A note on the product of element orders of finite abelian groups

A note on the product of element orders of finite abelian groups

Marius T˘ arn˘ auceanu April 24, 2012

Abstract

Given a finite groupG, we denote byψ⁰(G) the product of element orders of G. Our main result proves that the restriction of ψ⁰ to abelian p-groups of order pⁿ is strictly increasing with respect to a natural order on the groups relating to the lexicographic order of the partitions of n. In particular, we infer that two finite abelian groups of the same order are isomorphic if and only if they have the same product of element orders.

MSC (2010): Primary 20K01; Secondary 20D60, 20D15.

Key words: finite abelian groups, product of element orders.

1 Introduction

Let Gbe a finite group and

ψ(G) =X

x∈G

o(x),

where o(x) denotes the order of x∈G. The starting point for our discussion is given by the papers [1, 2] which investigate the minimum/maximum of ψ on groups of the same order. Other properties of the function ψ have been studied in [10] for finite abelian groups.

In the current note we will focus on the function ψ⁰(G) = Y

x∈G

o(x).

In contrast withψ, this is not multiplicative, as shown by the following result.

Proposition A. Let G₁, G₂, ..., G_k be finite groups having coprime orders.

Then

ψ⁰(

×

G_i) =

k

Y

i=1

ψ⁰(G_i)ⁿⁱ,

where n_i =

k

Y

j=1 j6=i

| G_j |, i = 1,2, ..., k. In particular, if G is a finite nilpotent group of order n = p^α₁¹p^α₂²· · ·p^α_k^k, G₁, G₂, ..., G_k are the Sylow subgroups of G and n_i =n/p^α_iⁱ, i= 1,2, ..., k, then

ψ⁰(G) =

k

Y

i=1

ψ⁰(Gi)ⁿⁱ.

By Proposition A we infer that the computation of ψ⁰(G) for nilpotent groups is reduced to p-groups and explicit formulas can be given in several particular cases. One of them consists of abelian groups, for which Corollary 4.4 of [9] leads to the following theorem.

Theorem B. Let G=

×

i=1 Zp^αi be a finite abelian p-group, where 1≤ α₁ ≤ α₂ ≤...≤α_k. Then

ψ⁰(G) =p^{αkpα1+α2+...+αk−Pαk}

−1 i=0 pⁱf_(α

1,α2,...,αk)(i)

, (1)

where

f(α1,α2,...,α_k)(i) =















p^(k−1)i, if 0≤i≤α₁ p^(k−2)i+α1, if α₁ ≤i≤α₂ p^{(k−3)i+α1+α2}, if α₂ ≤i≤α₃ ...

p^{α1+α2+...+αk−1}, if i≥αk−1(whereαk−1 = 1if k= 1).

We exemplify (1) by computing ψ⁰(G) for cyclic p-groups and for rank two abelian p-groups.

Example. We have:

1. ψ⁰(Zp^α) = p^αpα

+1−(α+1)pα+1

p−1 ;

2. ψ⁰(Zp^α×Zp^β) =p

βpα+β+2−pα+β+1−(β+1)pα+β+p2α+1+1

p2−1 .

Given a positive integer n, it is well-known that there is a bijection between the set of types of abelian groups of order pⁿ and the set P_n = {(x₁, x₂, ..., x_n)∈Nⁿ|x₁ ≥x₂ ≥...≥x_n, x₁+x₂+...+x_n =n}of partitions of n. Namely,

×

i=1 Zp^αi(with 1≤α₁ ≤α₂ ≤...≤α_k and

k

X

i=1

α_i =n)7→(α_k, ..., α₁, 0, ...,0

| {z }

n−kpositions

).

Moreover, recall that P_n is totally ordered under the lexicographic order, where

(x₁, x₂, ..., x_n)≺(y₁, y₂, ..., y_n)⇐⇒







x₁ =y₁, ..., x_m =y_m and

xm+1<ym+1 for somem ∈ {0,1, ..., n−1}. This induces a total order on the set of types of abelian p-groups of order pⁿ. Our next theorem shows that the restriction of ψ⁰ on this set is strictly increasing.

Theorem C. Let G =

×

i=1 Z^pαi and H =

×

j=1 Z_p^βj be two finite abelian p-groups of order pⁿ. Then

ψ⁰(G)< ψ⁰(H)⇐⇒(α_k, ..., α₁, 0, ...,0

| {z }

n−kpositions

)≺(β_r, ..., β₁, 0, ...,0

| {z }

n−rpositions

). (2)

Since a strictly increasing function is injective, by Theorem C we infer the following corollary.

Corollary D. Two finite abelian p-groups of order pⁿ are isomorphic if and only if they have the same product of element orders.

This can be extended to arbitrary finite abelian groups, according to Proposition A.

Theorem E. Two finite abelian groups of the same order are isomorphic if and only if they have the same product of element orders.

Remark. The above two results are not true for arbitrary finite abelian groups. For example, we haveψ⁰(Z4×Z²3) =ψ⁰(Z⁴2×Z3), but obviously the groups Z⁴×Z²3 and Z⁴2×Z³ are not isomorphic.

Finally, we associate to a finite (abelian) group G = {x₁, x₂, ..., x_n} the polynomial

P_G=

n

Y

i=1

(X−o(x_i))∈Z[X].

Recall that ifGandHare two finite abelian groups for whichP_G=P_H (that is,GandH have the same element orders with the same multiplicities), then G ∼= H by Theorem 5 of [8]. In order to improve this result, we construct the quantities

ψ_k(G) = X

1≤i₁<i2<...<ik≤n

o(x_i1)o(x_i2)· · ·o(x_ik), k = 1,2, ..., n.

Obviously, explicit formulas for ψ_k(G), k = 1,2, ..., n, can be given by using Corollary 4.4 of [9]. We also observe thatψ₁(G) = ψ(G) andψ_n(G) =ψ⁰(G).

Inspired by Conjecture 6 of [10] and the above Theorem E, we came up with the following conjecture, which we have verified by GAP for many values of k and n.

Conjecture F. Let G and H be two finite abelian groups of order n. Then for every k∈ {1,2, ..., n}, we have G∼=H if and only if ψ_k(G) =ψ_k(H).

Most of our notation is standard and will not be repeated here. For basic notions and results of group theory we refer the reader to [4, 5]. Other interesting papers on the above topic are [3, 6, 7].

2 Proof of Theorem C

We first remark that it suffices to verify (2) only for consecutive partitions of n, becauseP_n is totally ordered.

Assume that (α_k, ..., α₁,0, ...,0)≺(β_r, ..., β₁,0, ...,0) and lets∈ {1,2, ..., r−1} such that β₁ = β₂ = · · · = β_s < β_s+1. We distinguish the following two cases.

Case 1. β₁ ≥2

Then (α_k, ..., α₁,0, ...,0) is of type (β_r, ..., β₂, β₁−1,1,0, ...,0), i.e. k =r+ 1, α₁ = 1,α₂ =β₁−1 andα_i =βi−1 fori= 3,4, ..., r+ 1. It is easy to see that f(α1,α2,...,α_k)(i) = f(β1,β2,...,βr)(i) for all i≥β1. One obtains

ψ⁰(G)< ψ⁰(H)⇐⇒

βr−1

X

i=1

pⁱf_{(β1,β2,...,βr)}(i)<

α_k−1

X

i=1

pⁱf_{(α1,α2,...,αk)}(i)

⇐⇒

β1−1

X

i=1

pⁱf_{(β1,β2,...,βr)}(i)<

β1−1

X

i=1

pⁱf_{(α1,α2,...,αk)}(i) and the last inequality is true because

f_{(β1,β2,...,βr)}(i)=p^(r−1)i<p^(r−1)i+1=p^(k−2)i+1=f_{(α1,α2,...,αk)}(i), i= 1,2, ..., β₁−1. Case 2. β₁ = 1

Then (α_k, ..., α₁,0, ...,0) is of type (β_r, ..., β_s+1 − 1, β_t⁰, β_t−1⁰ , ..., β⁰₁,0, ...,0), whereβ_s+1−1≥β_t⁰ ≥β_t−1⁰ ≥...≥β₁⁰ ≥1 andβ_t⁰+β_t−1⁰ +...+β₁⁰ =s+ 1. We infer that f_{(α1,α2,...,αk)}(i) =f_{(β1,β2,...,βr)}(i) for all i≥β_s+1. So, we can suppose that s =r−1, i.e.

(α_k, ..., α₁,0, ...,0) = (β_r−1, β_t⁰, β_t−1⁰ , ..., β₁⁰,0, ...,0).

One obtains

ψ⁰(G)<ψ⁰(H)⇐⇒(β_r−1)pⁿ−

βr−2

X

i=0

pⁱf_{(α1,α2,...,αk)}(i)<β_rpⁿ−

βr−1

X

i=0

pⁱf_{(β1,β2,...,βr)}(i)

⇐⇒pⁿ−

βr−1

X

i=0

pⁱf_{(β1,β2,...,βr)}(i) +

βr−2

X

i=0

pⁱf_{(α1,α2,...,αk)}(i)>0.(∗) Since

f_{(β1,β2,...,βr)}(i) =

p^(r−1)i, if 0≤i≤1 p^r−1, if i≥1, we easily get

βr−1

X

i=0

pⁱf_{(β1,β2,...,βr)}(i) = 1 +p^r−1p^βr −p

p−1 = 1 + pⁿ−p^r p−1 < pⁿ and therefore (∗) is true.

Conversely, assume that (α_k, αk−1, ..., α₁,0, ...,0)(β_r, βr−1, ..., β₁,0, ...,0).

If these partitions are equal, then G ∼= H, so ψ⁰(G) = ψ⁰(H). Other- wise, (αk, αk−1, ..., α1,0, ...,0)(βr, βr−1, ..., β1,0, ...,0), and the first part of the proof implies ψ⁰(G)> ψ⁰(H), as desired.

Acknowledgements. The author is grateful to the reviewers for their re- marks which improve the previous version of the paper.

References

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Marius T˘arn˘auceanu Faculty of Mathematics

“Al.I. Cuza” University Ia¸si, Romania

e-mail: [email protected]