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 💡