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If $G$ is a finite group of order (size) $n$ then, for any $g \in G$, the order (period) of $g$ is a divisor of $n$.

Proof: $g$ must have finite order since $G$ is finite. If the order (period) of $g$ is $m$ then the order (size) of the cyclic subgroup $\left<g\right>$ it generates is also $m$. Since $G$ has a subgroup of size $m$, Lagrange's Theorem tells us that $m$ is a divisor of $n$.

What does this proof mean by "$g$ must have finite order since $G$ is finite". Firstly, are we talking about the order (period) or order (size) and either way why does $G$ being finite mean $g$ has finite order?

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  • $\begingroup$ $g$ must have finite order since $G$ is finite means there must exist $n \in \mathbb N$ such that $g^n = e$ where $e$ is the identity of $G$. If $g$ did not have finite order, that would mean that there does not exist $k \in \mathbb N$ such that $g^k = e$, we write $\vert g \vert = \infty$ in that case. $\endgroup$ – Robert Cardona Dec 23 '13 at 20:20
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For your first question: The period of $g$.

For your second question: Suppose for a contradiction that g does not have a finite period. Then it will generate an infinite number of elements, resulting in an group with infinite number of elements which is a contradiction to the supposition that $G$ is finite.

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  • $\begingroup$ 1) Ok. Can a group have an order and can a groups element have a size? - 2) So are we supposing g is cyclic then? $\endgroup$ – user2850514 Dec 23 '13 at 20:34
  • $\begingroup$ I refer to the size of the group as the order of the group $G$ and the period/order of the element in the group $G$ as the smallest number $n$ such that $g^n=1$ where 1 is the identity. An element of the group will not have a 'size' as it is not a set and does not contain anything. $\endgroup$ – ireallydonknow Dec 23 '13 at 20:37
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    $\begingroup$ @user2850514 $g$ is not assumed cyclic, but the group $\langle g\rangle$ it generates clearly is cyclic (i.e. it is generated by a single element, namely $g$) $\endgroup$ – Hagen von Eitzen Dec 23 '13 at 20:38
  • $\begingroup$ Ok, but then surely we are talking about a subgroup of $G$, namely $\left< g \right>$ for $g\in G$, rather than the actual group $G$? $\endgroup$ – user2850514 Dec 23 '13 at 20:41
  • $\begingroup$ Yes, since the key idea in the proof is the group $\left< g \right>$ generated by the element $g$. $\endgroup$ – ireallydonknow Dec 23 '13 at 20:43
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If $G$ is finite, and $g \in G$, look at the sequence $g^i$ for integer $i \ge 1$, i.e. $g^i = \{ g, g^2, . . . , g^k, g^{k + 1}, . . . , \text{etc.} \}$ Since $G$ is finite, this sequence of elements of $G$ must repeat itself at some point; thus we must have $g^l = g^k$ for $l > k \ge 1$. If we then choose $k$ to be the least integer for which such repetition occurs for some $l$, and $l$ to be the first integer after $k$ for which it does occur, we must have $g^{l - k} = e$, the identity of $G$, and $g^m \ne e$ for any $m < l - k$. The order of $g$ is thus $l - k$. Thus $G$ finite implies the order of any $g \in G$ finite, and now the argument based on Lagrange's theorem shows that the order of $g$ must divide that of $G$.

Hope this helps. Happy Holidays,

and as always,

Fiat Lux!!!

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It depends on the definition of order of an element.

My preferred definition is

Let $G$ be a group and $g\in G$. The order of $g$ is the number of elements of $\langle g\rangle$, if finite; otherwise we say the order is infinite.

Given $g\in G$, we can define $\varphi_g\colon\mathbb{Z}\to G$ by $$ \varphi_g(n)=g^n \qquad(n\in\mathbb{Z}) $$ and $\varphi_g$ is easily seen to be a homomorphism; the image of $\varphi_g$ is exactly $\langle g\rangle$.

The homomorphism theorem then says that $\varphi_g$ induces an isomorphism between $\mathbb{Z}/\ker\varphi_g$ and $\langle g\rangle$. A known fact about subgroups of $\mathbb{Z}$ gives that $\ker\varphi_g=k\mathbb{Z}$, for a unique $k\ge0$. If $k=0$, $\varphi_g$ is injective and so $g$ has infinite order.

Otherwise $k$ is exactly the order of $g$. In this case, $g^k=\varphi_g(k)=1$ and, for $0<n<k$, $g^n\ne1$, because $n\notin k\mathbb{Z}=\ker\varphi_g$. Another consequence, in this case, is that if $g^n=1$, then $k\mid n$.

Of course, if $G$ is finite, also $\langle g\rangle$ is finite and so the order of $g$ divides $|G|$ by Lagrange's theorem.


Defining the order of $g$ as the least positive integer $k$, if it exists, such that $g^k=1$, forces proving again, but in a complicated way, that the subgroups of $\mathbb{Z}$ are of the form $n\mathbb{Z}$. I see no reason for not using the homomorphism theorem, which is one of the most basic tools in group theory.

However, as already said in other answers or comments, when $G$ is finite there are two distinct positive integers $m<n$ such that $g^m=g^n$, so $g^{n-m}=1$. Therefore the order of $g$ is finite, according to the other definition.

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Suppose $g$ is an element of the group $G$ with infinite order. What is the order of the group $G$ then? Can it be finite?

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