## You are here

Homeexponent

## Primary tabs

# exponent

Let $G$ be a group with the property that there exists a positive integer $n$ such that, for every $g\in G$, $g^{n}=e_{G}$. The exponent of $G$, denoted $\operatorname{exp}~{}G$, is the smallest positive integer $m$ such that, for every $g\in G$, $g^{m}=e_{G}$. Thus, for every finite group $G$, $\operatorname{exp}~{}G$ divides $|G|$. Also, for every group $G$ that has an exponent and for every $g\in G$, $|g|$ divides $\operatorname{exp}~{}G$.

The concept of exponent for finite groups is similar to that of characterisic for finite fields.

If $G$ is a finite abelian group, then there exists $g\in G$ with $|g|=\operatorname{exp}~{}G$. As a result of the fundamental theorem of finite abelian groups, there exist $a_{1},\ldots,a_{n}$ with $a_{i}$ dividing $a_{{i+1}}$ for every integer $i$ between 1 and $n$ such that $G\cong{\mathbb{Z}}_{{a_{1}}}\oplus\cdots\oplus{\mathbb{Z}}_{{a_{n}}}$. Since, for every $c\in G$, $c^{{a_{n}}}=e_{G}$, then $\operatorname{exp}~{}G\leq a_{n}$. Since $|(0,\ldots,0,1)|=a_{n}$, it follows that $\operatorname{exp}~{}G=a_{n}$.

Following are some examples of exponents of finite nonabelian groups.

Since $|(12)|=2$, $|(123)|=3$, and $|S_{3}|=6$, it follows that $\operatorname{exp}~{}S_{3}=6$.

In $Q_{8}=\{1,-1,i,-i,j,-j,k,-k\}$, the ring of quaternions of order eight, since $|i|=|-i|=|j|=|-j|=|k|=|-k|=4$ and $1^{4}=(-1)^{4}=1$, it follows that $\operatorname{exp}~{}Q_{8}=4$.

Since the order of a product of two disjoint transpositions is 2, the order of a three cycle is 3, and the only nonidentity elements of $A_{4}$ are three cycles and products of two disjoint transpositions, it follows that $\operatorname{exp}~{}A_{4}=6$.

Since $|(123)|=3$ and $|(1234)|=4$, $\operatorname{exp}~{}S_{4}\geq 12$. Since $S_{4}$ has no elements of order 8, it cannot have an element of order 24. It follows that $\operatorname{exp}~{}S_{4}=12$.

Following are some examples of exponents of infinite groups.

Clearly, any infinite group that has an element of infinite order does not have an exponent. On the other hand, just because an infinite group has the property that every element has finite order does not mean that the group has an exponent. As an example, consider $G=\mathbb{Q}/\mathbb{Z}$, which is a group under addition. Despite that all of its elements have finite order, $G$ does not have an exponent. This is because, for every positive integer $n$, $G$ has an element of order $n$, namely $\displaystyle\frac{1}{n}+\mathbb{Z}$.

On the other hand, some infinite groups have exponents. For example, let $\mathbb{F}_{2}$ denote the field having two elements. Then $\mathbb{F}_{2}[x]$, the ring of all polynomials in $x$ with coefficients in $\mathbb{F}_{2}$, is an abelian group under addition. Moreover, it is an infinite group; however, every nonzero element has order $2$. Thus, $\operatorname{exp}~{}\mathbb{F}_{2}[x]=2$.

## Mathematics Subject Classification

20A99*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff

## Recent Activity

new question: numerical method (implicit) for nonlinear pde by roozbe

new question: Harshad Number by pspss

Sep 14

new problem: Geometry by parag

Aug 24

new question: Scheduling Algorithm by ncovella

new question: Scheduling Algorithm by ncovella

## Attached Articles

## Corrections

lots of typos by owenjonesuk ✓

typo by pahio ✓

exponents of infinite groups by yark ✓