If $G$ is a finite group, show that there exists a positive integer $N$ such that $a^N = e$ for all $a \in G$.

I know that there are numerous threads in the forum about this topic and each and everyone of it (more or less) asks for a proof. However, I have a question regarding it, so, I am posting it. I was recently reading about group theory when I encountered this question.

I was given a definition about the order of an element: The order of an element $a$ in a group $G$ is the least positive integer $n$ such that $a^n=e$, where $e$ is the identity element in $G$. If no such $n$ exists, we call the element of infinite order. For example: If $G={1,\omega, \omega^2}$ under the usual multiplication as the binary operation forms a group. Now, here order of $1$ is $1$, the order of $\omega$ is $3$ and the order of $\omega^2$ is $3$. So each and every element in the group is of finite order. Now, if we consider the group $\mathbb{Z}$ under the usual addition as the binary operation then, $1\in\mathbb{Z}$ is of infinite order as the identity element $e=0$ and for any $n\in\mathbb{N}$, $$1^n=\underbrace{1+1+1+...+1}_{\textit{n times}}\neq 0.$$

So, $1$ is of infinite order and $1\in\mathbb{Z}$. Now, in the above question how is the above thing possible at all... I mean the element in the group may be of infinite order and then we may never have such an $n\in \mathbb{N}$. How is then the above thing true?

I am not quite getting it.

  • 1
    $\begingroup$ It’s only true if $G$ is a finite group. $\endgroup$ Oct 9, 2022 at 5:02
  • $\begingroup$ @Shinrin-Yoku Thank you! So it means that if $G$ is a finite group then all the elements in $G$ have a finite order , right?....is this a lemma (or a theorem )?... $\endgroup$
    – Arthur
    Oct 9, 2022 at 5:05
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    $\begingroup$ Yes that’s correct, it’s a lemma. $\endgroup$ Oct 9, 2022 at 5:18
  • $\begingroup$ You can have an element of an infinite group of infinite order but you can't have a member of a finite group have infinite order. There are infinitely many natural numbers $m$ but only a finite number of elements $a^m$ may be equal to. That means there must be an $m<n$ where $a^m = a^n$. Apply $a^{-1}$ to each side $m$ times to get $a^{m-n} = e$. That means $a$ has a finite order that is at most $m-n$. $\endgroup$
    – fleablood
    Oct 9, 2022 at 7:11

2 Answers 2


Any element $g$ of a finite group generates a cyclic subgroup, $\langle g\rangle $, which must also be finite, of order $n=\lvert g\rvert $.

By Lagrange's theorem if $N=\lvert G\rvert $, then $n\mid N$. Hence $N$ does the trick.

The smallest such number is called the exponent of the group.

  • 1
    $\begingroup$ Btw you don't need Lagrange to show that some such N exists, e.g. the product of orders of all elements with do as well. $\endgroup$
    – lisyarus
    Oct 9, 2022 at 7:43

Alternative approach:

The problem, as stated in the title, can be solved with virtually no knowledge of Group theory, using only the fact that the group has a finite number of elements.

Take any element $a$ in the group.

Consider the infinite sequence:

$a^1, a^2, a^3, \cdots.$

Since the group is finite, there must exist two positive integers $i$ and $j$ such that $i < j$ and $a^i = a^j.$

This implies that $a^{(j-i)} = e.$

Further, for any element $(a)$ in the group, if $k$ is a positive integer such that $a^k = e,$ then any multiple of $k$, of form $(r \times k) ~: ~r \in \Bbb{Z^+}$ will be such that $a^{(r \times k)}$ must equal $e^r = e.$

So, to each element $a$ in the finite group, there exists at least one positive integer $N(a)$ such that $a^{[N(a)]}$ is equal to $(e)$.

Further, for each such element $(a)$, any multiple $[r \times N(a)]$ will also be such that $a^{[r \times N(a)]} = e.$

So, given that the complete list of distinct elements in the finite group is represented by the list
$a_1, a_2, \cdots, a_s$ in the finite group $G$, you can form the sequence of numbers

$$N(a_1), N(a_2), \cdots, N(a_s). \tag1 $$

By all of the discussion in this answer, simply set $N$ equal to the $~\color{red}{\text{least common multiple}}~$ (or any common multiple) of the positive integers referred to in (1) above.

One explicit example would be

$$N = \prod_{i=1}^s N(a_i).$$


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