If $G$ is a finite group, we define its commutativity degree $P(G)$ as the probability that two elements of $G$ commute, that is

$\displaystyle P(G)= \frac{1}{| G \times G|} \cdot | \{ (g,h) \in G \times G \mid [g,h]=1 \} |$.

It is surprising that the value of $P(G)$ may give strong information about $G$. In particular, we prove here the two following results:

Theorem 1: Let $G$ be a finite group. If $P(G) > 5/8$ then $G$ is abelian.

Theorem 2: Let $G$ be a finite group. If $P(G)>1/2$ then $G$ is nilpotent.

In fact, the two theorems above will follow from the upper degree equation bound:

Theorem 3: Let $G$ be a finite group and let $G'$ denote its commutator subgroup. Then

$\displaystyle P(G) \leq \frac{1}{4} \left( 1+ \frac{3}{|G'|} \right)$.

Indeed, if $G$ is non-abelian then $|G'| \geq 2$, and the upper degree equation bound implies $P(G) \leq 5/8$; therefore, theorem 1 is proved. Then, if $P(G) >1/2$ then the upper degree equation bound implies $|G'| < 3$. If $|G'|=1$, $G$ is abelian and there is nothing to prove. Otherwise, $|G'|=2$ ie. $G' = \{1,c \}$ for some $c \neq 1$. For all $g \in G$, $[c,g] \in G'$, so either $[c,g]=1$ or $[c,g]=c$ which implies $c=1$; therefore, $c \in Z(G)$. We proved that $G' \subset Z(G)$, so $[G,G']= \{1\}$; in particular $G$ is nilpotent.

In order to prove Theorem 3, we first notice the following link between commutativity degree and the number of conjugacy classes of $G$:

Lemma 1: Let $G$ be a finite group. Then $\displaystyle P(G)= \frac{k(G)}{|G|}$ where $k(G)$ denotes the number of conjugacy classes of $G$.

Proof. If $x \in G$, let $C(x)$ denote the centralizer of $x$ and $C_1,\dots, C_{k(G)}$ denote the conjugacy classes of $G$; for convenience, let $x_i$ be an element of $C_i$. Then

Finally, the lemma follows. $\square$

Therefore, computing $P(G)$ is equivalent to computing $|G|$ and $k(G)$. But we know that the number of conjugacy classes is quite related to representation theory, and indeed we will prove Theorem 3 thanks to representation theory.

Definitions: Let $G$ be a finite group. A (complex) representation is a morphism $G \to \mathrm{Aut}(V)$ for some complex vector space $V$ (of finite dimension); the degree of the representation is the dimension of $V$. By extension, we often say that $V$ is a representation.

Any finite group admits a representation. For example, if $\mathbb{C}[G]$ denotes the complex vector space with $G$ as a basis, the left multiplication extends to a representation, namely the regular representation of $G$.

If $V$ is a representation, a subrepresentation is a subspace $W$ stable under the action of $G$. A representation is irreducible if it does not contain a nontrivial subrepresentation (ie. different from itself and $\{0\}$).

If $V,W$ are two representations, we define the direct sum $V \oplus W$ as the representation defined by the action $g \cdot (v \oplus w) = (g \cdot v) \oplus (g \cdot w)$ for all $g \in G$, $v \in V$ and $w \in W$.

Two representations $V$ and $W$ are isomorphic if there exists a $G$-invariant isomorphism $V \to W$.

Our first result is:

Property 1: (Maschke) Any representation is a direct sum of irreducible representations.

Proof. Let $G$ be a finite group and $V$ be a representation. If $V$ is irreducible, there is nothing to prove, so we suppose that $V$ contains a nontrivial representation $W$. Then, if $( \cdot , \cdot)$ is any scalar product on $V$, it is possible to define a scalar product $\langle \cdot, \cdot \rangle$ invariant under $G$, for example by

$\displaystyle \langle u,v \rangle = \sum\limits_{g \in G} ( g \cdot u, g \cdot v)$.

Therefore, because $W$ is stable under $G$, the orthogonal $W^{\perp}$ for $\langle \cdot , \cdot \rangle$ so is, and $V = W \oplus W^{\perp}$. Finally, Property 1 follows by induction on the dimension of $V$. $\square$

Of course, in the decomposition given by Property 1, the same representation may appear several times (up to isomorphism). Thanks to a theorem due to Frobenius, this number may be characterized.

Definitions: Let $G$ be a finite group. Let $R(G)$ denote the set of central functions $G \to \mathbb{C}$, that is the functions $f : G \to \mathbb{C}$ satisfying $f(ghg^{-1})=f(h)$ for all $g,h \in G$. We endow $R(G)$ with the scalar product

$\displaystyle \langle f , g \rangle = \frac{1}{ |G| } \sum\limits_{x \in G } \overline{ f(x) } g(x)$.

Clearly, the dimension of $R(G)$ is $k(G)$, the number of conjugacy classes of $G$ (a basis of $R(G)$ is given by the set of functions taking the value $1$ on a fixed conjugacy class and $0$ otherwise).

Associated to a representation $\rho : G \to V$, we define a character $\chi_V : G \to \mathbb{C}$ defined by $\chi_V(g)= \mathrm{tr}( \rho(g))$. A character is said irreducible if the associated representation so is.

Property 2: (Frobenius) Let $G$ be a finite group. The set of irreducible characters defines an orthonormal basis of $R(G)$.

Proof. We first prove that the family of irreducible characters is orthonormal. Of course, two isomorphic representations $\rho_i : G \to \mathrm{Aut}(V_i)$ are conjugated, so their characters are equal; therefore, it is sufficient to prove that two irreducible representations $V,W$ satisfy $\langle \chi_V, \chi_W \rangle =1$ if $V$ and $W$ are isomorphic and $\langle \chi_V, \chi_W \rangle=0$ otherwise.

We define the representation $\mathrm{Hom}(V,W)$ as the representation defined by the action $g \cdot f(v) = g \cdot f(g^{-1} \cdot v)$ for every $f \in \mathrm{Hom}(V,W)$, $g \in G$ and $v\in V$. Now, let $\mathrm{Hom}_G(V,W)$ denote the set of fixed points of the action of $G$ on $\mathrm{Hom}(V,W)$.

Claim 1: (Schur) $\mathrm{dim} ~ \mathrm{Hom}_G(V,W)=1$ if $V$ and $W$ are isomorphic, and $0$ otherwise.

A nontrivial element of $\mathrm{Hom}_G(V,W)$ defines a $G$-invariant isomorphism $V \to W$, hence $\mathrm{dim} ~ \mathrm{Hom}_G(V,W)=0$ if $V$ and $W$ are not isomorphic. From now on, suppose that $V$ and $W$ are isomorphic, and let $j : V \to W$ be a $G$-invariant isomorphism.

Let $f \in \mathrm{Hom}_G(V,W)$. Then $j^{-1} \circ f \in \mathrm{Hom}_G(V,V)$. Let $\lambda \in \mathbb{C}$ be an eigenvalue of $j^{-1} \circ f$, so that $g := j^{-1} \circ f - \lambda \mathrm{Id}_V$ is not invertible. Notice that $\mathrm{ker}(g)$ is a subrepresentation of $V$, so $\mathrm{ker}(g)=V$ because $V$ is irreducible and $g$ is not invertible. We deduce that $g \equiv 0$ that is $f = \lambda \cdot j$.

We just proved that $\mathrm{Hom}_G(V,W)$ is one-dimensional. $\square$

Therefore, it is sufficient to prove that $\langle \chi_V, \chi_W \rangle =\mathrm{dim} ~ \mathrm{Hom}_G(V,W)$. We left as an exercise that $\chi_{\mathrm{Hom}(V,W)} = \overline{\chi_V} \cdot \chi_W$. In particular, we deduce that $\langle \chi_V, \chi_W \rangle = \frac{1}{|G|} \sum\limits_{g \in G } \chi_{\mathrm{Hom}(V,W)} (g)$. Now,

Claim 2: For any representation $\rho : G \to \mathrm{Aut}(V)$, $\mathrm{dim} ~ V^G = \frac{1}{|G|} \sum\limits_{g \in G} \chi_V(g)$ where $V^G$ denotes the set of fixed points.

Let $f = \frac{1}{ |G| } \sum\limits_{g \in G} \rho(g) \in \mathrm{Aut}(V)$. It is easy to show that $\rho(h) \circ f= f$ for all $h \in G$, so that we deduce that $f \circ f = f$. Therefore, $V = \mathrm{ker}(f) \oplus f(V)$. Now we notice that $f(V)=V^G$. Indeed, if $u=f(v) \in f(V)$ then $g \cdot u = \rho(g) \circ f(v) = f(v)=u$ for all $g \in G$, hence $u \in V^G$; conversely, if $u \in V^G$, $u = \frac{1}{|G|} \sum\limits_{g \in G} \rho(g)(u) =f(u)$ hence $u \in f(V)$. We conclude that

$\displaystyle \mathrm{dim} ~ V^G = \mathrm{dim} ~ f(V) = \mathrm{tr}(f) = \frac{1}{|G|} \sum\limits_{g \in G} \chi_V(g)$. $\square$

Finally, from Claim 1 and Claim 2, we deduce that the family of irreducible characters is orthonormal. Now, if $E$ denotes the subspace of $R(G)$ spanning by the irreducible characters, we want to prove that $E= R(G)$ or equivalently that $E^{\perp}= \{ 0 \}$.

So let $\varphi \in E^{\perp}$. We want to prove that, for any representation $\rho_V : G \to \mathrm{Aut}(V)$, the function $f = \sum\limits_{g \in G} \overline{ \varphi(g) } \rho_V(g)$ is zero. According to Property 1, $V$ is a direct sum of irreducible representations $V_1 \oplus \cdots \oplus V_n$; in particular, $\rho_V= \rho_{V_1} \oplus \cdots \oplus \rho_{V_n}$. Therefore, we may suppose without loss of generality that $V$ is irreducible.

Now, it is easy to show that $\rho(g) \circ f \circ \rho(g)^{-1} =f$ for every $g \in G$, so that $f \in \mathrm{Hom}_G(V,V)$. But $\mathrm{Id}_V \in \mathrm{Hom}_G(V,V)$ and $\mathrm{dim} ~ \mathrm{Hom}_G(V,V)=1$ according to Claim 1, so $f = \lambda \cdot \mathrm{Id}_V$ for some $\lambda \in \mathbb{C}$. Now,

$\lambda \cdot \dim (V)= \mathrm{tr}(\lambda \cdot \mathrm{Id}_V) = \mathrm{tr}(f)= \sum\limits_{g \in G} \overline{ \varphi(g) } \chi_V(g)= |G| \cdot \langle \varphi, \chi_V \rangle = 0$,

hence $f=0$. Now, we apply this result for the regular representation $\mathbb{C}[G]$: for every $h\in G$,

$\displaystyle 0= \sum\limits_{g \in G} \overline{ \varphi(g) } \rho_V(g) \cdot h = \sum\limits_{g \in G} \overline{ \varphi(g) } \cdot gh$,

hence $\varphi(g)=0$. We just prove that $\varphi=0$, so $E^{\perp} = \{ 0 \}$. $\square$

We now are able to precise Property 1:

Property 3: Let $G$ be a finite group and let $W_1,\dots, W_n$ denote its irreducible representations up to isomorphism. If $V$ is a representations of $G$, then

$\displaystyle V \simeq \bigoplus\limits_{i=1}^n W_i^{\langle \chi_V, \chi_{W_i} \rangle }$.

Proof. According to Property 1, $V$ may be written as a direct sum of $W_i$, say $W_{i_1} \oplus \cdots \oplus W_{i_r}$. Then $\langle \chi_V, \chi_{W_i} \rangle = \sum\limits_{k=1}^r \langle \chi_{W_k}, \chi_{W_i} \rangle$ because $\chi_V= \chi_{W_{i_1}} + \cdots + \chi_{W_{i_r}}$. But we saw during the proof of Property 2 that $\langle \chi_V, \chi_W \rangle=1$ if $V$ and $W$ are isomorphic and $0$ otherwise. Therefore, $\langle \chi_V, \chi_{W_i} \rangle$ is the number of representations in the decomposition isomorphic to $W_i$. So Property 3 follows. $\square$

Property 3 is our main result from representation theory, and now we are able to prove the two following corollaries:

Corollary 4: Let $G$ be a finite group. The number of conjugacy classes is equal to the number of irreducible representations up to isomorphism.

Proof. We already noticed that isomorphic representations lead to equal characters. Conversely, Property 3 implies that two representations with the same character are isomorphic. Therefore, $k(G)$ is equal to the dimension of $R(G)$, $\mathrm{dim} ~ R(G)$ is equal to the number of irreducible characters according to Property 2, and now the number of irreducible characters is equal to the number of irreducible representations. $\square$

Corollary 5: Let $G$ be a finite group. Then $\displaystyle |G|= \sum\limits_{W \ \mathrm{irr.}} \deg(W)^2$ where the sum is taken over the set of irreducible representations up to isomorphism.

Proof. Let $V= \mathbb{C}[G]$ be the regular representation. Let $V= \bigoplus\limits_{W \ \mathrm{irr.}} W^{ \langle \chi_V, \chi_W \rangle }$ be the decomposition given by Property 3. Then

$\displaystyle |G| = \dim(V)= \chi_V(1) = \sum\limits_{W \ \mathrm{irr.}} \langle \chi_V, \chi_W \rangle \cdot \chi_W(1)$.

Of course, $\chi_W(1)= \dim(W)$ so it is sufficient to prove that $\langle \chi_V, \chi_W \rangle = \dim(W)$. Notice that $\rho_V(g)$ is a permutation matrix, so that $\mathrm{tr}(\rho_V(g))$ is the number of fixed points of the function $h \mapsto gh$ from $G$ to itself; therefore, $\mathrm{tr}(\rho_V(g))=|G|$ if $g=1$ and $0$ otherwise. Hence

$\displaystyle \langle \chi_V, \chi_W \rangle = \frac{1}{ |G| } \sum\limits_{g \in G} \overline{ \chi_V (g) } \cdot \chi_W (g) = \chi_W (1) = \dim(W)$.

Corollary 5 follows. $\square$

Now we are ready to prove Theorem 3:

Proof of Theorem 3. Let $W_1, \dots, W_{k(G)}$ denote the irreducible representations of $G$ up to isomorphism. Representations of degree 1 are only morphisms from $G$ to $\mathbb{C}$, so the number of such representations is equal to the number of morphisms from the abelian group $G/G'$ to $\mathbb{C}$.

But this number turns out to be $[G:G']$ because the number of morphisms from a finite abelian group $A$ to $\mathbb{C}$ is precisely $|A|$; we leave this statement as an exercise (hint: argue by induction on the number of factors in the decomposition of $A$ as a direct sum of cyclic groups).

Therefore, the formula given by Corollary 5 may be written as

$\displaystyle |G| = [G:G'] + \sum\limits_{i=[G:G']+1}^{k(G)} \dim(W_i)^2$,

where $\dim(W_i) \geq 2$. Therefore,

$|G| \geq [G:G'] + 4 ( k(G) - [G:G'] )$

and Theorem 3 follows. $\square$

In fact, the relation

$\displaystyle |G| = [G:G'] + \sum\limits_{i=[G:G'] +1}^{k(G)} \dim(W_i)^2$

proved above may be quite useful to compute $P(G)$ for some small groups. For example, if $Q_8$ denotes the quaternion group, its commutator subgroup is $\{ \pm 1 \}$, so the relation above becomes $8= 1+1+1+1+2^2$; therefore, according to Corollary 4, $Q_8$ has five conjugacy classes, hence $P(Q_8)=5/8$. (Noticing that $Q_8$ is not abelian, we deduce that Theorem 1 is optimal.)

Another example is the symmetric group $S_3$: its commutator subgroup is $\{ \mathrm{Id}, \ (123), \ (132) \}$, so the relation above becomes $6= 1+1+2^2$; therefore, $S_3$ has three conjugacy classes, hence $P(S_3)=1/2$. (Noticing that $S_3$ is not nilpotent, since $[S_3, [S_3,S_3]]= [S_3,S_3]=A_3$, we deduce that Theorem 2 is optimal; however, a lot a nilpotent groups satisfies $P(G) \leq 1/2$ because we saw that a group satisfying $P(G) >1/2$ has nilpotent-length at most two.)

For more information on commutativity degree of finite groups, see Anna Castelaz’s thesis Commutativity degree of finite groups. In particular, an elementary proof of Theorem 1 based on conjugacy class equation can be found there, as well as a complete classification of groups satisfying $P(G) \geq 1/2$ (see Proposition 5.1.4):

Theorem: Let $G$ be a finite group. If $P(G) \geq 1/2$, then either

• $G$ is abelian,
• or $G \simeq P \times A$ for some abelian group $A$ and some $2$-group $P$ (in this case, $|G'|=2$),
• or $G \simeq G_m \times A$ where $A$ is an abelian group, $m \geq 1$ and

$G_m= \langle a,b \mid a^3=b^{2^m}=1, \ bab^{-1}=a^{-1} \rangle$.