The order of a group versus its cyclic subgroup

Question

I read a statement in my textbook, which was taken as a premise, that for some element $g$ of a finite group $G$, that the order of the element $g$ is the same as the cyclic subgroup, $<g>$.

I think I might be confusing the order of an element with the order of a group. If I'm not mistaken, $<g>$ is equal to the powers of $g$. Since $G$ is finite, we can write these powers as $g^0, g_1, \ldots, g_m$ where $g$ has order $m$. (More commonly, I think we'd just write the identity element as $1$.) The reason it has order $m$, I believe, is because $g^m$ is equal to the identity, i.e., if we apply $g$ to itself $m$ times we get back $1$, so we need not continue to write higher, or lower, negative powers of $g$ because we'll simply be repeating ourselves.

Is the argument here, then, that $<g>$ has $m$ elements by definition, provided $m$ is the order of $g$? I think this question might be a bit trivial, but it would be great to get these concepts down.

Thanks in advance.

Answer 1

It's a question of how you define the order of an element.

Definition 1 The order of the element $g$ in the finite group $G$ is the least positive integer $m$ such that $g^m=1$.

Such a least integer exists, because there surely are $p$ and $q$ distinct positive integers such that $g^p=g^q$, by the pigeonhole principle. We can assume $p>q$, so that $g^{p-q}=1$. Then a positive integer $n$ with the property $g^n=1$ exists and therefore also the least one exists.

Now we can show that the elements $g^0,g^1,\dots,g^{m-1}$ are pairwise distinct (easy). Moreover if $n$ is any integer, we can write $n=mq+r$, with $0\le r<m$ and $g^n=g^r$ is one of the previously listed elements. Hence $\langle g\rangle=\{g^0=1,g^1=g,\dots,g^{m-1}\}$ has indeed $m$ elements.

Definition 2 The order of the element $g$ in the finite group $G$ is the cardinality of $\langle g\rangle=\{g^n:n\in\mathbb{Z}\}$.

Consider the homomorphism $\varphi_g\colon\mathbb{Z}\to G$ defined by $\varphi_g(n)=g^n$. The image of $\varphi_g$ is, by definition, $\langle g\rangle$. By the homomorphism theorem, $$ \langle g\rangle\cong \mathbb{Z}/\ker\varphi_g $$ Since $\ker\varphi_g$ is a subgroup of $\mathbb{Z}$, it is of the form $\ker\varphi_g=m\mathbb{Z}$ for a unique $m>0$ (it cannot be $m=0$, because otherwise $\mathbb{Z}/\ker\varphi_g$ would be infinite. Then $$ \langle g\rangle\cong \mathbb{Z}/\ker\varphi_g=\mathbb{Z}/m\mathbb{Z} $$ and so $m$ equals the order of $g$ (as defined here). Since $m$ is the least positive integer in $m\mathbb{Z}=\ker\varphi_g$, it is the least positive integer such that $g^m=1$.

Bonus for definition 2: the fact that the order of $g$ divides $|G|$ is a consequence of Lagrange’s theorem.

Answer 2

By definition the order of $<g>$ is the cardinality of this subgroup. This subgroup consists of all elements of the form $g^k$, for $k\in \mathbb{Z}$.

Now, because $g$ is of order $m$, we see that if $k=l \, \operatorname{mod}(m)$, then $g^k=g^l$. It follows that every element of the form $g^k$ equals an element of the form $g^l$ for some $l\in \{ 0,1,\ldots ,m-1 \}$.

Moreover, it is also true that for any two integers $l, l' \in \{ 0,1,\ldots ,m-1 \}$, we have $g^l=g^{l'}$ if and only if $l=l'$. Indeed, otherwise, assuming that $l< l'$ without loss of generality, we would have $g^{l'-l}=1$ with $0< l-l' < m$, which comes as a contradiction of the definition of the order of an element (as the least positive integer $m$ such that $g^m=1$).

Hence, we deduce that the cardinality (order) of $<g>$ really is $m$, the order of the element $g$.

NB: To justify it in a little more elaborate way, we can argue that we have a group morphism $\mathbb{Z}\rightarrow <g>$ sending an integer $k$ to $g^k$, which is surjective and has kernel $m\mathbb{Z}$. Hence, we have an isomorphism $<g>\simeq \mathbb{Z}/m\mathbb{Z}$, proving that this group has cardinality (order) $m$.