# Abstract Nonsense

## Review of Group Theory: Cyclic Groups and Cyclic Subgroups (Pt. II)

Point of post: This post is a continuation of this one.

Our last theorem we wish to prove is the most important of all. Indeed, it will completely classify all subgroups of  cyclic groups. Unsurprisingly this theorem is called the fundamental theorem of cyclic groups.

Theorem: Let $G=\left\langle g\right\rangle$. Then:

\displaystyle \begin{aligned}&\mathbf{(1)}\quad \textit{Every subgroup of a cyclic group is cyclic}\\ &\mathbf{(2})\quad \textit{If }|G|=\infty\textit{ then the distinct subgroups of }G\textit{ are precisely }\left\langle g^n\right\rangle,\text{ }n\in\mathbb{N}\cup\{0\}\\ &\mathbf{(3)}\quad \textit{If }|G|=n<\infty\textit{ then there is precisely one subgroup of order }d\textit{ for each }\text{ }d\mid n\textit{ and it's }\left\langle g^{\frac{n}{d}}\right\rangle\end{aligned}

Proof:

$\mathbf{(1)}$: To do this let $H\leqslant G$. If $H=\{e\}$ this is trivial, so assume not and let $\alpha=\min\left\{n\in\mathbb{N}:g^n\in H\right\}$. We claim that $H=\left\langle g^\alpha\right\rangle$. To do this note that for any $g^\beta\in H$ we have( by the division algorithm) that $\beta=z\alpha+r$ with $z\in\mathbb{Z}$ and $0\leqslant r<\alpha$. Note though that since $g^\alpha\in H$ and $H$ is a subgroup we have that $g^{-z\alpha}\in H$ and thus $g^{\beta}g^{-z\alpha}=g^r\in H$. But, if $r>0$ we would have that $r\in\left\{n\in\mathbb{N}:g^n\in H\right\}$ and $r<\alpha$, but this contradicts the minimality of $\alpha$. It follows that $\beta=z\alpha$. Thus, since $\beta$ was arbitrary it follows that $H\subseteq \left\{g^{z\alpha}:z\in\mathbb{Z}\right\}$ and the opposite containment follows from closure of $H$. Thus, $H=\left\{g^{z\alpha}:z\in\mathbb{Z}\right\}$ as required.

$\mathbf{(2)}:$ We first claim that if $n,m\in\mathbb{N}\cup\{0\}$ are distinct then $\left\langle g^n\right\rangle \ne\left\langle g^m\right\rangle$. But, this  is clear. If $n$ or $m$ is zero this is trivial, so assume that $n,m\in\mathbb{N}$. Then if $\left\langle g^m\right\rangle=\left\langle g^n\right\rangle$ then we must have that $g^m\in\left\langle g^n\right\rangle$ and so there exists $z\in\mathbb{Z}$ such that $g^{nz}=g^m$ or that $g^{nz-m}=e$ but this is true if and only if $nz-m=0$ and so it clearly follows that $n\mid m$. Applying the exact same argument is reverse shows that $m\mid n$ and by assumption this can only be true if $n=m$. But, by $\mathbf{(1)}$ and noticing that $\left\langle g^z\right\rangle=\left\langle g^{|z|}\right\rangle$ (clearly) we know that these are the only such subgroups from where the conclusion follows.

$\mathbf{3)}$: Let $d\mid n$. There is clearly at least one subgroup of order $d$ since $\left\langle g^{\frac{n}{d}}\right\rangle\leqslant G$ and

$\displaystyle \left|\left\langle g^{\frac{n}{d}}\right\rangle\right|=\frac{n}{\left(n,\frac{n}{d}\right)}=\frac{n}{\frac{n}{d}}=d$

To see that this is the only such subgroup suppose that $H\leqslant G$ and $\displaystyle H=\left\langle g^m\right\rangle$. Note then that

$\displaystyle \frac{n}{\frac{n}{d}}=d=\left|H\right|=\left|\left\langle g^m\right\rangle\right|=\left|g^m\right|=\frac{n}{(n,m)}$

comparing the two we find that $\displaystyle \frac{n}{d}=(n,m)$ and thus $\displaystyle \frac{n}{d}\mid m$. It clearly follows then that every multiple of $m$ is a multiple of $\displaystyle \frac{n}{d}$ and so

$\displaystyle \left\langle g^m\right\rangle\subseteq\left\langle g^{\frac{n}{d}}\right\rangle$

but since the two sets have the same cardinality it follows that $\left\langle g^m\right\rangle=\left\langle g^{\frac{n}{d}}\right\rangle$ as desired.

Thus, it remains to show that if $\{e\} (we clearly don’t need to consider the trivial subgroup, since the proof follows immediately) and $|H|=k$ then $k\mid n$. To do this note that by $\mathbf{(1)}$ we have that $H=\left\langle g^m\right\rangle$ where $m=\min\left\{n\in\mathbb{N}:g^n\in H\right\}$. Note though that

$\displaystyle k=|H|=\left|\left\langle g^m\right\rangle\right|=\left|g^m\right|$

But, from the division algorithm we know that there exists $q,r\in\mathbb{Z}$ with $0\leqslant r such that $n=qk+r$. We see then that

$e=g^{mn}=g^{qkr}g^{rm}=g^{rm}$

and since $k=\left|g^m\right|$ and $0\leqslant r it follows that $r=0$ so that $n=qk$ as desired. The conclusion follows $\blacksquare$

References:

1.  Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.