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Task is:

Let $G$ be an abelian group and let $H$ and $K$ be finite cyclic subgroups with $|H| = r$ and $|K| = s$.
Show that if $r$ and $s$ are relatively prime then $G$ contains a cyclic subgroup of order $rs$.

I'm thinking about how to solve this.

My intuition is that because the orders are coprime and the groups are cyclic, the generators must be $r$ and $s$, by Lagrange's theorem. That would mean that the elements of the groups are distinct and since the group is abelian, we can simply count the pairs, giving us a group of $rs$ order.

Is this correct?

Aryaman Maithani
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user569685
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  • Instead of saying generators must be $r$ and $s$, I think you wanted to say the order of the generators of the respective subgroups are $r$ and $s$. Your thought process is fine, but you have to prove things that you are claiming, for example why would the elements necessarily be distinct in $H$ and $K$. – Anurag A Jun 17 '20 at 17:59
  • Note that $H\cap K$ must be trivial. – Maryam Jun 17 '20 at 18:03

2 Answers2

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Hint: How many any elements does $\{ hk : h \in H, k \in K \}$ have? Note that $hk=h'k'$ iff $h^{-1}h'= k(k')^{-1}$ and $h^{-1}h'= k(k')^{-1} \in H \cap K$.

lhf
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generators must be $r$ and $s$

The above statement does not make much sense since $r$ and $s$ are not the group elements. You mean to say "generators must be $x$ and $y$ where $x$ and $y$ are the elements of order $r$ and $s$."

Your idea is correct. You would, however, need to prove that $\langle x, y\rangle$ does indeed have order $rs$. To do this, you would need to show that if $$x^a = y^b$$ for $0 \le a < r$ and $0 \le b < s$, then it is forced that $a = 0 = b$.

This will then tell you that the set $\{x^ay^b \mid 0 \le a < r,\;0 \le b < s\}$ has cardinality no less than $rs$. (Why?)

Alternately, you could try to show that the element $xy$ itself has order $rs$. (The proof will be almost identical to the above one.) Then, you can simply consider the subgroup $\langle xy\rangle$.

EDIT: Details on how to prove that $a = 0 = b$ claim.
We can note that $x^a \in H$ and $y^b \in K$. Since the two elements are assumed to be equal, we see that $x^a \in H \cap K$.

However, note that $H \cap K$ is a subgroup of $H$ and of $K$. (Why? Show that the intersection of subgroups is again a subgroup.)

Thus, the order of $H \cap K$ must divide that of $H$ and $K$. This gives us that $|H\cap K| = 1$ and thus, its only element is the identity. Thus, $x^a = 1 = y^b$. Conclude.

Aryaman Maithani
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  • Since the subgroups have order r and s, $x^a$ for some xEH must generate the entire subgroup H, since H has no non-trivial proper subgroups. Hence $x^a = y^b$ cannot be correct for anything other than $a=id=b$ as the orders of the subgroups are coprime. Hence the elements are distict and the number of pairs is rs. Am i correct? – user569685 Jun 17 '20 at 18:18
  • Why does $H$ have no nontrivial proper subgroups? Consider $G = \Bbb Z/8 \Bbb Z$, $x = \bar{2}$. Then $H \cong \Bbb Z/4 \Bbb Z$ and your claim is incorrect – Aryaman Maithani Jun 17 '20 at 18:25
  • I had assumed r and s were prime, which is not necessarily true. My bad. – user569685 Jun 17 '20 at 18:40
  • Since you know Lagrange's theorem, I have edited and added more details on how to prove the claim using that. – Aryaman Maithani Jun 17 '20 at 18:46
  • I understand the answer ! Thank you! – user569685 Jun 17 '20 at 19:02
  • Great! :) $\phantom{padding}$ – Aryaman Maithani Jun 17 '20 at 19:04