# Character Tables

The most coveted piece of information about a group is its character table, a tabulation of the value of its irreducible characters. Many character tables are tabulated in the Atlas of finite groups, by Conway, Curtis, Norton, Parker and Wilson, and a free software program called GAP is available that can compute many character tables. GAP is bundled with SAGE, which may be found at:

http://www.sagemath.org

http://www.gap-system.org

Let us compute the character table of $S_3$. There are three conjugacy classes, so there are three irreducible representations. Two linear characters present themselves, namely the trivial character $\chi_1$, and the alternating character, which we call $\chi_2$. There must be exactly one other character $q_3$, and since $\sum d_i^2 = |G| = 6$, we must have $d_3 = 2$. We may tabulate the information that we have in the following table: $\begin{array}{|l|l|l|l|} \hline & 1 & \left( 123) \right. & (12)\\ \hline \chi_1 & 1 & 1 & 1\\ \hline \chi_2 & 1 & 1 & - 1\\ \hline \chi_3 & 2 & & \\ \hline \end{array}$ Here we have chosen representatives for the three conjugacy classes, and labeled the columns with them, and left blank the two entries that we haven't computed yet. These can be determined by column orthogonality: we have $0 = \chi_1 (1) \chi_1 (g) + \chi_2 (1) \chi_2 (g) + \chi_3 (1) \chi_3 (g)$ when $g \neq 1$, and this allows us to finish the character table: $\begin{array}{|l|l|l|l|} \hline S_3 & 1 & \left( 123) \right. & (12)\\ \hline \chi_1 & 1 & 1 & 1\\ \hline \chi_2 & 1 & 1 & - 1\\ \hline \chi_3 & 2 & - 1 & 0\\ \hline \end{array}$

We can understand $\chi_3$ directly as an induced representation. We will induce from the subgroup $H = \left\langle (123) \right\rangle$. This group is cyclic of order 3, and it has two nontrivial linear characters; both would produce the same induced representation. Let us take $\theta : H \longrightarrow \mathbb{C}$ to be the character whose values are as follows: $\begin{array}{|l|l|l|l|} \hline h & 1 & (123) & (132)\\ \hline \theta (h) & 1 & \zeta & \zeta^{- 1}\\ \hline \end{array}$ where $\zeta = e^{2 \pi i / 3}$ is a primitive cube root of unity. Let us choose our coset representatives to be $x_1 = 1$, $x_2 = (12)$. Then $\theta^G (g) = \dot{\theta} (g) + \dot{\theta} ((12) g (12)^{- 1} .$ From this we compute easily that \begin{eqnarray*} \theta^G (1) & = & 1 + 1 = 2,\\ \theta^G ((123)) & = & \zeta + \zeta^{- 1} = - 1,\\ \theta^G ((12)) & = & 0 + 0 = 0, \end{eqnarray*} and so this induced character coincides with $\chi_3$.

But if we construct $\chi_3$ this way, how do we know that it is irreducible? A general answer to when induced representations are irreducible will be given later in terms of Mackey theory, but for the time being, all we know is that inducing an irreducible character from a subgroup may produce an irreducible, or it may not. However it is easy enough to check.

Lemma 2.7.1: Let $\chi$ be a character of the finite group $G$. Then $\chi$ is irreducible if and only if $\left\langle \chi, \chi \right\rangle = 1$.

Proof. (Click to Expand/Collapse)

Decompose $\chi$ in terms of the irreducible characters $\chi_1, \cdots, \chi_h$: write $\chi = \sum d_i \chi_i$, where the $d_i$ are nonnegative integers. Since the $\chi_i$ are orthonormal, we have $\left\langle \chi, \chi \right\rangle = \sum d_i^2$. Thus $\left\langle \chi, \chi \right\rangle = 1$ if and only if one $d_i$ is 1, the others 0. That means $\chi = \chi_i$ for some $i$.

In the case at hand, we know that $1$ has 1 conjugate, $(123)$ has 2 conjugates and (12) has 3. We have to take these numbers into account in computing the inner product $\left\langle \theta^G, \theta^G \right\rangle$. We find $\left\langle \theta^G, \theta^G \right\rangle = \frac{1}{6} \left( | \theta^G (1) |^2 + 2| \theta^G ((123)) |^2 + 3| \theta^G ((12)) |^2 \right) = \frac{1}{6} (1 \cdot 2^2 + 2 (- 1)^2 + 3 \cdot 0^2) = 1,$ so the character is indeed irreducible.

Our construction of the induced character did not produce an explicit representation, though we promised a more representation-theoretic construction in the next chapter. So let us give an alternative construction of $\chi_3$ that actually produces a representation. We note that $S_3 \cong D_6$ is a semidirect product, and we can describe it by generators and relations. Let $x = (123)$ and $y = (12)$. The $S_3 = \left\langle x, y|x^3 = y^2 = 1, y x y^{- 1} = x^{- 1} \right\rangle .$ Referring to the discussion of (1.8.1), this means that the following universal property is satisfied:

• For any group $H$ with generators $\xi$ and $\eta$ that satisfy the same relations, that is \begin{equation} \xi^3 = \eta^2 = 1, \hspace{2em} \eta \xi \eta^{- 1} = \xi^2 \end{equation} there is a unique group homomorphism $f : G \longrightarrow H$ with $f (x) = \xi$ and $f (y) = \eta$.
Now let $H = \operatorname{GL}_2 (\mathbb{C})$, and let $\xi = \left(\begin{array}{cc} \zeta & \\ & \zeta^{- 1} \end{array}\right), \hspace{2em} \eta = \left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right), \hspace{2em} \zeta = e^{2 \pi i / 3} .$ We see easily that (2.7.1) is satisfied, and so there is a homomorphism that we will call $\pi_3 : G \longrightarrow \operatorname{GL}_2 (\mathbb{C})$ such that $\pi_3 ((123)) = \left(\begin{array}{cc} \zeta & \\ & \zeta^{- 1} \end{array}\right), \hspace{2em} \pi_3 ((12)) = \left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right) .$ Observe that $\chi_3 (g) = \operatorname{tr} \; \pi_3 (g)$ when $g = 1, (123)$ or $(12)$, so $\chi_3$ is the character of $\pi_3$. One can detect in this description the character $\theta$ of the group $H = \left\langle (123) \right\rangle$ that we induced in our first description of the induced representation, and we hope that the reader will get some intuition for the induced representation underlying the induced character in advance of the formal discussion that will come later; we have proven that the induced representation exists, but rather indirectly.

We will describe two other methods by which the values of $\chi_3$ can be obtained.

• Since the action of $S_3$ on $\{1, 2, 3\}$ is doubly transitive, we can subtract 1 from the permutation character and get an irreducible character by Theorem 2.6.1. Thus we count the number of fixed points and subtract 1 to get these values:
$\begin{array}{|l|l|l|l|} \hline & 1 & \left( 123) \right. & (12)\\ \hline \chi_3 & 2 & - 1 & 0\\ \hline \end{array}$
• By Theorem 2.6.2, we have $\chi_{\operatorname{reg}} = \chi_1 + \chi_2 + 2 \chi_3$, and the values of $\chi_{\operatorname{reg}}$, $\chi_1$ and are known, so $\chi_3 = \frac{1}{2} (\chi_{\operatorname{reg}} - \chi_1 - \chi_2)$. By this method, if we have computed all but one of the irreducible characters of $G$, we get the last one for free!
Generally, we can find the linear characters – those irreducibles with $\chi_i (1) = 1$ very easily. They all factor through a certain computable quotient of $G$, the abelianization. Let $G'$ be the subgroup of $G$ generated by all commutators, which are elements of the form $[x, y] = x y x^{- 1} y^{- 1}$. The group $G'$ is called he derived group or commutator subgroup of $G$

Proposition 2.7.1: The derived group $G'$ is normal. Let $G^{\operatorname{ab}} = G / G'$, and let $p : G \longrightarrow G^{\operatorname{ab}}$ be the projection map. The group $G^{\operatorname{ab}}$ is abelian, and if $\phi : G \longrightarrow A$ is any homomorphism of $G$ into an abelian group, then $\ker (\phi) \supseteq G'$.

Proof. (Click to Expand/Collapse)

To see that the derived subgroup is normal, we note the identity $z [x, y] z^{- 1} = [z x z^{- 1}, z y z^{- 1}] .$ Thus every conjugate of a commutator is a commutator. Since the set of commutators is invariant under conjugation, the group $G'$ they generate is invariant under conjugation, which is to say that $G' \vartriangleleft G$.

To see that $G^{\operatorname{ab}}$ is abelian, we observe that any two typical elements may be written $p (x)$ and $p (y)$. Now $p (x) p (y) p (x)^{- 1} p (y)^{- 1} = p ([x, y]) = 1$ in $G^{\operatorname{ab}}$, so $p (x) p (y) = p (y) p (x)$.

Now given any homomorphism $\phi : G \longrightarrow A$ into an abelian group, $\phi ([x, y]) = \phi (x) \phi (y) \phi (x)^{- 1} \phi (y)^{- 1} = 1$ because $A$ is abelian, so $\ker (\phi)$ contains all commutators and hence the group $G'$ they generate.

Proposition 2.7.2: Since $G^{\operatorname{ab}}$ is abelian, all its irreducible characters are linear, and if $\chi : G^{\operatorname{ab}} \longrightarrow \mathbb{C}^{\times}$ is a linear character, then $\chi \circ p$ is a linear character of $G$. Every linear character of $G$ is of this form, so $|G^{\operatorname{ab}} |$ is the total number of linear characters of $G$.

Proof. (Click to Expand/Collapse)

It is obvious that if $\chi : G^{\operatorname{ab}} \longrightarrow \mathbb{C}^{\times}$ is a linear character, then $\chi \circ p$ is a linear character of $G$. What requires proof is that every linear character of $G$ is of this form. If $\theta : G \longrightarrow \mathbb{C}^{\times}$ is a linear character, that is, a homomorphism, it is a homomorphism into an abelian group and so $\ker (\theta) \supset G = \ker (p)$, which means that we can define $\chi : G^{\operatorname{ab}} \longrightarrow \mathbb{C}^{\times}$ by $\chi (p (x)) = \theta (x)$, and this $\chi$ is well-defined; and since $\theta = \chi \circ p$, the theorem is proved.

Exercise 2.7.1: Prove that any homomorphism of $G$ into an abelian group "factors through'' $G^{\operatorname{ab}}$. In other words, if $A$ is abelian and $f : G \longrightarrow A$ is a homomorphism, there is a unique homomorphism $F : G^{\operatorname{ab}} \longrightarrow A$ such that $f = F \circ p$. Observe that this is a universal property, and deduce that the abelianization can be made into a functor from the category of groups to the category of abelian groups.

To see Proposition 2.7.1 in action, let us compute the character table of the dihedral group $D_8 = \left\langle x, y|x^4 = y^2 = 1, y x y^{- 1} = x^{- 1} \right\rangle .$ A first step is to compute the commutator subgroup. We have $[x, y] = x y x^{- 1} y^{- 1} = x (y x y^{- 1})^{- 1} = x (x^{- 1})^{- 1} = x^2,$ so the commutator subgroup contains $x^2$, and $G' \supseteq \left\langle x^2 \right\rangle$. On the other hand, $x^2$ lies in the center, and the group $G / \left\langle x^2 \right\rangle$ is abelian, since if $\xi$ and $\eta$ are the images of $x$ and $y$ in $G / \left\langle x^2 \right\rangle$, we have $\xi^2 = \eta^2 = 1$, $\eta \xi \eta^{- 1} = \xi^{- 1} = \xi$, so $\xi$ and $\eta$ commute. Thus the projection $G \longrightarrow G / \left\langle x^2 \right\rangle$ is a homomorphism into an abelian group, and so by Proposition 2.7.1 its kernel contains $G'$; thus $\left\langle x^2 \right\rangle \supseteq G'$. We have proved that $G' = \left\langle x^2 \right\rangle$. It has order 2, so $G^{\operatorname{ab}}$ has order 4, and in fact is generated by $\xi$ and $\eta$ which both have order 2, so $G^{\operatorname{ab}} = \left\langle \xi, \eta | \xi^2 = \eta^2 = 1, \xi \eta = \eta \xi \right\rangle \cong Z_2 \times Z_2 .$ We can write its character table down: $\begin{array}{|l|l|l|l|l|} \hline & 1 & \xi & \eta & \xi \eta\\ \hline \theta_1 & 1 & 1 & 1 & 1\\ \hline \theta_2 & 1 & 1 & - 1 & - 1\\ \hline \theta_3 & 1 & - 1 & 1 & - 1\\ \hline \theta_4 & 1 & - 1 & - 1 & 1\\ \hline \end{array}$ We pull these characters back to obtain four linear characters of $D_8$, and we can be sure we've exhausted the linear characters.

We can also use the generators and relations to describe the conjugacy classes. Computing the conjugacy classes is always a step in computing the character table of a group, since one needs to know the columns as well as the rows. In this case, we find five classes: $\begin{array}{c} \{1\}\\ \{x, x^{- 1} \}\\ \{x^2 \}\\ \{y, x^2 y\}\\ \{x y, x^{- 1} y\} \end{array}$ To see that this is correct, use the generators and relations to compute the effect of conjugation on each element. For example, to see that $\{y, x^2 y\}$ is a conjugacy class, we note that conjugation by $y$ has no effect on either element, but conjugation by $x$ interchanges $y$ and $x^2 y$, and so this is a conjugacy class, and similar considerations apply to each. (With practice, this is easy.)

Now we know this much of the character table: $\begin{array}{|l|l|l|l|l|l|} \hline & 1 & x & x^2 & y & x y\\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1\\ \hline \chi_2 & 1 & 1 & 1 & - 1 & - 1\\ \hline \chi_3 & 1 & - 1 & 1 & 1 & - 1\\ \hline \chi_4 & 1 & - 1 & 1 & - 1 & 1\\ \hline \chi_5 & & & & & \\ \hline \end{array}$ There are five conjugacy classes, hence five irreducible characters and since $\sum d_i^2 = 8$ we must have $d_5 = \chi_5 (1) = 2$.

Now there are three methods by which we can compute the character table:

• We may use column orthogonality to determine $\chi_5$;
• We may use $\chi_5 = \frac{1}{2} (\chi_{\operatorname{reg}} - \chi_1 - \chi_2 - \chi_3 - \chi_4)$;
• We may construct $\chi_5$ as an induced character from the subgroup $\left\langle x \right\rangle$. Induce the character such that $x \longmapsto i$.
In each case we get the same values: $\begin{array}{|l|l|l|l|l|l|} \hline D_8 & 1 & x & x^2 & y & x y\\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1\\ \hline \chi_2 & 1 & 1 & 1 & - 1 & - 1\\ \hline \chi_3 & 1 & - 1 & 1 & 1 & - 1\\ \hline \chi_4 & 1 & - 1 & 1 & - 1 & 1\\ \hline \chi_5 & 2 & 0 & - 2 & 0 & 0\\ \hline \end{array}$

Exercise 2.7.2: Use each of the above methods to compute $\chi_5$ for this character table. I hope each method gives the same $\chi_5$.

Exercise 2.7.3: Use similar methods to compute the character table of the quaternion group $Q = \left\langle x, y|x^4 = y^4 = 1, x^2 = y^2, y x y^{- 1} = x^{- 1} \right\rangle .$

Now let us give find the character table for $S_4$. We find that there are five conjugacy classes, so we expect five irreducible characters. We know two linear ones already, namely the trivial character and the alternating character. We call these $\chi_1$ and $\chi_2$. So far we have: $\begin{array}{|l|l|l|l|l|l|} \hline & 1 & (123) & (12) (34) & (12) & (1234)\\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1\\ \hline \chi_2 & 1 & 1 & 1 & - 1 & - 1\\ \hline \end{array}$ (I've ordered the conjugacy classes so that the even ones come first.) Now the action of $S_4$ on $\{1, 2, 3, 4\}$ is doubly transitive, so we can get another character by taking the number of fixed points and subtracting 1. $\begin{array}{|l|l|l|l|l|l|} \hline & 1 & (123) & (12) (34) & (12) & (1234)\\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1\\ \hline \chi_2 & 1 & 1 & 1 & - 1 & - 1\\ \hline \chi_4 & 3 & 0 & - 1 & 1 & - 1\\ \hline \end{array}$ We're calling this character $\chi_4$ so that when we finish the table, the characters will be ordered by degree. Now we get one more character for free, but let us take a moment to explain why.

We will eventually prove that if $\chi, \theta$ are characters then $\chi \theta$ is a character (though usually not irreducible, even if $\chi$ and $\chi'$ both are). Here is a special case:

Proposition 2.7.3: If $\chi$ is a character, and $\theta$ is a linear character, then $\chi \theta$ is a character, and is irreducible if and only if $\chi$ is.

Proof. (Click to Expand/Collapse)

Let $\chi$ be the character of $\pi : G \longrightarrow \operatorname{GL} (V)$. We define another representation $\pi'$ on the same subspace by $\pi' (g) = \theta (g) \pi (g) .$ We note that since $\theta (g)$ is just a complex number, this makes sense, and since $\theta (g g') = \theta (g) \theta (g')$ it is clear that this is a representation. The character is $\operatorname{tr} \left( \theta (g) \pi (g) \right) = \theta (g) \cdot \operatorname{tr} \left( \pi (g) \right) = \theta (g) \chi (g),$ again since $\theta (g)$ is just a complex number.

If $U$ is invariant subspace for $\pi$ it is also obviously invariant for $\pi'$, so the statement about irreducibility is obvious.

Now returning to $S_4$, $\chi_2$ is a linear character, so $\chi_2 \chi_4$ is an irreducible character. We call it $\chi_5$. We now have: $\begin{array}{|l|l|l|l|l|l|} \hline & 1 & (123) & (12) (34) & (12) & (1234)\\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1\\ \hline \chi_2 & 1 & 1 & 1 & - 1 & - 1\\ \hline \chi_4 & 3 & 0 & - 1 & 1 & - 1\\ \hline \chi_5 & 3 & 0 & - 1 & - 1 & 1\\ \hline \end{array}$ The last character $\chi_3$ can be found by various methods. For example, we have already pointed out that by using the regular representation, one can get the last character for free if one has all but one irreducible characters. Alternatively, one may construct a homomorphism $S_4 \longrightarrow S_3$. One way to do this is to observe that $S_4$ acts on its conjugacy class $\{(12) (34), (13) (24), (14) (23)\}$ of order 3, giving rise to such a homomorphism. Composing $S_3$'s character of degree 2 with this map $S_4 \longrightarrow S_3$ gives us $\chi_3$. Either way, here's the completed character table. $\begin{array}{|l|l|l|l|l|l|} \hline S_4 & 1 & (123) & (12) (34) & (12) & (1234)\\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1\\ \hline \chi_2 & 1 & 1 & 1 & - 1 & - 1\\ \hline \chi_3 & 2 & - 1 & 2 & 0 & 0\\ \hline \chi_4 & 3 & 0 & - 1 & 1 & - 1\\ \hline \chi_5 & 3 & 0 & - 1 & - 1 & 1\\ \hline \end{array}$

Exercise 2.7.4: Compute the character table of $A_4$. (Hint: there are 4 conjugacy classes; not all 3-cycles are in the same class.)

Next, let us consider a nonabelian group of order 39. The group is the semidirect product of a group of order 13 by a group of order 3; it has generators and relations $\left\langle x, y|x^{13} = y^3 = 1, \; y x y^{- 1} = x^3 \right\rangle .$ Note that $3$ has order 3 in $(\mathbb{Z}/ 13\mathbb{Z})^{\times}$, since $3^3 = 27 \equiv 1$ modulo 3, so the automorphism $x \longmapsto x^3$ of $Z_{13}$ has order 3 in $\operatorname{Aut} (Z_{13})$.

Let us compute the conjugacy classes. We do this by picking an element of $G$, conjugating it by $x$ and $y$, and following it around to determine its orbit. It is useful to write each element in a standard form $x^i y^j$ where $0 \le i < 13$ and $0 \le j < 3$. For example, $x y x^{- 1} = x (y x y^{- 1})^{- 1} y = x^4 y.$ Repeating this conjugation, $x^i y x^{- i} = x^{4 i} y$, so $\{x^i y| 0 \le i < 13\}$ is a conjugacy class of order 13. The conjugacy classes are: $\begin{array}{c} \{1\}\\ \{x, x^3, x^9 \}\\ \{x^2, x^6, x^5 \}\\ \{x^4, x^{12}, x^{10} \}\\ \{x^7, x^8, x^{11} \}\\ \{y, x y, x^2 y, \cdots, x^{12} y\}\\ \{y^2, x y^2, x^2 y^2, \cdots, x^{12} y^3 \} \end{array}$ Thus there are 7 conjugacy classes and we therefore expect seven irreducible characters. We first note that the commutator subgroup is the normal subgroup $\left\langle x \right\rangle$ of order 13. Indeed, $x^4 = x y x^{- 1} y^{- 1}$ is a commutator, and it generates this subgroup, so $\left\langle x \right\rangle \subseteq G'$. On the other hand, $G / G'$ is abelian (of order 3) and so $G' \subseteq \left\langle x \right\rangle$. We get 3 linear characters by pulling back the linear characters of $G / G'$. If $\rho = e^{2 \pi i / 3}$, we have this much of the character table: $\begin{array}{|l|l|l|l|l|l|l|l|} \hline & 1 & x & x^2 & x^4 & x^7 & y & y^2\\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline \chi_2 & 1 & 1 & 1 & 1 & 1 & \rho & \rho^2\\ \hline \chi_3 & 1 & 1 & 1 & 1 & 1 & \rho^2 & \rho\\ \hline \end{array}$ There must be four remaining characters, all of degrees $d_i > 1$, and $\sum_{i = 1}^7 d_i^2 = 39$, so (with $d_1 = d_2 = d_3 = 1$) we have $\sum_{i = 4}^7 d_i^2 = 36$. Later we will prove that $d_i$ divides $|G|$, but we don't know that yet. However, there is only one way we can write $36$ as a sum of four numbers chosen from the set $\{4, 9, 16, 25, \cdots\}$ which is $9 + 9 + 9 + 9$. So there are four characters of degree 3. $\begin{array}{|l|l|l|l|l|l|l|l|} \hline & 1 & x & x^2 & x^4 & x^7 & y & y^2\\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline \chi_2 & 1 & 1 & 1 & 1 & 1 & \rho & \rho^2\\ \hline \chi_3 & 1 & 1 & 1 & 1 & 1 & \rho^2 & \rho\\ \hline \chi_4 & 3 & & & & & & \\ \hline \chi_5 & 3 & & & & & & \\ \hline \chi_6 & 3 & & & & & & \\ \hline \chi_7 & 3 & & & & & & \\ \hline \end{array}$ Now, since $G$ has a subgroup $\left\langle x \right\rangle$ of index 3, we can obtain some characters of degree 3 by inducing characters of $\left\langle x \right\rangle$. We need a 13-th root of unity, so let $\zeta = e^{2 \pi i / 13}$; let $\theta$ be the character $\theta (x^i) = \zeta^i$. We have 3 handy coset representatives at our disposal, namely $1, y, y^2$. We have the formula $\theta^G (g) = \dot{\theta} (g) + \dot{\theta} (y g y^{- 1}) + \dot{\theta} (y^2 g y^{- 2}) .$ Note that if $g \notin \left\langle x \right\rangle$ then $\theta^G (g) = 0$. (Exercise 2.7.5.)

Exercise 2.7.5: Explain why if $N \vartriangleleft G$ and $\theta$ is a character of $N$ then $\theta^G (g) = 0$ when $g \notin N$.

We now get the following values for $\theta^G$: $\begin{array}{|l|l|l|l|l|l|l|l|} \hline & \begin{array}{l} \small{1}\\ 1 \end{array} & \begin{array}{l} \small{3}\\ x \end{array} & \begin{array}{l} \small{3}\\ x^2 \end{array} & \begin{array}{l} \small{3}\\ x^4 \end{array} & \begin{array}{l} \small{3}\\ x^7 \end{array} & \begin{array}{l} \small{13}\\ y \end{array} & \begin{array}{l} \small{13}\\ y^2 \end{array}\\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline \chi_2 & 1 & 1 & 1 & 1 & 1 & \rho & \rho^2\\ \hline \chi_3 & 1 & 1 & 1 & 1 & 1 & \rho^2 & \rho\\ \hline \chi_4 & 3 & \zeta + \zeta^3 + \zeta^9 & \zeta^2 + \zeta^6 + \zeta^5 & \zeta^4 + \zeta^{12} + \zeta^{10} & \zeta^7 + \zeta^8 + \zeta^{11} & 0 & 0\\ \hline \chi_5 & 3 & & & & & & \\ \hline \chi_6 & 3 & & & & & & \\ \hline \chi_7 & 3 & & & & & & \\ \hline \end{array}$ The tiny numbers that have been added are the sizes of the conjugacy classes, since we will need these momentarily. To check that $\chi_4$ is irreducible, we need to compute its character $\left\langle \chi_4, \chi_4 \right\rangle$, and this can certainly be done directly. But we will postpone this point, display the rest of the character table, then settle the irreducibility by a trick. To get the rest of the character table, we induce the characters $\theta^2$, $\theta^4$ and $\theta^7$ and we get the table: $\begin{array}{|l|l|l|l|l|l|l|l|} \hline & \begin{array}{l} \small{1}\\ 1 \end{array} & \begin{array}{l} \small{3}\\ x \end{array} & \begin{array}{l} \small{3}\\ x^2 \end{array} & \begin{array}{l} \small{3}\\ x^4 \end{array} & \begin{array}{l} \small{3}\\ x^7 \end{array} & \begin{array}{l} \small{13}\\ y \end{array} & \begin{array}{l} \small{13}\\ y^2 \end{array}\\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline \chi_2 & 1 & 1 & 1 & 1 & 1 & \rho & \rho^2\\ \hline \chi_3 & 1 & 1 & 1 & 1 & 1 & \rho^2 & \rho\\ \hline \chi_4 & 3 & \zeta + \zeta^3 + \zeta^9 & \zeta^2 + \zeta^6 + \zeta^5 & \zeta^4 + \zeta^{12} + \zeta^{10} & \zeta^7 + \zeta^8 + \zeta^{11} & 0 & 0\\ \hline \chi_5 & 3 & \zeta^2 + \zeta^6 + \zeta^5 & \zeta^4 + \zeta^{12} + \zeta^{10} & \zeta^7 + \zeta^8 + \zeta^{11} & \zeta + \zeta^3 + \zeta^9 & 0 & 0\\ \hline \chi_6 & 3 & \zeta^4 + \zeta^{12} + \zeta^{10} & \zeta^7 + \zeta^8 + \zeta^{11} & \zeta + \zeta^3 + \zeta^9 & \zeta^2 + \zeta^6 + \zeta^5 & 0 & 0\\ \hline \chi_7 & 3 & \zeta^7 + \zeta^8 + \zeta^{11} & \zeta + \zeta^3 + \zeta^9 & \zeta^2 + \zeta^6 + \zeta^5 & \zeta^4 + \zeta^{12} + \zeta^{10} & 0 & 0\\ \hline \end{array}$ Now let us show that $\left\langle \chi_4, \chi_4 \right\rangle = \left\langle \chi_5, \chi_5 \right\rangle = \left\langle \chi_6, \chi_6 \right\rangle = \left\langle \chi_7, \chi_7 \right\rangle = 1$. We note that since $\zeta + \zeta^2 + \zeta^3 + \zeta^4 + \zeta^5 + \zeta^6 + \zeta^7 = - 1,$ the character $\chi_4 + \chi_5 + \chi_6 + \chi_7$ has these values: $\begin{array}{|l|l|l|l|l|l|l|l|} \hline \begin{array}{l} \small{\text{group of}}\\ \small{\text{order 39}} \end{array} & \begin{array}{l} \small{1}\\ 1 \end{array} & \begin{array}{l} \small{3}\\ x \end{array} & \begin{array}{l} \small{3}\\ x^2 \end{array} & \begin{array}{l} \small{3}\\ x^4 \end{array} & \begin{array}{l} \small{3}\\ x^7 \end{array} & \begin{array}{l} \small{13}\\ y \end{array} & \begin{array}{l} \small{13}\\ y^2 \end{array}\\ \hline \chi_4 + \chi_5 + \chi_6 + \chi_7 & 12 & - 1 & - 1 & - 1 & - 1 & 0 & 0\\ \hline \end{array}$ From this we compute $\left\langle \chi_4 + \chi_5 + \chi_6 + \chi_7, \chi_4 \right\rangle = \frac{1}{39} \left[ 12 \cdot 3 - 3 (\zeta + \zeta^2 + \zeta^3 + \zeta^4 + \zeta^5 + \zeta^6 + \zeta^7) \right] = 1.$ Thus $\left\langle \chi_4, \chi_4 \right\rangle + \left\langle \chi_5, \chi_4 \right\rangle + \left\langle \chi_6, \chi_4 \right\rangle + \left\langle \chi_7, \chi_4 \right\rangle = 1.$ Now each of these inner products is nonnegative; indeed the inner product of two characters is always nonnegative, and $\left\langle \chi_4, \chi_4 \right\rangle$ is a positive integer. (Exercise 2.7.6). It follows that $\left\langle \chi_4, \chi_4 \right\rangle = 1$ and so $\chi_4$ is an irreducible character. We have succeeded in computing the character table.

Exercise 2.7.6: Explain why the inner product of two characters is a nonnegative integer, and if $\chi$ is a nonzero character, $\left\langle \chi, \chi \right\rangle$ is a positive integer.

Exercise 2.7.7: Construct the character tables of the nonabelian groups of orders 21 and 55.

The group $\operatorname{SL}_2 (\mathbb{F}_3)$ is a nonabelian group of order 24. It is somewhat related to the group $S_4$, which is another nonabelian group of order 24, in that we have short exact sequences: $1 \longrightarrow Z_2 \longrightarrow \operatorname{SL}_2 (\mathbb{F}_3) \longrightarrow A_4 \longrightarrow 1,$ $1 \longrightarrow A_4 \longrightarrow S_4 \longrightarrow Z_2 \longrightarrow 1.$

Exercise 2.7.8: Explain where these come from. (Hint: The second is easy; for the first, see Table 1.9.2 and the discussion of the isomorphism $\operatorname{PGL}_2 (\mathbb{F}_3) \cong S_4$ that follows it.)

However computing the character table of $\operatorname{SL}_2 (\mathbb{F}_3)$ is harder than computing the character table of $S_4$. Here are the conjugacy classes. We give the order of each conjugacy class, and a representative. $\begin{array}{|l|l|l|l|l|l|l|} \hline \begin{array}{l} \scriptsize{1}\\ \scriptsize{\left(\begin{array}{cc} 1 & \\ & 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{1}\\ \scriptsize{\left(\begin{array}{cc} - 1 & \\ & \!\!- 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{6}\\ \scriptsize{\left(\begin{array}{cc} & 1\\ - 1 & \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} 1 & 1\\ & 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} 1 & - 1\\ & 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} - 1 & - 1\\ & - 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} - 1 & 1\\ & - 1 \end{array}\right)} \end{array}\\ \hline \end{array}$

Using the homomorphism $\operatorname{SL}_2 (\mathbb{F}_3) \longrightarrow A_4$, we can pull back the four irreducible characters of $A_4$ and obtain the following four characters. Here $\rho = e^{2 \pi i / 3}$; we assume you've done Exercise 2.7.4 $\begin{array}{|l|l|l|l|l|l|l|l|} \hline & \begin{array}{l} \scriptsize{1}\\ \scriptsize{\left(\begin{array}{cc} 1 & \\ & 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{1}\\ \scriptsize{\left(\begin{array}{cc} - 1 & \\ & - 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{6}\\ \scriptsize{\left(\begin{array}{cc} & 1\\ - 1 & \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} 1 & 1\\ & 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} 1 & - 1\\ & 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} - 1 & - 1\\ & - 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} - 1 & 1\\ & - 1 \end{array}\right)} \end{array}\\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline \chi_2 & 1 & 1 & 1 & \rho & \rho^2 & \rho & \rho^2\\ \hline \chi_3 & 1 & 1 & 1 & \rho^2 & \rho & \rho^2 & \rho\\ \hline \chi_4 & 3 & 3 & - 1 & 0 & 0 & 0 & 0\\ \hline \end{array}$

The remaining 3 characters are surprisingly hard to construct, and we will introduce some new concepts in order to get them. First, we note the concept of the central character of a representation.

Proposition 2.7.4: Let $(\pi, V)$ be an irreducible representation of the finite group $G$. Then there exists a linear character $\omega$ of the center $Z (G)$ such that if $z \in Z (G)$ then $\pi (z)$ is the scalar linear transformation $\omega (z) \cdot I$. (Here $I : V \longrightarrow V$ is the identity map.) So if $\chi = \chi_{\pi}$ we have $\chi (z) = d \chi (z)$ for all $z \in Z (G)$, where $d = \dim (V)$.

Proof. (Click to Expand/Collapse)

Since $z \in Z (G)$ we have $z g = g z$ for all $g \in G$, so $\pi (z) \pi (g) = \pi (g) \pi (z)$. This means that $\pi (z) \in \operatorname{Hom}_{\mathbb{C}[G]} (V, V)$. So by Schur's Lemma (Lemma 2.3.3 (ii)) there is a scalar $\omega (z)$ such that $\pi (z) v = \omega (z) v$ for all $v \in V$. We note that if $z, z' \in Z (G)$ and $0 \neq v \in V$ is any nonzero element we have $\omega (z z') v = \pi (z z') v = \pi (z) \pi (z') v = \omega (z) \omega (z') v,$ so $\omega (z z') = \omega (z) \omega (z')$, and $\omega$ is a linear character of $Z (G)$.

In the case at hand, $Z (G) =\{\pm I\}$ has order 2, and a glance at the characters we've constructed shows that they all have trivial central character. Our search for characters to induce can be narrowed with this in mind. We will seek subgroups $H \subset G$ such that $H$ contains $Z (G)$, and the characters that we will induce will have to have nontrivial central characters.

Unfortunately, we won't be able to make do with a single subgroup $H$. We'll use two different ones, then take linear combinations of the characters that we induce in order to get an irreducible one. The first $H$ that we will consider is a 2-Sylow subgroup. There are 3 such 2-Sylow subgroups, but let us take this one: $Q = \left\langle x, y \right\rangle, \hspace{2em} x = \left(\begin{array}{cc} 0 & 1\\ - 1 & 0 \end{array}\right), \hspace{1em} y = \left(\begin{array}{cc} 1 & 1\\ 1 & - 1 \end{array}\right) .$ We have $x^2 = y^2 = - I$ and $y x y^{- 1} = x^{- 1}$, so this Sylow subgroup is isomorphic to the quaternion group. Its character table is as follows: $\begin{array}{|l|l|l|l|l|l|} \hline Q & 1 & x & x^2 & y & x y\\ \hline \theta_1 & 1 & 1 & 1 & 1 & 1\\ \hline \theta_2 & 1 & 1 & 1 & - 1 & - 1\\ \hline \theta_3 & 1 & - 1 & 1 & 1 & - 1\\ \hline \theta_4 & 1 & - 1 & 1 & - 1 & 1\\ \hline \theta_5 & 2 & 0 & - 2 & 0 & 0\\ \hline \end{array}$ (The resemblence to the character table of $D_8$ is uncanny, but the two groups are nonisomorphic.) If we are to have any hope of obtaining an induced representation of $G$ that has nontrivial central character, we must induce a character with nontrivial central character of $Q$. (Luckily $Q$ and $G$ have the same center $\{\pm I\}$.) There is only one candidate, so we try inducing $\theta_5$ to $G$. We choose 3 arbitrary right coset representatives $x_1, \cdots, x_3$; then $\sum \dot{\theta}_5 (x_i g x_i^{- 1})$ is easy to compute, because it must vanish unless $g = \pm I$, since $\dot{\theta_5}$ vanishes on the noncentral elements. However if $g = \pm I$ it is central, so $\theta_5^G (g) = \sum_{i = 1}^3 \dot{\theta}_5 (x_i g x_i^{- 1}) = 3 \dot{\theta}_5 (6),$ and we have the following characters: $\begin{array}{|l|l|l|l|l|l|l|l|} \hline & \begin{array}{l} \scriptsize{1}\\ \scriptsize{\left(\begin{array}{cc} 1 & \\ & 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{1}\\ \scriptsize{\left(\begin{array}{cc} - 1 & \\ & - 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{6}\\ \scriptsize{\left(\begin{array}{cc} & 1\\ - 1 & \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} 1 & 1\\ & 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} 1 & - 1\\ & 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} - 1 & - 1\\ & - 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} - 1 & 1\\ & - 1 \end{array}\right)} \end{array}\\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline \chi_2 & 1 & 1 & 1 & \rho & \rho^2 & \rho & \rho^2\\ \hline \chi_3 & 1 & 1 & 1 & \rho^2 & \rho & \rho^2 & \rho\\ \hline \chi_4 & 3 & 3 & - 1 & 0 & 0 & 0 & 0\\ \hline \theta_5^G & 6 & - 6 & 0 & 0 & 0 & 0 & 0\\ \hline \end{array}$ This representation is, unfortunately, not irreducible, but we will find an irreducible representation inside it. Before we can do that, we take another candidate for $H$, namely the cyclic subgroup of order 6 generated by $u = \left(\begin{array}{cc} - 1 & - 1\\ & - 1 \end{array}\right) .$ We induce the linear character $\psi (u^i) = (- 1)^i$, and we obtain $\begin{array}{|l|l|l|l|l|l|l|l|} \hline & \begin{array}{l} \scriptsize{1}\\ \scriptsize{\left(\begin{array}{cc} 1 & \\ & 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{1}\\ \scriptsize{\left(\begin{array}{cc} - 1 & \\ & - 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{6}\\ \scriptsize{\left(\begin{array}{cc} & 1\\ - 1 & \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} 1 & 1\\ & 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} 1 & - 1\\ & 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} - 1 & - 1\\ & - 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} - 1 & 1\\ & - 1 \end{array}\right)} \end{array}\\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline \chi_2 & 1 & 1 & 1 & \rho & \rho^2 & \rho & \rho^2\\ \hline \chi_3 & 1 & 1 & 1 & \rho^2 & \rho & \rho^2 & \rho\\ \hline \chi_4 & 3 & 3 & - 1 & 0 & 0 & 0 & 0\\ \hline \theta_5^G & 6 & - 6 & 0 & 0 & 0 & 0 & 0\\ \hline \psi^G & 4 & - 4 & 0 & 1 & 1 & - 1 & - 1\\ \hline \end{array}$ Now let us consider $\theta^G_5 - \psi^G$. We will show that this is an irreducible character of $G$. At the moment, we don't even know that it is a character, since we've taken two characters and subtracted them! How can we show that it is an irreducible character when we don't even know that it is a character? There is a way.

We define a generalized character of a finite group $G$ to be a linear combination of characters with integer coefficients (possibly negative). If $\chi_1, \cdots, \chi_h$ are the irreducible characters, then a generalized character is any class function $\sum_{i = 1}^h d_i \chi_i, \hspace{2em} d_i \in \mathbb{Z}.$ This generalized character is a character if and only if the $d_i$ are all nonnegative. The set of generalized characters is closed under both addition and subtraction. So although we don't know that $\theta^G_5 - \psi^G$ is a character, at least it is a generalized character.

Lemma 2.7.2: Suppose that $\phi$ is a generalized character. If $\left\langle \phi, \phi \right\rangle = 1$ and $\phi (1) > 0$, then $\theta$ is an irreducible character.

Proof. (Click to Expand/Collapse)

Write $\phi = \sum d_i \chi_i$ as a linear combination of the irreducible characters. Since the $\chi_i$ are orthonormal, $\left\langle \phi, \phi \right\rangle = \sum d_i^2$. Since the $d_i$ are inegers and $\left\langle \phi, \phi \right\rangle = 1$, all but one of the $d_i$ are zero, and one is $\pm 1$. We have to eliminate the possibility that this nonvanishing $d_i$ is $- 1$. But it is impossible that $\phi = - \chi_i$ since both $\phi (1)$ and $\chi_i (1)$ are positive, so $\phi = \chi_i$.

In the case at hand, $\theta^G_5 - \psi^G$ satisfies the hypothesis of the Lemma, so let us call it $\chi_5$. Then two more degree 2 irreducibles can be obtained using Proposition 2.7.3. We thus have the completed character table of $\operatorname{SL}_2 (\mathbb{F}_3)$: $\begin{array}{|l|l|l|l|l|l|l|l|} \hline & \begin{array}{l} \scriptsize{1}\\ \scriptsize{\left(\begin{array}{cc} 1 & \\ & 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{1}\\ \scriptsize{\left(\begin{array}{cc} - 1 & \\ & \!\!- 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{6}\\ \scriptsize{\left(\begin{array}{cc} & 1\\ - 1 & \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} 1 & 1\\ & 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} 1 & - 1\\ & 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} - 1 & - 1\\ & - 1 \end{array}\right)} \end{array} & \begin{array}{l} \scriptsize{4}\\ \scriptsize{\left(\begin{array}{cc} - 1 & 1\\ & - 1 \end{array}\right)} \end{array}\\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline \chi_2 & 1 & 1 & 1 & \rho & \rho^2 & \rho & \rho^2\\ \hline \chi_3 & 1 & 1 & 1 & \rho^2 & \rho & \rho^2 & \rho\\ \hline \chi_4 & 3 & 3 & - 1 & 0 & 0 & 0 & 0\\ \hline \chi_5 & 2 & - 2 & 0 & - 1 & - 1 & 1 & 1\\ \hline \chi_6 & 2 & - 2 & 0 & - \rho & - \rho^2 & \rho & \rho^2\\ \hline \chi_7 & 2 & - 2 & 0 & \rho & \rho^2 & - \rho & - \rho^2\\ \hline \end{array}$

Exercise 2.7.9: Compute the character tables of $S_5$ and $A_5$.