The Engel graph of a finite group

Theorem 1.1.

If G is a finite group, then Γ ⁢ ( G ) is weakly connected and its undirected diameter is at most 10.

Theorem 1.2.

Suppose that G / Z ∞ ⁢ ( G ) is not an almost simple group. Then Γ ⁢ ( G ) is strongly connected if and only if G / Z ∞ ⁢ ( G ) is not a Frobenius group.

Theorem 1.3.

Suppose that G is soluble and G / Z ∞ ⁢ ( G ) is not a Frobenius group. Then diam ⁡ Γ ⁢ ( G ) ≤ 4 . Furthermore, there exists a soluble group G such that diam ⁡ Γ ⁢ ( G ) = 4 .

Lemma 2.1.

The graph Γ ⁢ ( G ) is weakly or strongly connected if and only if Γ ⁢ ( G / Z ∞ ⁢ ( G ) ) is weakly or strongly connected.

Proof.

If y is adjacent to x in Γ ⁢ ( G ) , then its image y ¯ in G / Z ∞ ⁢ ( G ) is adjacent to x ¯ in Γ ⁢ ( G / Z ∞ ⁢ ( G ) ) . Conversely, if y ¯ ∈ G / Z ∞ ⁢ ( G ) is adjacent to x ¯ in Γ ⁢ ( G / Z ∞ ⁢ ( G ) ) , then [ y , n x ] ∈ Z ∞ ( G ) = Z m ( G ) for some n and m. Thus [ y , n + m x ] = [ [ y , n x ] , m x ] = 1 and y is adjacent to x in Γ ⁢ ( G ) . ∎

Lemma 2.2.

Let x , y ∈ G .

1. If [ y , n x ] = [ y , m x ] ≠ 1 for some 0 ≤ n < m , then y is not adjacent to x in Γ ⁢ ( G ) .

2. If y ∈ F ⁢ ( G ) , then x ↦ y for every x ∈ G .

3. If y ∈ N G ⁢ ( 〈 x 〉 ) , then y ↦ 2 x .

Proof.

(1) Assume that 1 ≠ [ y , n x ] = [ y , m x ] for some 0 ≤ n < m and let z = [ y , n x ] . Then z = [ z , m - n x ] and so z = [ z , l ⁢ ( m - n ) x ] for every l ≥ 0 . If [ y , N x ] = 1 , then 1 = [ y , n + N x , N ⁢ ( m - n - 1 ) x ] = [ z , N ⁢ ( m - n ) x ] = z , which yields a contradiction.

(2) It follows from the fact that the Fitting subgroup F = F ⁢ ( G ) coincides with the set of the left Engel elements of G.

(3) If y ∈ N G ⁢ ( 〈 x 〉 ) , then [ y , x ] ∈ 〈 x 〉 and it commutes with x, so [ y , 2 x ] = 1 . ∎

Proof of Theorem 1.1.

By Lemma 2.1, we may assume that Z ∞ ⁢ ( G ) = 1 . If the soluble radical R ⁢ ( G ) is nontrivial, then F ⁢ ( G ) ≠ 1 and Γ ⁢ ( G ) is weakly connected by Lemma 2.2 (2). So we may assume R ⁢ ( G ) = 1 . Let x 1 , x 2 be two distinct non-identical elements of G. If 〈 x 1 , x 2 〉 is soluble, then there exists 1 ≠ f ∈ F ⁢ ( 〈 x 1 , x 2 〉 ) and x 1 ↦ f , x 2 ↦ f . Since the soluble graph of G is connected with diameter at most 5 (see [3]), we deduce that Γ ⁢ ( G ) is weakly connected and its undirected diameter is at most 10, as claimed. ∎

Theorem 2.3 ([6, Theorem 10.2.1]).

If a finite group G admits a fixed-point-free automorphism of prime order, then G is nilpotent.

Lemma 2.4 ([10, Lemma 2.1]).

Suppose X is a Frobenius complement. Then every Sylow subgroup of X is cyclic or generalized quaternion. If X has odd order, then any two elements of prime order commute and X is metacyclic. If X has even order, then X contains a unique involution.

Lemma 3.1.

If G = K ⋊ H is a Frobenius group with kernel K and complement H, then Γ ⁢ ( G ) is not strongly connected. In particular, for every nontrivial k ∈ K there is no element g ∈ G ∖ K with k ↦ g .

Proof.

Let k be a nontrivial element of K and assume that there exists g ∈ G ∖ K with k ↦ g . Thus [ k , n g ] = 1 for some integer n. Take g ∈ G ∖ K such that n is minimal. Since g ∈ G ∖ K = ⋃ k ∈ K H k , without loss of generality we can assume that g belongs to H. So z = [ k , n - 1 g ] belongs to K, since K is normal, and it belongs to H, since it centralizes g ∈ H and G is Frobenius. Therefore z = 1 , a contradiction to the minimality of n. ∎

Lemma 3.2.

Let N be a non-nilpotent normal subgroup of a finite group G. If Δ ⁢ ( N ) is strongly connected, or, more generally, if N ∖ { 1 } is contained in a strong component of Δ ⁢ ( G ) , then Δ ⁢ ( G ) is strongly connected.

Proof.

Let Ω be the strong component of Δ ⁢ ( G ) containing N ∖ { 1 } . Let x be a nontrivial element of G and let x ~ be an element of prime order in 〈 x 〉 . Since N is not nilpotent, it follows from Theorem 2.3 that there exists y ∈ N ∖ { 1 } such that [ y , x ~ ] = 1 . Clearly [ x , x ~ ] = 1 , and so x ∈ Ω . ∎

Lemma 3.3.

Suppose that N = G 1 ⁢ G 2 is the central product of two nontrivial finite groups. Then Δ ⁢ ( N ) is strongly connected and diam ⁡ ( N ) ≤ 3 .

Proof.

Let g 1 = x 1 ⁢ y 1 , g 2 = x 2 ⁢ y 2 be two nontrivial elements of N with x 1 , x 2 ∈ G 1 , y 1 , y 2 ∈ G 2 . We may assume x 1 ≠ 1 . If y 2 ≠ 1 , then

is a path in Δ ⁢ ( N ) . Suppose y 2 = 1 . This implies x 2 ≠ 1 . If y 1 ≠ 1 , then

is a path in Δ ⁢ ( N ) . We remain with the case y 1 = y 2 = 1 . Then, let 1 ≠ z ∈ G 2 . In this case

is a path in Δ ⁢ ( N ) . ∎

Lemma 3.4.

If G is not an almost simple group and F * ⁢ ( G ) > F ⁢ ( G ) , then Δ ⁢ ( G ) is strongly connected.

Proof.

Recall that F * ⁢ ( G ) = E ⁢ ( G ) ⁢ F ⁢ ( G ) is the central product of the Fitting subgroup of G with the layer E ⁢ ( G ) of G. Since E ⁢ ( G ) is the central product of the components, either F * ⁢ ( G ) is a non-abelian simple group or Δ ⁢ ( F * ⁢ ( G ) ) is strongly connected by Lemma 3.3. In the first case G is almost simple, in the second Δ ⁢ ( G ) is strongly connected by Lemma 3.2. ∎

Lemma 3.5.

Let x , y be two distinct nontrivial elements of a finite group G. If Z ∞ ⁢ ( G ) = 1 , F ⁢ ( G ) = F * ⁢ ( G ) and G is not a Frobenius group, then the following hold:

1. If d ⁢ ( x , y ) > 3 , then N G ⁢ ( 〈 y 〉 ) ⁢ F ⁢ ( G ) (and consequently C G ⁢ ( 〈 y 〉 ) ⁢ F ⁢ ( G ) ) is a Frobenius group.

2. If d ⁢ ( x , y ) > 4 and y r is a power of y of order a prime r , then C G ⁢ ( y r ) has odd order and it is metacyclic.

Proof.

Assume d ⁢ ( x , y ) > 3 . If F ⁢ ( G ) ∩ N G ⁢ ( 〈 y 〉 ) ≠ 1 , then there exists a nontrivial element g ∈ F ⁢ ( G ) ∩ N G ⁢ ( 〈 y 〉 ) , so by Lemma 2.2 (2) and (3),

a contradiction. So F ⁢ ( G ) ∩ N G ⁢ ( 〈 y 〉 ) = 1 . Assume, by contradiction, that N G ⁢ ( 〈 y 〉 ) ⁢ F ⁢ ( G ) is not a Frobenius group. Then there exists an element 1 ≠ g ∈ C F ⁢ ( G ) ⁢ ( k ) for some 1 ≠ k ∈ N G ⁢ ( 〈 y 〉 ) and

against d ⁢ ( x , y ) > 3 .

Now assume d ⁢ ( x , y ) > 4 . In particular, d ⁢ ( x , y r ) > 3 . By (1), C G ⁢ ( y r ) is a Frobenius complement. Assume that | C G ⁢ ( y r ) | is even and take an involution z ∈ C G ⁢ ( y r ) . As every nontrivial element of C G ⁢ ( y r ) acts fixed-point-freely on F, z is the unique involution of C G ⁢ ( y r ) . But then, for every g ∈ C G ⁢ ( y r ) , we have z g ∈ C G ⁢ ( y r ) , hence z g = z . Since z acts fixed-point-freely on F ⁢ ( G ) , it inverts the elements of F ⁢ ( G ) and F ⁢ ( G ) is abelian. In particular, F ⁢ ( G ) = [ F ⁢ ( G ) , z ] as a set. We claim:

Indeed, let g ∈ G . As z g acts as involution on the Fitting subgroup F ⁢ ( G ) , it follows that [ z , g ] centralizes F. Since C G ⁢ ( F * ⁢ ( G ) ) ≤ F * ⁢ ( G ) = F ⁢ ( G ) , we have [ z , g ] ∈ F ⁢ ( G ) = [ z , F ⁢ ( G ) ] . Thus there exists an element f ∈ F ⁢ ( G ) such that [ z , g ] = [ z , f ] . So z g = z f , hence g ∈ C G ⁢ ( z ) ⁢ F ⁢ ( G ) . It follows that G = C G ⁢ ( z ) ⁢ F ⁢ ( G ) is not a Frobenius group. Since C G ⁢ ( z ) ∩ F ⁢ ( G ) = 1 , there exists a nontrivial element g ∈ C G ⁢ ( z ) such that C F ⁢ ( g ) ≠ 1 . Thus, for some 1 ≠ f ∈ C F ⁢ ( g ) , since y ∈ C G ⁢ ( y r ) centralizes z, we have

a contradiction. Thus C G ⁢ ( y r ) is a Frobenius complement with odd order and our conclusion follows from Lemma 2.4. ∎

Corollary 3.6.

Assume that Z ∞ ⁢ ( G ) = 1 and G is neither a Frobenius group nor an almost simple group. Then all the elements whose centralizer has even order belong to the same strong component of Γ ⁢ ( G ) .

Proof.

It suffices to prove that if | x | = | y | = 2 , then x ↠ y . This follows from Lemmas 3.4 and 3.5. ∎

Lemma 3.7.

Let X be a finite group. Assume that F * ⁢ ( X ) = F ⁢ ( X ) and that X / F ⁢ ( X ) ≅ S t , where S is a finite non-abelian simple group. Then Δ ⁢ ( X ) is strongly connected.

Proof.

The statement is trivially true if Z ∞ ⁢ ( G ) ≠ 1 . So we may assume Z ∞ ⁢ ( G ) = 1 .

It follows from the Brauer–Suzuki theorem that there is no simple group with generalized quaternion Sylow 2-subgroups. In particular, by Lemma 2.4, X / F ⁢ ( X ) cannot be isomorphic to a Frobenius complement and consequently X is not a Frobenius group. By Corollary 3.6, all the involutions of X belong to the same strong component, say Ω, of X. Moreover, if f is a nontrivial element of F ⁢ ( X ) and z is an involution of X, then f ↠ z by Lemma 3.5 and z ↦ f by Lemma 2.2, so F ⁢ ( X ) ∖ { 1 } ⊆ Ω .

We claim that for any odd prime p dividing | S | , Ω contains all the elements of X with order p. If t > 1 , then a Sylow p-subgroup of S t cannot act fixed-point-freely on F ⁢ ( X ) , so any Sylow p-subgroup of X contains an element in Ω, and, consequently, all the elements of this Sylow subgroup belong to Ω (using the fact that the center of a Sylow subgroup is nontrivial). Assume t = 1 and that, by contradiction, there exists no element of order p in Ω and let p be the smallest (odd) prime with this property. Let P be a Sylow p-subgroup of S. Since P acts fixed-point-freely on F ⁢ ( X ) , it follows that P = 〈 u 〉 is cyclic. By Burnside’s Theorem [13, Theorem 10.1.8], as S is not p-nilpotent, we have that P ≰ Z ⁢ ( N S ⁢ ( P ) ) , so there exists an element v ∈ N S ⁢ ( P ) ∖ C S ⁢ ( P ) . In particular, v ⁢ C S ⁢ ( P ) is a nontrivial element of N S ⁢ ( P ) / C S ⁢ ( P ) ≤ Aut ⁢ ( P ) , so its order divides φ ⁢ ( p n ) = p n - 1 ⁢ ( p - 1 ) and is coprime to p (since P ≤ C G ⁢ ( P ) ). But then there is a prime q dividing ( | v | , p - 1 ) and so the minimality of p implies that v is contained in Ω. Since v normalizes 〈 u 〉 , we have v → u . On the other hand, u → f for any f ∈ F ⁢ ( X ) , which means that u ∈ Ω and we have reached a contradiction. ∎

Lemma 3.8.

Assume Z ∞ ⁢ ( G ) = 1 , F ⁢ ( G ) = F * ⁢ ( G ) and there exist x , y ∈ G with d ⁢ ( x , y ) > 4 . Let J / F ⁢ ( G ) = F ⁢ ( G / F ⁢ ( G ) ) . Then | J / F ⁢ ( G ) | is prime to | F ⁢ ( G ) | .

Proof.

Let s be a prime divisor of | F ⁢ ( G ) | and let S be a Sylow s-subgroup of J. Take a power of y of prime order, say y r . As F ⁢ ( G ) ⁢ S is normal in G, either y r acts fixed-point-freely on F ⁢ ( G ) ⁢ S , and so F ⁢ ( G ) ⁢ S is nilpotent hence contained in F ⁢ ( G ) , or C F ⁢ ( G ) ⁢ S ⁢ ( y r ) ≠ 1 . Let g ∈ C F ⁢ ( G ) ⁢ S ⁢ ( y r ) be a nontrivial element of prime power order. Thus either g ∈ F ⁢ ( G ) , and

or g is an s-element and it is contained in a Sylow s-subgroup containing S ∩ F ⁢ ( G ) ≠ 1 . Thus g is contained in a nilpotent subgroup having nontrivial intersection with F ⁢ ( G ) , hence there exists a nontrivial element f ∈ F ⁢ ( G ) such that [ f , n g ] = 1 . It follows that

against the assumptions. ∎

Lemma 3.9.

Assume Z ∞ ⁢ ( G ) = 1 , F ⁢ ( G ) = F * ⁢ ( G ) and there exist x , y ∈ G with d ⁢ ( x , y ) > 4 . Let J / F ⁢ ( G ) = F ⁢ ( G / F ⁢ ( G ) ) , J * / F ⁢ ( G ) = F * ⁢ ( G / F ⁢ ( G ) ) and Z / F ⁢ ( G ) = Z ⁢ ( J / F ⁢ ( G ) ) . Let y r be a power of y of prime order r. If G is not Frobenius and J = J * , then y r ∈ Z .

Proof.

Suppose y r ∉ Z . As J / F ⁢ ( G ) = F * ⁢ ( G / F ⁢ ( G ) ) , we have y r ∉ C G / F ⁢ ( G ) ⁢ ( J / F ⁢ ( G ) ) , and so y r does not centralize a Sylow t-subgroup of J / F ⁢ ( G ) .

Assume by contradiction t = r and let R be a Sylow r-subgroup of J ⁢ 〈 y r 〉 containing y r . As y r does not centralize an r-subgroup of J / F ⁢ ( G ) , it follows that y r is not central in R and the center of R provides another element of order r that centralizes y r . Since, by Lemma 3.5, the Sylow subgroups of C G ⁢ ( y r ) are cyclic, we get a contradiction. Thus we have t ≠ r .

We now claim that there exists a Sylow t-subgroup T of J which is y r -invariant. Notice that y r acts on J = ∏ p T p ⁢ F ⁢ ( G ) and centralizes the r-Sylow subgroups. If r does not divide | F ⁢ ( G ) | , then the action is coprime, so there exists an invariant t-Sylow. If r divides | F ⁢ ( G ) | , then y r belongs to a r-Sylow subgroup of F ⁢ ( G ) ⁢ 〈 y r 〉 , and there exists a nontrivial element f of F ⁢ ( G ) such that [ f , n y r ] = 1 . Thus

a contradiction.

Let T be a Sylow t-subgroup of J which is y r -invariant. Note that y r acts faithfully on T.

Let V be a minimal normal subgroup of G. Thus V ≤ F ⁢ ( G ) . We claim that C G ⁢ ( V ) ≤ F ⁢ ( G ) . Otherwise y r could not act fixed-point-freely on C G ⁢ ( V ) , and consequently there would exist g ∈ C G ⁢ ( V ) ∩ C G ⁢ ( y r ) and thus, for any 1 ≠ v ∈ V ,

Moreover, V is a faithful Q ⁢ 〈 y r 〉 -module for every 〈 y r 〉 -invariant subgroup Q of T. Indeed, C G ⁢ ( V ) ≤ F ⁢ ( G ) and | F ⁢ ( G ) | is prime to the order of T ⁢ 〈 y r 〉 .

Since C V ⁢ ( 〈 y r 〉 ) is trivial, by [12, Lemma 2.5] we deduce that T is a 2-group and that every abelian characteristic subgroup of T is centralized by y r . Since C G ⁢ ( y r ) has odd order, we get a contradiction. ∎

Proposition 3.10.

Assume Z ∞ ⁢ ( G ) = 1 , F ⁢ ( G ) = F * ⁢ ( G ) and there exist x , y ∈ G with d ⁢ ( x , y ) > 4 . Let J / F ⁢ ( G ) = F ⁢ ( G / F ⁢ ( G ) ) and J * / F ⁢ ( G ) = F * ⁢ ( G / F ⁢ ( G ) ) . If J = J * , then G is a Frobenius group.

Proof.

We use the notation introduced in the previous lemma. Since y r ∈ Z , in particular y r belongs to the center of a Sylow r-subgroup R of J. Thus R ≤ C G ⁢ ( y r ) is cyclic, and 〈 y r 〉 is a characteristic subgroup of R. Now, R ⁢ F ⁢ ( G ) is normal in G and 〈 y r 〉 ⁢ F ⁢ ( G ) is a characteristic subgroup of R ⁢ F ⁢ ( G ) . Thus 〈 y r 〉 ⁢ F ⁢ ( G ) is normal in G. By the Frattini argument, as y r has order prime to F ⁢ ( G ) ,

hence G is Frobenius by Lemma 3.5. ∎

Corollary 3.11.

If G is soluble and G / Z ∞ ⁢ ( G ) is not a Frobenius group, then Γ ⁢ ( G ) is strongly connected and has diameter at most 4.

Proof of Theorem 1.2.

Let G be a finite group such that the factor G / Z ∞ ⁢ ( G ) is not an almost simple group. By Lemma 2.1, we can assume that Z ∞ ⁢ ( G ) = 1 . In Lemma 3.1 we have seen that if G is a Frobenius group, then Γ ⁢ ( G ) is not strongly connected. So we are left to prove that if G is neither almost simple nor Frobenius, then Γ ⁢ ( G ) is strongly connected. By Lemma 3.4 and Proposition 3.10 we are reduced to the case where F * ⁢ ( G ) = F ⁢ ( G ) and J ≠ J * . If J = F ⁢ ( G ) , then Δ ⁢ ( J * ) is connected by Lemma 3.7. In this case Γ ⁢ ( G ) is strongly connected by Lemma 3.2.

So we are left with the case J > F ⁢ ( G ) . Since J is not nilpotent, by Lemma 3.2 it is sufficient to prove that J ∖ { 1 } is contained in a strong component of Γ ⁢ ( G ) .

If J is not a Frobenius group, then Γ ⁢ ( J ) is strongly connected, by Proposition 3.10, and we are done. So J is a Frobenius group. Let P / J be a Sylow 2-subgroup of J * / J . Note that P / F is a central product of a Sylow 2-subgroup of E ⁢ ( G / F ⁢ ( G ) ) and J / F ⁢ ( G ) .

We distinguish two cases:

1. 2 divides | F ⁢ ( G ) | . In this case P is neither Frobenius nor almost simple, so, by Proposition 3.10, Γ ⁢ ( P ) is strongly connected. As a consequence J ∖ { 1 } is contained in a strong component of Γ ⁢ ( G ) .

2. 2 does not divide | F ⁢ ( G ) | . In this case F ⁢ ( G ) has a complement L in P and any element of L has centralizer of even order, since P / F ⁢ ( G ) is a central product of a Sylow 2-subgroup of E ⁢ ( G / F ⁢ ( G ) ) and J / F ⁢ ( G ) . As G is neither Frobenius nor almost simple, from Corollary 3.6 it follows that J ∖ { 1 } is contained in a strong component of Γ ⁢ ( G ) , as claimed.∎

Theorem 3.12.

Suppose Z ∞ ⁢ ( G ) = 1 . If Γ 2 ⁢ ( G ) is not strongly connected, then G is either almost simple or Frobenius.

Proof.

The proof of Theorem 1.2 relies on the following observations:

1. If [ x , y ] = 1 , then x ↦ y .

2. If x ∈ N G ⁢ ( 〈 y 〉 ) , then [ x , 2 y ] = 1 , hence x ↦ y .

3. If y ∈ F ⁢ ( G ) , then x ↦ y .

Clearly if either [ x , y ] = 1 or x ∈ N G ⁢ ( 〈 y 〉 ) , then x ↦ 2 y . If z ∈ Z ⁢ ( F ⁢ ( G ) ) , then [ x , 2 z ] = 1 , so x , z , y is a path in Γ 2 ⁢ ( G ) for any y ∈ F ⁢ ( G ) . ∎

Remark 3.13.

The previous theorem is no more true if the assumption Z ∞ ⁢ ( G ) = 1 is removed. Consider for example G = GL ⁡ ( 2 , 3 ) . We have Z ∞ ⁢ ( G ) = Z ⁢ ( G ) ≅ C 2 and G / Z ⁢ ( G ) ≅ S 4 . By Lemma 3.11, Γ ⁢ ( G ) is strongly connected. However, if x has order 3 or 6 and x ↦ 2 y , then y has order 3 or 6. Indeed, there are nine (non-central) elements g ∈ G with x ↦ g : the three elements of order 3 or 6 different from x in the unique cyclic subgroup of G of order 6 containing x, and six elements of order 4. If g is one of these six elements of order 4, then [ x , 3 g ] = 1 but [ x , 2 g ] ≠ 1 . So Γ 3 ⁢ ( G ) is connected, but Γ 2 ⁢ ( G ) is not.

Lemma 4.1.

The element x r has order r, commutes with c and acts fixed-point-freely on F. Moreover, c x = c q ≠ c .

Proof.

We have

Since | e | = r 2 and r 2 does not divides q r - 1 q - 1 , it follows that x r has order r and acts fixed-point-freely on F. The other information in the statement can easily be verified. ∎

Lemma 4.2.

The following statements hold:

1. C G ⁢ ( x ) = 〈 x 〉 .

2. C G ⁢ ( x r ) = D .

3. Let C = 〈 c 〉 . Then [ F , C ] is a nontrivial subgroup of F normalized by D and [ F , C ] ⁢ D is a Frobenius group.

Proof.

(1) Let g = f ⁢ d ∈ C G ⁢ ( x ) for some f ∈ F and d ∈ D . So, 1 = [ f ⁢ d , x ] = [ f , x ] d ⁢ [ d , x ] , and therefore [ f , x ] = 1 and [ d , x ] = 1 . Since x acts fixed-point-freely on F, we deduce f = 1 . Moreover, we have c x = c q ≠ c , hence C G ⁢ ( x ) = C D ⁢ ( x ) = 〈 x 〉 .

(2) Arguing as before, since x r acts fixed-point-freely on F, we deduce C G ⁢ ( x r ) = C D ⁢ ( x r ) = D .

(3) Since C ⁢ ⊴ ⁢ D , it follows that [ F , C ] is normalized by D. Notice that [ F , C ] is precisely the eigenspace of c corresponding to the eigenvalue f. In particular, [ F , C ] is a 1-dimensional 𝔽 -subspace of F. Let 1 ≠ d ∈ D . If r divides | d | , then x r ∈ 〈 d 〉 , so C [ F , C ] ⁢ ( d ) ≤ C [ F , C ] ⁢ ( x r ) ≤ C F ⁢ ( x r ) = 1 . If r does not divide | d | , then 〈 d 〉 = 〈 c 〉 , so C [ F , C ] ⁢ ( d ) = C [ F , C ] ⁢ ( c ) = 1 . ∎