The intersection of cyclic groups is cyclic, source of claim
My attempt at proof:
Definition of a cyclic subgroup:
A cyclic group is a group which is equal to one of its cyclic subgroups: $G = \langle g\rangle $ for some element $g$, called a generator.
If one proves for the statement that "intersection of two cyclic subgroup is cyclic", then the statement naturally extends through induction for countable number of cyclic subgroup.(*)
We know that intersection of subgroup is a subgroup. What is left is to prove that this subgroup can be generated out of some element.
Let $A$ and $B$ be subgroups of a group $G$ generated by elements $\langle a\rangle$ and $\langle b \rangle $ respectively. Intuition tells me that this element which generates the cyclic subgroup must have order $\gcd( |a|,|b|)$ , but how do I show an element with such an order exists so I can generate the whole intersection subgroup out of it?
*: Not sure how to show it's true for uncountable case.
Best Answer
All subgroups of any cyclic group are cyclic. The intersection $H$ of cyclic groups $(G_i)_{i\in I}$ for any index set $I$ is a subgroup of each $G_i$. Hence $H$ is cyclic.