1)$ for all irreducible representations $\pi$. We will show that $g \in Z (G)$. We need the following fact

**Lemma 3.2.1:***
( The "Converse to the triangle inequality'') Let $x_1, \cdots,
x_n$ be vectors in $\mathbb{R}^k$ such that $|x_1 + \cdots + x_n | = |x_1
| + \ldots + |x_n |$. Then $x_1, \cdots, x_n$ are proportional. That is,
there exists a vector $\xi$ and nonnegative real numbers $w_1, \cdots,
w_n$ such that $x_i = w_i \xi$.
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Perhaps this is obvious, but if not, we recall the "law of cosines'' in the form \[ |x_1 + x_2 | = \sqrt{|x_1 |^2 + |x_2 |^2 + 2| x_1 | |x_2 | \cos (\theta)} \] where $\theta$ is the angle between the vectors $x_1$ and $x_2$, so if this equals $|x_1 | + |x_2 |$ then $\theta = 0$. This proves the case $n = 2$, and the general case follows by an induction: if the statement is true for $n - 1$ (when $n \ge 3$) then by the usual triangle inequality \[ |x_1 + \ldots + x_n | \le |x_1 + \ldots + x_{n - 1} | + |x_n | \le |x_1 | + \ldots + |x_n |. \] If $|x_1 + \cdots + x_n | = |x_1 | + \ldots + |x_n |$ then the inequalities here are equalities. Thus we may subtract $|x_n |$ to obtain \[ |x_1 + \ldots + x_{n - 1} | \le |x_1 | + \ldots + |x_{n - 1} | \] and invoke induction. This proves the Lemma.

Now if $\omega_1, \cdots, \omega_d$ are the eigenvalues of $\pi (g)$, then $| \omega_i | = 1$ since $\omega_i$ is a root of unity. It follows from our assumption that $| \chi_{\pi} (g) | = | \omega_1 + \ldots + \omega_d |$ equals $d = | \omega_1 | + \ldots + | \omega_d |$ that the complex numbers $\omega_1, \cdots, \omega_d$ are proportional, so the eigenvalues of $\pi (g)$ are all equal – say $\omega$. Since $\pi (g)$ is diagonalizable, we see that $\pi (g)$ is a scalar. In particular $\pi (g) \pi (h) = \pi (h) \pi (g)$ for all $h \in G$ (since the scalar linear transformation $\pi (g)$ commutes with an obvious linear transformation $\pi (h)$.

We've proved that $\pi (g) \pi (h) = \pi (h) \pi (g)$ for all $h \in G$ when $\pi$ is any irreducible representation. It follows that $\pi (g h) = \pi (h g)$ where $(\pi, V)$ is any representation of $G$, because we can decompose $V$ into invariant irreducible subspaces.

For example, we can take $\pi$ to be the regular representation. Then $\pi (g h) = \pi (h g)$ implies that $h g = g h$, and so $g$ is in the center.

**Exercise 3.2.1:***
Let $(\pi, V)$ be a representation of $G$
(irreducible or not) with character $\chi$. Prove that $\chi (g) = \chi (1)$
for some $g \in G$ if and only if $g \in \ker (\pi)$. ( Hint:
use the converse to the triangle inequality!)
*