Note on structural properties of graphs

Theorem 2.1

A finite group G contains an abelian subgroup of order c if and only if G G contains a c -clique.

Corollary 2.2

A finite group G contains an abelian subgroup A of order c if and only if G G is a weakly perfect graph with χ ( G G ) = c .

Proof

If G contains an abelian subgroup of order c , then c ≤ ω ( G G ) ≤ χ ( G G ) ≤ c by Theorem 2.1, and hence χ ( G G ) = c .

For the converse, since ω ( G G ) = χ ( G G ) = c , the graph G G contains a c -clique, which implies that G contains an abelian subgroup A of order c .□

Corollary 2.3

A finite group G contains an abelian subgroup A of order c if and only if κ ( G ) = Λ ( G ) = c .

Proof

If G G is the commuting graph of G , then κ ( G ) = ω ( G G ) , Λ ( G ) = χ ( G G ) and Theorem 2.2 applies.□

Proposition 2.4

Let G be a finite group. Then the following statements are equivalent.

1. G contains a commuting set with c elements.

2. G contains an abelian subgroup of order c .

3. G G contains a c -clique.

4. G G is a weakly perfect graph and χ ( G G ) = c .

5. G G contains a complete subgraph on c vertices.

6. χ ( G G ) = κ ( G ) = c .

7. ω ( G G ) = Λ ( G ) = c .

Proof

1. Note that if X is a commuting set of G , then ⟨ X ⟩ is a complete subgraph of G G on c vertices. ⟨ X ⟩ is in fact a c -clique and we have already seen that this implies that X is an abelian subgroup.

2. If G contains a commuting set with c elements, G G contains a c -clique. By Theorem 2.1, G contains an abelian subgroup of order c .

3. If G G contains a c -clique, G contains an abelian subgroup of order c , by Corollary 2.2 G G is weakly perfect and χ ( G G ) = c .

4. G G weakly perfect implies ω ( G G ) = χ ( G G ) , thus there is a c -clique.

5. G G can be c -colored and does not contain a complete subgraph in more than c vertices and therefore G does not contain a commutative set with more than c vertices. From (5), G G cannot be k -colored for k < c and G contains a commutative set with c elements. It follows at once that χ ( G G ) = κ ( G ) = c .

6. Suppose χ ( G G ) = κ ( G ) = c , since κ ( G ) = ω ( G G ) and Λ ( G ) = χ ( G G ) we get the result.

7. ω ( G G ) = c implies the existence of a c -clique, thus, there is commuting set of cardinality c .□

Lemma 2.5

Let G be a graph with chromatic number χ ( G ) = q + 1 and without any ( q + 1 ) -cliques. Let T = { x ∈ G : d ( x ) > q } , then

Proposition 2.6

Let G be a finite group. If χ ( G G ) = c and the set S = { x ∈ G : ∣ C G ( x ) ∣ > c } satisfies

then G contains an abelian subgroup of order c .

Proof

If G does not contain such an abelian subgroup, Theorem 2.1 says that G G does not contain any c -clique, hence we may apply Lemma 2.5 with q = c − 1 , if T = { x ∈ V ( G G ) : d ( x ) > c − 1 } , then

But d ( x ) = ∣ C G ( x ) ∣ − 1 for each x . Hence, S = T and inequality (4) becomes

Proposition 2.7

Let G be a finite group. If G satisfies the hypothesis of Proposition 2.6, then G is abelian of order > 3 .

Proof

If S = ∅ , then left hand side of inequality (3) is 0 and therefore we must have c > 3 . Moreover, note that

and G is abelian (otherwise some pair of vertices of G G could be colored with the same color).

If S is non-empty, then 1 ∈ G must be in S and hence ∣ C G ( 1 ) ∣ − c = ∣ G ∣ − c < c − 3 . This implies that ∣ G ∣ < 2 c − 3 . On the other hand, since G contains an abelian subgroup of order c , we have ∣ G ∣ = c k for some integer k and c ( k − 2 ) < − 3 . But this inequality holds just for k = 1 and c > 3 , in particular we must have k = 1 and hence G is abelian of order c > 3 .□

Theorem 2.8

Let G be a finite group of order > 3 with Λ ( G ) = c and consider S = { x ∈ G : ∣ C G ( x ) ∣ > c } . Then G is abelian if and only if

Proof

If G is abelian, then ∣ C G ( x ) ∣ = ∣ G ∣ = Λ ( G ) = c for all x ∈ G , hence the left hand side in inequality (3) is 0, while the right hand side is ∣ G ∣ − 3 , a number greater than 0.

For the converse, suppose that G is a non-abelian group of order > 3 . According to Proposition 2.7, the hypothesis of Proposition 2.6 is not satisfied by G , so we must have χ ( G G ) ≠ c or

Since χ ( G G ) = Λ ( G ) = c , the result follows.□

Theorem 2.9

Let p be a prime and G a group of order p 3 , then G has an abelian subgroup of order c .

Proof

If G is abelian, we are done. If not, it is known that ∣ Z ( G ) ∣ = p and the number N of conjugacy classes is

where the sum ∑ ∣ C G ( x ) ∣ is taken over the non-central elements.

Note that ∣ C G ( x ) ∣ = p 2 for each non-central element, otherwise there exists some non-central elements such that ∣ C G ( x ) ∣ = p . But Z ( G ) ≤ C G ( x ) also has order p ; hence, Z ( G ) = C G ( x ) . Since g ∈ C G ( x ) , x ∈ Z ( G ) and this is a contradiction. Therefore, the number of conjugacy classes is N = p + ( p 2 − 1 ) ; p with just one element and ( p 2 − 1 ) with p elements.

Finally, if m = 2 p − 1 we have c = 1 + 1 + ⋯ + 1 + p + ⋯ + p = p 2 . Note that G contains a subgroup of order p 2 , which is necessarily an abelian subgroup.□

Theorem 2.10

If p is a prime number and P is a non-abelian group of order p 3 then, P × Z p contains an abelian subgroup of order c .

Proof

First note that Z ( P × Z p ) = Z ( P ) × Z ( Z p ) = Z ( P ) × Z p . Now, since P is non-abelian, ∣ Z ( P ) ∣ = p and therefore ∣ Z ( P × Z p ) ∣ = p 2 . Also, we have C P × Z p ( x , y ) = C P ( x ) × C Z p ( y ) = C P ( x ) × Z p for each ( x , y ) ∈ P × Z p . If x ∈ P is a non-central element, we have already seen in the proof of Theorem 2.9 that ∣ C P ( x ) ∣ = p 2 , in particular, C P ( x ) is an abelian subgroup of P . Hence, if ( x , y ) ∈ P × Z p is a non-central element, then C P × Z p ( x , y ) is an abelian subgroup of order p 3 .

According to Burnside’s lemma, there are p 2 conjugacy classes of order 1 and p 3 − p conjugacy classes of order p . Hence, m = 2 p 2 − p in Theorem 1.2 and consequently c = p 2 + ( p 2 − p ) p = p 3 . If x ∈ P − Z ( P ) and y ∈ Z p is any element, then ( x , y ) is a non-central element of P × Z p and C P × Z p ( x , y ) is an abelian subgroup of order c .□

Lemma 2.11

If G is a group of order p n q , where p > q are primes, then G contains a unique Sylow p -subgroup P .

Lemma 2.12

If G is a group which is not abelian, then G / Z ( G ) is not cyclic. In particular, [ G : Z ( G ) ] can never be a prime number.

Theorem 2.13

Let G be a non-abelian group of order p n q , where p and q are prime numbers:

1. If p > q and the unique Sylow p -subgroup of G is abelian, then c = p n and G contains an abelian subgroup of order c . In particular, for n = 1 , 2 the group G contains an abelian subgroup of order c .

2. If p < q and n = 2 , then either c = 4 , c = p q or c = q . In all these cases G contains an abelian subgroup of order c .

Proof

(1) Let P be the unique Sylow p -subgroup of G , note that all elements of this Sylow p -subgroup have the smallest conjugacy classes with one or q elements and note also that actually there exist elements x ∈ P with ∣ [ x ] ∣ = q (otherwise Z ( G ) = G or ∣ Z ( G ) ∣ = p n contradicting hypothesis or Lemma 2.12 in any case).

Finally, let [ x 1 ] , [ x 2 ] , … , [ x k ] be the conjugacy classes of the elements in P ordered increasingly by size. Since P is the union of these conjugacy classes we have

If n = 1 , 2 the unique Sylow p -subgroup of G is abelian.

(2) First, we examine the possible numbers of Sylow q -subgroups and Sylow p -subgroups. Let n q and n p be the numbers of Sylow q -subgroups and Sylow p -subgroups, respectively. Since q > p , q ∤ ( p − 1 ) , it follows that n q ≠ p . Now, if n q = p 2 , then necessarily q ∣ ( p + 1 ) , and hence q = p + 1 . But the only prime numbers satisfying this condition are p = 2 and q = 3 . Consequently, ∣ G ∣ = 12 and then G ≅ A 4 , D 12 , Z 3 ⋊ Z 4 . In the first case c = 4 and in the other cases c = 6 . A glance to the subgroups of G shows the existence of an abelian subgroup of order c .

The other possibility is n q = 1 , let Q be such a Sylow q -subgroup. Let us now consider the possible values for n p . Note that such a number cannot be equal to 1, otherwise G is the direct product of these Sylow subgroups, and being each abelian, G must be abelian. This yields to n p = q .

For analyzing the order of Z ( G ) , Lemma 2.12 ensures that ∣ Z ( G ) ∣ ≠ p q , p 2 . On the other hand, if Z ( G ) has order q , then this is the unique Sylow q -subgroup, so all the elements of the different Sylow p -subgroups, except for the identity, are non-central elements. Hence, their centralizers have order p 2 . But ∣ Z ( G ) ∣ divides the order of such centralizers, which is impossible. The only remaining possible values for ∣ Z ( G ) ∣ are p or 1. Suppose that ∣ Z ( G ) ∣ = p . Since each Sylow p -subgroup is a centralizer, Z ( G ) is contained in each Sylow p -subgroup. Thus, the sum of ∣ Z ( G ) ∣ and the number of non-central elements in each Sylow ( p or q )-subgroup is p + q ( p 2 − p ) + ( q − 1 ) = p 2 q − p q + p + q − 1 . Subtracting this sum from p 2 q we get p q − p − q + 1 , which corresponds to the number of non-central elements of G that do not belong to any Sylow ( p or q )-subgroup.

Now, the non-central elements of Sylow p -subgroups and Sylow q -subgroups have centralizers of order p 2 and p q , respectively. The remaining p q − p − q + 1 elements have centralizers of order p q . Applying Burnside’s lemma we get:

where p , q − 1 and p ( p − 1 ) are the number of conjugacy classes of cardinally 1, p and q , respectively. Thus, the sum of the cardinals of the first p conjugacy classes and the following q − 1 conjugacy classes whose cardinals are p is c = p q . If x is an element not belonging to any Sylow ( p or q )-subgroup, then ∣ ⟨ x ⟩ ∣ = p q .

Suppose ∣ Z ( G ) ∣ = 1 , for x ∈ Q , x ≠ 1 , we may observe that Q ≤ C G ( x ) , thus, ∣ C G ( x ) ∣ = q or p q . If ∣ C G ( x ) ∣ = p q there exists y ∈ C G ( x ) of order p and, therefore, it must belong to a Sylow p -subgroup P , which implies that ∣ P ∩ Q ∣ ≥ 2 , a contradiction. Hence, ∣ C G ( x ) ∣ = q and Q = C G ( x ) . Moreover, the intersection of all Sylow p -subgroups is trivial, otherwise we may obtain a subgroup of order p contained in all of them and the centralizer of a non-trivial element of this subgroup would have order > p 2 (it commutes with every element of these Sylow p -subgroups). Now, if we take the sum over the number of elements in all Sylow subgroups we obtain q ( p 2 − 1 ) + ( q − 1 ) + 1 = p 2 q , which means that G is the union of its Sylow subgroups. Finally, calculating N we obtain

So G has one conjugacy class with a unique element, q − 1 p 2 conjugacy classes each with p 2 elements and p 2 − 1 conjugacy classes each with q elements. This yields c = q and the order of the unique Sylow q -subgroup attains the bound. Notice that this case is not possible if p 2 does not divide q − 1 .□