Number of cyclic subgroups in an Abelian group $G$

Question

I tried to work out which cyclic subgroups an Abelian group $G$ contains if it contains an element of order $|a|$ and an element of order $|b|$.

I think the answer is that $G$ contains a cyclic subgroup for every divisor of $\text{lcm}(|a|,|b|)$. My resoning is that $G$ has to contain all the cyclic subgroups for divisors of $|a|$ and $|b|$. In addition to that it can contain subgroups generated by elements of the form $a^n b^m$. But all of these cases correspond to divisors of $\text{lcm}(|a|,|b|)$.

Please can you tell me if this is correct?

Does this answer your question? Let $G$ be abelian, $H$ and $K$ subgroups of orders $n$, $m$. Then G has subgroup of order $\operatorname{lcm}(n,m)$. — Clemens Bartholdy, Mar 10 '22 at 22:42

score 0 · Answer 1 · answered Nov 24 '15 at 08:26

0

You have just shown that some cyclic subgroups will be of order dividing $lcm(|a|,|b|)$. What you need to show is something stronger :

there exists an element of order $lcm(|a|,|b|)$

which is actually not so easy (but it is true).

answered Nov 24 '15 at 08:26

Clément Guérin

11,721

Number of cyclic subgroups in an Abelian group $G$

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