Remark on subgroup intersection graph of finite abelian groups

Theorem 2.1

[11, Chapter I, Theorem 8.1] Let G be a torsion abelian group. Then G is the direct sum of its subgroups G ( p ) for all primes p such that G ( p ) ≠ 0 .

Theorem 2.2

[11, Chapter I, Theorem 8.2] Every finite abelian p-group is isomorphic to a product of cyclic p-groups. If it is of type ( p r 1 , … , p r s ) with

then the sequence of integer ( r 1 , … , r s ) is uniquely determined.

Theorem 2.3

[11, Chapter I, Theorem 8.5] Let G be a finitely generated abelian group. Then the torsion subgroup G tor is finite, G G tor is free and there exists a free subgroup of H of G such that G is the direct sum of G tor and H.

Lemma 3.1

Let G be a finite abelian p-group of exponent p e . Then

where m i = | { x ∈ G | x p i = 1 G } | and c i = 1 p − 1 m i + 1 m i − m i + 2 m i + 1 . In particular, c e − 1 > 0 .

Proof

Let x ∈ G be an element with order p and let H = { 1 G , x , x 2 , … , x p − 1 } be the cyclic group generated by x. Since z is adjacent to some identity element of H if and only if H ⊂ 〈 z 〉 , so the connected component containing x, say L x , is a complete graph, whose vertices are { z ∈ G : ∃ k ≥ 0 , z p k ∈ { x , … , x p − 1 } } .

Let n k = | { z ∈ G : z p k = x } | . Then there exists i > 0 such that n j = m j for j ≤ i and n j = 0 for j > i . Note that | { z ∈ G : z p i = x r } | = n i for given r ∈ { 1 , … , p − 1 } . It follows that | L x | = ( p − 1 ) ( 1 + m 1 + ⋯ + m i ) vertices and there are exactly ( p − 1 ) m t vertices in L x with order p t for t ∈ { 1 , … , i + 1 } . Therefore, the number of elements in G with order p i + 1 is

and we obtain that c i = 1 p − 1 m i + 1 m i − m i + 2 m i + 1 . This completes the proof.□

Theorem 3.1

Γ ( G 1 ) ≃ Γ ( G 2 ) if and only if G 1 ≃ G 2 for any two abelian p-groups.

Proof

Suppose that the type of G 1 is ( p r 1 , … , p r s ) and the type of G 2 is ( p t 1 , … , p t l ) , where 1 ≤ r 1 ≤ ⋯ ≤ r s , 1 ≤ t 1 ≤ ⋯ ≤ t l . Then we have m i = | { x ∈ G 1 | x p i = 1 G 1 } | = p ∑ j = 1 s min { i , r j } , n i = | { x ∈ G 2 | x p i = 1 G 2 } | = p ∑ j = 1 l min { i , t j } , c i = 1 p − 1 m i + 1 m i − m i + 2 m i + 1 and d i = 1 p − 1 n i + 1 n i − n i + 2 n i + 1 . Note that c r s − 1 > 0 and d t l − 1 > 0 . So r s = t l = e and c i = d i for any 0 ≤ i ≤ e − 1 . By a direct computation, we obtain

Combining with that m i + 1 m i = p | { j : r j = i + 1 , 1 ≤ j ≤ s } | and n i + 1 n i = p | { j : t j = i + 1 , 1 ≤ j ≤ l } | , one has s = l , r i = t i and G 1 ≃ G 2 . □

Lemma 4.1

Let G be a finite group. Then β ( Γ ( G ) ) is equal to the number of subgroups of prime order of G. An element of G is contained in an independent set of maximal size if and only if o ( x ) is a power of a prime.

Proof

Let A = { H i } i ∈ I be the collection of all cyclic subgroups of prime order of G and let X be a maximal independent set of Γ ( G ) . We claim that: for any given H ∈ A , there exists exactly one x ∈ X such that H ⊆ 〈 x 〉 . Suppose that H ⊈ 〈 x 〉 for any x ∈ X . Choose a non-identity element z ∈ H . Since H is a cyclic group of prime order, one has | 〈 z 〉 ∩ 〈 x 〉 | = | H ∩ 〈 x 〉 | > 1 if and only if H ⊆ 〈 x 〉 . So z is not adjacent to any point in X and X ∪ { z } is also an independent set. This contradicts to the choice of X. If H ⊆ 〈 x 〉 and H ⊆ 〈 y 〉 for x ≠ y , then H ⊂ 〈 x 〉 ∩ 〈 y 〉 and x and y are adjacent. This proves the claim.

We define a map σ : A → X such that σ ( H ) = x if H ⊆ 〈 x 〉 . By the claim, σ is well-defined and surjective. Therefore, β ( Γ ( G ) ) ≤ | A | . On the other hand, choose y i ≠ 0 in H i , then 〈 y i 〉 = H i and Y = { y i } i ∈ I is an independent set with | Y | = β ( Γ ( G ) ) . This shows that β ( Γ ( G ) ) = | A | . If x ∈ G and o ( x ) is a power of a prime, then there exists unique y ∈ Y such that 〈 y 〉 ⊂ 〈 x 〉 . Thus, Y ∪ { x } \ { y } is also an independent set with size β ( Γ ( G ) ) .

Now suppose that a ∈ X and o ( a ) is not a power of a prime. Then 〈 a 〉 contains at least two distinct subgroups of prime order. So σ : A → X is not injective in this case and | X | < | A | . The proof is finished.□

Lemma 4.2

Let G be a group and let V 1 , V 2 be the set of non-identity elements of finite and infinite order of G, respectively. Let Γ i be the induced subgraph on V i . Then Γ ( G ) = Γ 1 ∪ Γ 2 and each connected component of Γ 2 is a complete graph of infinite order.

Proof

If x , y ∈ G with o ( x ) < ∞ and o ( y ) = ∞ , then each element in 〈 x 〉 has finite order and each non-identity element in 〈 y 〉 has infinite order. So x and y are not adjacent and thus Γ = Γ 1 ∪ Γ 2 .

Now suppose o ( x ) , o ( y ) , o ( z ) are all infinite and ( x , y ) , ( x , z ) ∈ E ( Γ ( G ) ) . Then exist non-zero integers m , n , k , l such that x n = y m and x k = z l . Hence, y m k = x n k = z ln and ( y , z ) ∈ E ( Γ ( G ) ) . It follows that the connected component containing x is a complete graph. This completes the proof.□

Lemma 4.3

Suppose that G is a finitely generated abelian group of rank r. Then

1. If r = 1 , then Γ ( G ) = Γ ( G tor ) ∪ K ∞ ;

2. If r ≥ 2 , then Γ ( G ) is the union graph of Γ ( G tor ) and infinite countable copies of K ∞ .

Proof

Keeping the same notations in Lemma 4.2. Since G is countable, Γ ( G ) is a union of Γ 1 = Γ ( G tor ) and some copies of K ∞ by Lemma 4.2. It is easy to prove that Γ 2 = K ∞ if r = 1 , and there are infinite connected component in Γ 2 if r ≥ 2 . □

Theorem 4.1

Suppose that G 1 , G 2 are finitely generated abelian groups of rank r 1 , r 2 , respectively. Then Γ ( G 1 ) ≃ Γ ( G 2 ) if and only if G 1 tor ≃ G 2 tor and one of the following conditions is satisfied.

1. r 1 = r 2 ≤ 1 ;

2. min { r 1 , r 2 } ≥ 2 .

Proof

It suffices to deal with the case that G i = G i tor by Lemma 4.3. Assume that Γ i = Γ ( G i ) and φ : Γ 1 → Γ 2 is an isomorphism of graphs. Hence, | G 1 | = | G 2 | = n . Let p be a prime divisor of n. Since the induced subgraph of Γ i on the subset G i ( p ) ⁎ is isomorphic to Γ ( G i ( p ) ) , it suffices to prove that φ ( G 1 ( p ) ⁎ ) = G 2 ( p ) ⁎ .

Suppose x 1 , x 2 ∈ G 1 ( p ) ⁎ . By Lemma 4.1, the order of y i = φ ( x i ) is also a power of a prime, say p i . We claim that p 1 = p 2 . If x 1 and x 2 are adjacent, then y 1 and y 2 are also adjacent. Hence, both o ( y 1 ) and o ( y 2 ) are powers of a same prime. Now assume that x 1 and x 2 are not adjacent and p 1 ≠ p 2 . Then o ( y 1 y 2 ) = o ( y 1 ) o ( y 2 ) and 〈 y i 〉 ⊂ 〈 y 1 y 2 〉 . It follows that ( y 1 y 2 , y 1 ) , ( y 1 y 2 , y 2 ) ∈ E ( Γ 2 ) and p 1 p 2 ∤ o ( y 1 y 2 ) . By Lemma 4.1, the order of x = φ − 1 ( y 1 y 2 ) is not a power of a prime. Since φ is an isomorphism of graphs, we have ( x , x 1 ) , ( x , x 2 ) ∈ E ( Γ 1 ) . Hence, the cyclic group generated by x contains two distinct subgroups of order p, which is impossible. So p 1 = p 2 .

Let n = ∏ i = 1 r p i e i be the prime factorization. According to the aforementioned arguments, there exists a permutation τ of { 1 , … , r } such that φ ( G 1 ( p i ) ⁎ ) = G 2 ( p τ ( i ) ) ⁎ . Hence, we obtain | G 1 ( p i ) | = p i e i = | G 2 ( p τ ( i ) ) | = p τ ( i ) e τ ( i ) and τ ( i ) = i . It follows that φ ( G 1 ( p ) ⁎ ) = G 2 ( p ) ⁎ . So Γ ( G 1 ( p ) ) ≃ Γ ( G 2 ( p ) ) . By Theorem 3.1, we have G 1 ( p ) ≃ G 2 ( p ) , and thus G 1 ≃ G 2 .□

Corollary 4.1

Let G 1 , G 2 be finite abelian groups. Then Γ ( G 1 ) ≃ Γ ( G 2 ) if and only if G 1 ≃ G 2 .