First of all, I am very new to group theory.

The order of an element $g$ of a group $G$ is the smallest positive integer $n: g^n=e$, the identity element. I understand how to find the order of an element in a group when the group has something to with modulo, for example, in the group $$U(15)=\text{the set of all positive integers less than } n \text{ and relatively prime to } n.$$

$$\text{ which is a group under multiplication by modulo }n=\{1,2,4,7,8,11,13,14\},$$ then $|2|=4$, because

\begin{align*} &2^1=2\\ &2^2=4\\ &2^3=8\\ &2^4=16\mod15=1\\ &\text{So } |2|=4. \end{align*}

However, I don't understand how this works for groups that don't have any relation to modulo. Take $(\mathbb{Z},+)$ for instance. If I wanted to find the order of $3$, then I need to find $n:3^n$ is equal to the identity, which in this case is $0$.

I suppose my question can be summarized as follows:

Does the order of an element only make sense if we are dealing with groups dealing with modulo?


3 Answers 3


Yes, it makes sense. The order of an element $g$ in some group is the least positive integer $n$ such that $g^n = 1$ (the identity of the group), if any such $n$ exists. If there is no such $n$, then the order of $g$ is defined to be $\infty$.

As noted in the comment by @Travis, you can take a small permutation group to get an example. For instance, the permutation $(1,2,3,4)$ in the symmetric group $S_4$ of degree $4$ (all permutations of the set $\{1,2,3,4\}$) has order $4$. This is because $$(1,2,3,4)^1 = (1,2,3,4)\neq 1,$$ $$(1,2,3,4)^2 = (1,3)(2,4)\neq 1,$$ $$(1,2,3,4)^3 = (1,4,3,2)\neq 1$$ and $$(1,2,3,4)^4 = 1,$$ so $4$ is the smallest power of $(1,2,3,4)$ that yields the identity.

For the additive group $\mathbb{Z}$ of integers, every non-zero element has infinite order. (Of course, here, we use additive notation, so to calculate the order of $g\in\mathbb{Z}$, we are looking for the least positive integer $n$ such that $ng = 0$, if any. But, unless $g = 0$, there is no such $n$, so the order of $g$ is $\infty$.)

  • $\begingroup$ Could you provide an example of a group that doesn't have to do with modulo, but the (finite) order of an element can be found? $\endgroup$ Oct 13, 2014 at 18:13
  • 2
    $\begingroup$ @SujaanKunalan It depends on what you mean by "doesn't have to do with modulo", but if you mean a group that is not a multiplicative subgroup of $\mathbb{Z}_n$, then any nonabelian group will certainly do. The smallest of these is $S_3$: the identity has order $1$, the three $2$-cycles have order $2$, and the two $3$-cycles have order $3$. $\endgroup$ Oct 13, 2014 at 18:21
  • 1
    $\begingroup$ Consider the group of symmetries of the square (the dihedral group $D_4$): four rotations through angles $0^\circ$, $90^\circ$, $180^\circ$, $270^\circ$, and four reflections, for a total of $8$ elements. The reflections each have order $2$, as does the $180^\circ$ rotation, two of the rotations have order $4$, and the identity (listed as a $0^\circ$ rotation) has order $1$. $\endgroup$ Oct 13, 2014 at 18:22

No, the notion makes sense for all groups (at least all finite groups, anyway as infinite groups can have elements with infinite order), and its definition is just the one you give. (All multiplicative subgroups of $\mathbb{Z}_n$, i.e., integers modulo $n$ are abelian, but not all groups are abelian.)


A group can have finite or infinite number of elements. When the group has finite number of elements, we see the least POSITIVE n i.e.(n>0) such that g^n gives the identity of the group (in case of multiplication) or n*g gives the identity (in case of addition). Here Z has an infinite number of elements. There does not exist any n>0 for which you obtain identity. Hence Z is of infinite order.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.