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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.

tryst with freedom
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1 Answers1

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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.

Shaun
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  • Actually now I think of it. You need to show that intersection of subgroups is a subgroup of the original subgroups intersected for applying theorem in linked. I only linked to intersection being subgroup of total group. It's a trivial thing to prove but I can't find the link... – tryst with freedom Mar 16 '22 at 01:36
  • Use the one-step subgroup test, @Buraian. – Shaun Mar 16 '22 at 01:38
  • Could you please remark on what the order of generator should be? – tryst with freedom Mar 16 '22 at 01:40
  • That's a separate question, @Buraian. If you cannot find it here, I suggest you ask it in a new post. – Shaun Mar 16 '22 at 01:42
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    https://math.stackexchange.com/questions/3395659/let-h-and-k-be-subgroups-of-a-finite-cyclic-group-g-prove-h-cap-k?rq=1 – tryst with freedom Mar 16 '22 at 02:43