Abstract Nonsense

Crushing one theorem at a time

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

Point of post: In this post we give a brief overview of cyclic groups. We include the Fundamental Theorem of Finite Cyclic Groups (every finite cyclic groups–namely that every cyclic group has precisely one subgroup of each divisor of the order of the cyclic group). We also include information about the orders of elements in arbitrary groups including the result that if |g|=n then \displaystyle |g^m|=\frac{n}{(m,n)}.


Cyclic groups are the simplest possible groups imaginable. They are the analogue of one-dimensional vector spaces. Thus, it makes sense to start studying them. We shall even see that they are embedded into every finite group and moreover that they form a sort of ‘building block’ for finite abelian groups in general.

Cyclic Groups

If you’ll recall from our last post we crudely laid out what it meant for a subgroup to be generated by a set. A cyclic group is nothing more than such a group but with only one generator. Namely a group G is cyclic if there exists some g\in G for which

\left\{g^z:z\in\mathbb{Z}\right\}=\left\langle g\right\rangle=G

Clearly we may then define a cyclic subgroup H of a group G to be a subgroup which is cyclic. If H=\langle h\rangle we say that H is the cyclic subgroup of G generated by h.

Our first result is a fundamental one:

Theorem: Let G be a cyclic group with G=\langle g\rangle. Then, if |G|=n<\infty then G=\{e,g,\cdots,g^{n-1}\}. If |G|=\infty then |g|=\infty and the map f:\mathbb{Z}\to G:z\mapsto g^z is injective.

Proof: First assume that G is finite and of order n. We note first for distinct i,j\in[n-1]\cup\{0\} that g^i\ne g^j. Indeed, if g^i=g^j with 0\leqslant i<j\leqslant n-1 then g^{j-i}=e and 0\leqslant j-i\leqslant n-1 but this contradicts that |g|=n. Thus, it follows that e,g,\cdots,g^{n-1} are all distinct. But, this implies that \{e,g,\cdots,g^{n-1}\}\subseteq G and \#\left(\{e,g,\cdots,g^{n-1}\}\right)=\#(G) and so \{e,g,\cdots,g^{n-1}\}=G as required.

Suppose next that |G|=\infty. We note then that |g|=\infty. Indeed, if g^n=e for some n\in\mathbb{N} we see then that for any z\in\mathbb{Z}; g^z=g^{|z|\text{ mod }n} and so |G|\leqslant n contradictory to our assumption. Thus, |g|=\infty. Moreover we see that the map f:\mathbb{Z}\to G:z\mapsto g^z is injective, for to suppose not would imply there exists distinct z,z'\in\mathbb{Z} for which g^z=g^{z'} and so g^{|z'-z|}=e which contradicts that |g|=\infty. The conclusion follows. \blacksquare


Our next theorem is a general theorem about orders in general groups

Theorem: Let G be a group and g\in G. If m,n\in\mathbb{Z} are such that g^m=g^n=e then g^{(m,n)}=e where (m,n) is the greatest common divisor. In particular, if g^n=e then |g| \mid m.

Proof: By Bézout’s identity there exists x,y\in\mathbb{Z} such that mx+ny=(m,n). Thus,



From this we see that if g^n=e then g^{(|g|,n)}=e but 0\leqslant (|g|,n)\leqslant |g| and by the definition of |g| this implies that |g|=(|g|,n). But, this is only possible if |g| \mid n. The conclusion follows. \blacksquare



We now wish to, in a general sense once again, classify how the order of a group element relates to the order of a power. So, for example if |g|=20 what is \left|g^{15}\right|? It turns out that we can figure this out entirely. In particular:

Theorem: Let G be a group and g\in G. If z\in\mathbb{Z}-\{0\} then

\displaystyle \begin{aligned}&\mathbf{(1)}\quad \textit{ If }|g|=\infty\textit{ then }\left|g^z\right|=\infty\\ &\mathbf{(2)}\quad \textit{If }|g|=n<\infty\textit{ then }\left|g^z\right|=\frac{n}{(n,z)}\end{aligned}

Proof: For part \mathbf{(1)} we must merely note that if \left|g^z\right|=n<\infty then by definition \left(g^z\right)^n=g^{nz}=e and so evidently g^{|nz|}=e. But, this contradicts that |g|=\infty from where it follows that \left|g^z\right|=\infty.

For \mathbf{(2)} we first note that


so that by our previous problem we have that \displaystyle \left|g\right| \mid \frac{n}{(n,z)}. Conversely, we know there exists integers a,b such that n=a(n,z) and z=b(n,z) and by construction (a,b)=1. We note then that


and thus by our previous lemma again we know that n\mid z|g^z| or, with our relabeling a(n,z)\mid b(n,z)|g^z|. It follows that a\mid b|g^z| and since (a,b)=1 we may conclude that a\mid |g^z|; but \displaystyle a=\frac{n}{(n,z)}. Thus it follows that \displaystyle |g^z|\mid \frac{n}{(n,z)} and \displaystyle \frac{n}{(n,z)}\mid |g^z| and since both \displaystyle \frac{n}{(n,z)} and |g^z| are positive integers this implies that \displaystyle |g^z|=\frac{n}{(n,z)} as required. \blacksquare



A question then arises as to which elements of a cyclic group generate the group? For example it’s easy to see that \mathbb{Z}_{20} is cyclic (this is the usual group where one adds residue classes). That said, we see that \mathbb{Z}_{20}=\left\langle \overline{3}\right\rangle and \mathbb{Z}_{20}=\left\langle 7\right\rangle. We can, in fact, decide precisely which elements of a cyclic group generate the full group.


Theorem: Let G=\langle g\rangle. Then:

\displaystyle \begin{aligned}&\mathbf{(1)}\quad \textit{If }|G|=\infty\textit{ then the generators of }G\textit{ are exactly }g\textit{ and }g^{-1}\\ &\mathbf{(2)}\quad \textit{If }|G|=n<\infty\textit{ then the set of generators of }G\textit{ are precisely }g^z\textit{ with }(n,z)=1\end{aligned}

Proof: For \mathbf{(1)} suppose that G=\left\langle g^z\right\rangle with |z|>1. Then, by assumption there exists some n\in\mathbb{Z} such that g^{zn}=g and so g^{zn-1}=e or, g^{|zn-1|}=e. Note though that |zn-1|>0 and thus |G|=|g|\leqslant |nz-1| which is a contradiction. Thus, the only possible generators of G are g and g^{-1}. Clearly though these both work (the second works since given any g'\in G we have there exists some z\in\mathbb{Z} such that g^z=g', then \left(g^{-1}\right)^{-z}=g').


Next, suppose the conditions of \mathbf{(2)}. Note first that it is, in fact, necessary for (n,z)=1 to have that \left\langle g^z\right\rangle =G . Indeed, note that \left|\left\langle g^z\right\rangle\right|=\left|g^z\right|, but by our previous theorem \displaystyle \left|g^z\right|=\frac{n}{(n,z)}. Thus, if n=\left|\left\langle g^z\right\rangle\right| then we must have that (n,z)=1. Conversely, note if (n,z)=1 then \left\langle g^z\right\rangle\leqslant G and

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

from where it evidently follows that \left\langle g^z\right\rangle=G. \blacksquare



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.


December 27, 2010 - Posted by | Algebra, Group Theory, Uncategorized | , , ,


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