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

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.

• It’s only true if $G$ is a finite group. Oct 9, 2022 at 5:02
• @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 )?... Oct 9, 2022 at 5:05
• Yes that’s correct, it’s a lemma. Oct 9, 2022 at 5:18
• 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 $m-n$. Oct 9, 2022 at 7:11

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.

• 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. 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

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