Let $G$ be an abelian group and let $H$ and $K$ be finite cyclic subgroups with $|H| = r$ and $|K| = s$. Show that $G$ contains a cyclic subgroup of order $lcm(r, s)$.

I have seen Let G be abelian, H and K subgroups of orders n, m. Then G has subgroup of order lcm(n,m). However, since it is required to show that $G$ contains a cyclic subgroup of order $lcm(r, s)$, I still have no clue.

This is an exercise problem on Fraleigh A First Course in Abstract Algebra (p.68, Exercise 56), after the section in which cyclic groups are introduced.



  1. If $H$ and $K$ are cyclic and their orders are coprime, then $H\times K$ is cyclic.
  2. $\dfrac{lcm(r,s)}r$ is coprime to r $\dfrac{lcm(r,s)}{s}$ and is a divisor of $s$.

Edit: Thanks to @lhf's comment, (2) is not necessary. We only need to note that if $$|H|=r=p_1^{a_1}\cdots p_k^{a_k}\text{ and } |K|=s=p_1^{b_1}\cdots p_k^{b_k}$$ then $H$ is the product of the corresponding cyclic groups of orders $p_1^{a_1},\dots, p_k^{a_k}$. Same goes for $K$.

On the other hand, $$lcm(r,s)=p_1^{\max(a_1,b_1)}\cdots p_k^{\max(a_k,b_k)}$$ and we can pick out the corresponding components from $H$ or $K$ depending on whether $a_k>b_k$ or not.

  • 2
    $\begingroup$ Counterexample to #2: $r=2$, $s=4$. $\endgroup$ – lhf Sep 22 '15 at 10:20
  • $\begingroup$ #2 still not right... $r=12=2^2\cdot 3$, $s=18=2\cdot 3^2$. $\endgroup$ – lhf Sep 22 '15 at 11:04
  • $\begingroup$ @lhf I think it's correct. But see edit. $\endgroup$ – Quang Hoang Sep 22 '15 at 12:46

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