# $a^n = e$ for $a \in G$ where $G$ is a group with finitely many elements.

Let $G$ be a group with a finite number of elements. Show that for any $a \in G$, there exists an $n \in \mathbb{Z}^+$ such that $a^n = e$, where $e$ is the identity and $a^n = a * a * a \space ... *\space a$ where $*$ is a binary operation.

Being a new student to algebra, I find this question very counterintuitive. Let, for example, $G$ be the group of the positive rational numbers over multiplication. Thus we have the identity $e = 1$ and inverse $\frac{1}{a}$ for any number $a \in G$. I claim that there is no positive integer $n$ where $2^n = 1$. How does one make sense of all this?

• "Group with a finite number of elements" is critical here. – user61527 Jan 19 '14 at 10:56
• How to the positive rationals "up to 1000" form a group under multiplication? What is $100\cdot 100$ in this "group"? – Hagen von Eitzen Jan 19 '14 at 10:58
• @HagenvonEitzen, yes, I missed that, still not used to this kind of thinking. – Andrew Thompson Jan 19 '14 at 10:59

If $a=e$, then the result is trivial. For $a\neq e$: Since the group is finite then $a^n$ can not be distinct elements for all $n$. Therefore, there exist $m_1$ and $m_2$ $\in \mathbb{Z}^+$ where $m_1\neq m_2$ such that $a^{m_1}=a^{m_2}$. Without loss of generality suppose $m_1>m_2$, so that $m_1-m_2 \in \mathbb{Z}^+$. Thus, $a^{m_1-m_2}=e$ which completes the proof. As an example in $\mathbb{Z}_4$: $0$ is the identity. $1+1+1+1=2+2=3+3+3+3=0$ in $\mathbb{Z_4}$.

• Dumbing this down: $a^n$ can not be distinct elements for all $n$ in the finite group, seeing as there is only a finite number of elements in the group, and at some point one would get the same element. I understand the rest. A concrete example would be highly appreciated. – Andrew Thompson Jan 19 '14 at 11:27
• I gave an example. You can try another one for yourself like the $S_n$ group or any other finite group that you know. – Spock Jan 19 '14 at 11:33

Hint: How many distinct elements can the sequence $$a, a^2, a^3, a^4, \ldots$$ have?

• is it the case that there is always an element $a^N = e$ where $N = |G|$? – Salihcyilmaz Oct 30 '18 at 17:12

Let $\exists a \in G$ such that $a^k \ne e$ $\forall k \in \mathbb{N}$

then we can write $k = k_1 - k_2$ where $k_1, k_2 \in \mathbb{Z}$

$\implies a^{k_1 - k_2} \ne e \implies a^{k_1} \ne a^{k_2} \forall k_1, k_2$ $\in \mathbb{Z}$

So all powers of $a$ (i.e. $a^1, a^2, a^3, a^4, \ldots$) are distinct.

$\implies G$ is an infinite group

Hence $\exists N \in \mathbb{N}$ satisfying $a^N = e$

• I would really appreciate if whoever downvoted the answer to please give a reason for that: Because if there is something wrong in the answer, i could improve upon that. – gxyd Oct 27 '15 at 6:24
• Perhaps it seems like my answer is similar to the one answered above. Is that the reason? – gxyd Oct 27 '15 at 10:04