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exponent of a group

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Definition

The exponent of a group GG, denoted exp(G)\exp(G), is equivalently

Here “least” should, by all rights, refer to the divisibility order on \mathbb{N}, where mnm \leq n means mm divides nn. Notice that with this poset structure, \mathbb{N} is a complete lattice with bottom element 11 and top element 00; under this convention, the exponent always exists as the least common multiple of the orders of elements, which is their join (supremum) in the lattice, and is 00 if (for example) GG is a torsion-free group.

Properties

If GG is a finite group of order nn, then exp(G)\exp(G) divides nn since g n=1g^n = 1 for all gGg \in G.

Proposition

If GG is a finite abelian group and exp(G)=ord(G)\exp(G) = ord(G), then GG is cyclic.

Proof

Of course ord(G)ord(G) can’t be 00, so e=exp(G)e = \exp(G) here has a prime factorization e=p 1 r 1p 2 r 2p k r ke = p_1^{r_1} p_2^{r_2} \ldots p_k^{r_k}. Since ee is the least common multiple of the orders of elements, the exponent r ir_i is the maximum multiplicity of p ip_i occurring in orders of elements; any element realizing this maximum will have order divisible by p i r ip_i^{r_i}, and some power y iy_i of that element will have order exactly p i r ip_i^{r_i}. Then y= iy iy = \prod_i y_i will have order e=ord(G)e = ord(G) by the following lemma and induction, so that powers of yy exhaust all ord(G)ord(G) elements of GG, i.e., yy generates GG as desired.

Lemma

If m,nm, n are relatively prime and xx has order mm and yy has order nn in an abelian group, then xyx y has order mnm n.

Proof

Suppose (xy) k=x ky k=1(x y)^k = x^k y^k = 1. For some a,ba, b we have ambn=1a m - b n = 1, and so 1=x kamy kam=y kam=y ky kbn=y k1 = x^{k a m} y^{k a m} = y^{k a m} = y^k y^{k b n} = y^k. It follows that nn divides kk. Similarly mm divides kk, so mn=lcm(m,n)m n = lcm(m, n) divides kk, as desired.

Proposition

If GG is a finite group, then the prime factors of exp(G)\exp(G) coincide with the prime factors of ord(G)ord(G).

Proof

Since exp(G)\exp(G) divides ord(G)ord(G), every prime factor of exp(G)\exp(G) is a factor of ord(G)ord(G). If pp is a prime factor of ord(G)ord(G), then by the Cauchy group theorem, GG has an element of order pp, and thus pp divides exp(G)\exp(G).

Examples

  • If exp(G)=2\exp(G) = 2, then GG is abelian and is a vector space over the field 𝔽 2\mathbb{F}_2.

  • Similarly, if GG is abelian and its exponent is a prime pp, then GG is a vector space over 𝔽 p\mathbb{F}_p.

  • A finitely generated group of exponent 33, 44, or 66 must be finite. On the other hand, it is not known whether a group of exponent 55 generated by 22 elements must be finite.

These last facts are part of the lore of the celebrated Burnside problem. The free Burnside group B(m,n)B(m, n) of exponent nn with mm generators is presented by x 1,,x m|w n=1forallwordswinx 1,,x m\langle x_1, \ldots, x_m\; |\; w^n = 1 \; for\; all\; words\; w\; in \; x_1, \ldots, x_m \rangle, and one formulation of the original Burnside problem was whether B(m,n)B(m, n) is finite. It is now known that the answer is negative for all m>1m \gt 1 and odd n>665n \gt 665, and there is a similar result for certain even nn as well.

  • For a fixed prime pp, a Tarski monster for pp is a finitely generated infinite group where every proper nontrivial subgroup is cyclic of order pp. Such monsters exist for sufficiently large pp, forming a class of dramatic counterexamples to the Burnside conjecture; they are of course of exponent pp.

  • It was shown by Efim Zelmanov that there are only finitely many finite groups with mm generators and exponent nn (restricted Burnside problem). It was largely on the strength of this work that he was awarded the Fields Medal (1994).

References

See also

Last revised on September 27, 2018 at 04:52:17. See the history of this page for a list of all contributions to it.