# Cyclic Groups

A cyclic group $$G$$ is a group that can be generated by a single element $$a$$, so that every element in $$G$$ has the form $$a^i$$ for some integer $$i$$. We denote the cyclic group of order $$n$$ by $$\mathbb{Z}_n$$, since the additive group of $$\mathbb{Z}_n$$ is a cyclic group of order $$n$$.

Theorem: All subgroups of a cyclic group are cyclic. If $$G = \langle a\rangle$$ is cyclic, then for every divisor $$d$$ of $$|G|$$ there exists exactly one subgroup of order $$d$$ which may be generated by $$a^{|G|/d}$$.

Proof: Let $$|G| = d n$$. Then $$1, a^n, a^{2 n},..., a^{(d-1)n}$$ are distinct and form a cyclic subgroup $$\langle a^n \rangle$$ of order $$d$$. Conversely, let $$H = \{1,a_1,...,a_{d-1}$$ be a subgroup of $$G$$ for some $$d$$ dividing $$G$$. Then for all $$i$$, $$a_i = a^k$$ for some $$k$$, and since every element has order dividing $$|H|$$, $$a_i^d = a^{k d} = 1$$. Thus $$k d = |G|m = n d m$$ for some $$m$$, and we have $$a_i = a^{n m}$$ so each $$a_i$$ is in fact a power of $$a^n$$. From above this means it must be one of the $$d$$ subgroups already described.

Theorem: Every group of composite order has proper subgroups.

Proof: Let $$G$$ be a group of composite order, and let $$1\ne a\in G$$. Then if $$\langle a \rangle \ne G$$ we are done, otherwise the subgroup $$\langle a^d \rangle \ne G$$ for every divisor $$d$$ of $$|G|$$.

Ben Lynn blynn@cs.stanford.edu 💡