G is finite group. Need to proof that exists natural k that $g^k = e$

Question

How do I prove that in a finite group G, for each element in G there is natural power (say $k$) which depends on g,such that $g^k=e$ ? I need to show the existence and the dependence on which $g$ I choose.

I tried write it that way, but I don't have any direction in the proof:

$$G\:=\:\left|n\right|\::\:G=\left\{e,\:g,\:g^2,\:...\:,\:g^{n-1}\right\}$$

Can anybody give me any direction of thinking ?

$G$ being finite doesn't necessarily mean that $G$ is cyclic. — Silvia Ghinassi, Aug 14 '15 at 14:36
Oh, I didnt mean it is. I just say here, that if it is a group, so it is closed, so I can write all the members of the group as pows of g, because g^2 is also in the group because of closure, and alo g^3 and so on until the order of the group minus one (because e is there eather). — Ilya.K., Aug 14 '15 at 14:39
If $k=0$, there is nothing to prove. Let $k\ne 0$. Set ${g, g^2, \ldots}$ must be finite. So...? — Michael Galuza, Aug 14 '15 at 14:42
@user183297 By writing $G=\left{e,:g,:g^2,:...:,:g^{n-1}\right}$ you are saying $G$ is cyclic. You can say that that is a subgroup of $G$. — Silvia Ghinassi, Aug 14 '15 at 15:55

score 2 · Answer 1 · answered Aug 14 '15 at 16:09

Notice that when you write $e,a,a^2,a^3,...$ you will encounter repetition at some point.

Why? Well, suppose not. Then all the elements in the infinite set $\{e,a,a^2,a^3,...\}$ are distinct. But all these elements are in $G$, as $G$ is a group! So a subset of $G$ being infinite forces $G$ to be infinite as well. Contradiction!

So there exist $m<n\in\mathbb N$ such that $a^m=a^n$ implying $a^{n-m}=e$. Choose the minimum $m$ and $n$ for which this is satisfied. Then precisely $n-m$ is the order of $a$. And of course, $n-m$ is finite!!

And to conclude, note that for different elements, the $m$ and $n$ may not necessarily be equal. Consider the group $\{0,1,2,3\}$ under $+$ modulo $4$. Note that $o(2)=2$ while $o(1)=4$, showing that order is element-dependent.

score 1 · Answer 2 · answered Aug 14 '15 at 15:05

Since $G$ is finite, $\;\langle\, g\,\rangle$ (the subgroup generated by $g$) is finite too. This implies the map: $\; \mathbf N\to G$, $n\mapsto g^n$ is not injective, i.e. there exists $r< s\in \mathbf N$ such that $g^r=g^s$, hence $g^{s-r}=e$.

Note: in addition , Lagrange's theorem say the order of $g$ ($\lvert\langle\, g\,\rangle\rvert$, i.e. the smallest $k$ such that $g^k=e$) is a divisor of $\lvert G\rvert$.

goblin GONE · Answer 3 · 2015-08-27T19:09:16.220

Just use:

Theorem. Let $X$ denote a set. Then $X$ is finite iff for all functions $f : X \leftarrow \mathbb{N},$ there exist $n,m \in \mathbb{N}$ such that $f(n)=f(m)$ and $n > m$.

So the proof goes like so:

Proof. Suppose $G$ is a finite group and that $g \in G$ is a fixed element.

Define a function $f : G \leftarrow \mathbb{N}$ by $g^n = f(n)$. Using the aforementioned theorem, find $n,m \in \mathbb{N}$ such that $f(n)=f(m)$ and $n>m$. Then $g^n = g^m$. So $g^{n-m}=1$. Now let $k \in \mathbb{Z}$ be defined by $k=n-m$. Then $g^k=1$. But since $n>m$, hence $n-m>0$, or in other words $k>0$.

So: there exists a positive integer $k$ such that $g^k=1$.

G is finite group. Need to proof that exists natural k that $g^k = e$

3 Answers3