Subgroups of Cyclic Groups

Theorem 1: Every subgroup of a cyclic group is cyclic.

Proof: Let G = \left\{ a \right\} be a cyclic group generated by a. Let H be a subgroup of G. Now every element of G, hence also of H, has the form {a^s}, with s being an integer. Let m be the smallest possible integer such that {a^m} \in H. We claim that H = \left\{ {{a^m}} \right\}. For this it is sufficient to show that {a^s} \in H, then s = mh for then {a^s} = {\left( {{a^m}} \right)^h}. Now, if m does not divide s, then there exist integers q and r such that

s = mq + r,\,\,\,0 \leqslant r < m


{a^s} = {a^{mq + r}} = {a^{mq}} \cdot {a^r}


{a^r} = {a^s} \cdot {\left( {{a^{mq}}} \right)^{ - 1}}

Since {a^m} \in H, it follows that {a^{mq}} \in H and hence its inverse {\left( {{a^{mq}}} \right)^{ - 1}} \in H.

But {a^s} \in H by supposition. Then from the above result it follows that {a^r} \in H, contrary to the choice of m since m was assumed to be the least positive integer such that {a^m} \in H. Therefore r = 0 and so s = mq. But then {a^s} = {\left( {{a^m}} \right)^q}

Thus every element {a^s} of H is of the form {\left( {{a^m}} \right)^q}. Hence H = \left\{ {{a^m}} \right\}.

Theorem 2: Every subgroup of an infinite cyclic group is infinite.

Proof: Let G = \left\{ a \right\} be an infinite cyclic group. Let H be a subgroup of G. Then by the preceding theorem, H = \left\{ {{a^m}} \right\} where m is the least positive integer such that {a^m} \in H. Now suppose, if possible, that H is finite.

This implies that {\left( {{a^m}} \right)^s} = e for some s > 0.

It follows that a is of finite order and this in turn implies that G is finite, contrary to the hypothesis. Hence H must be an infinite cyclic subgroup of G.