Is there a cyclic group $|\langle hk\rangle|= L$ or $|\langle hk\rangle|\ne L$?

Question

$G$ is Abelian, $H=\langle h\rangle$ and $K=\langle k\rangle$ are distinct cyclic subgroups of $G$, $|H|=r, |K|=s$, let $L$ be the least common multiple of $r$ and $s$. I am wondering if $\langle hk\rangle$ is a cyclic subgroup with order $|\langle hk\rangle|=L$, or if this idea is incorrect then what is the counter-example?

Here is my attempt:

Let $r=r_1\sigma, s=s_1\sigma$, where $\sigma\in \mathbb{N}$, and $r_1, s_1$ are coprimes, hence $L=r_1\sigma s_1$. Since $G$ is Abelian, it is easy to check $(hk)^L=e$. Now assume $|\langle hk\rangle|=q, q\le L$, so we have $(hk)^q=e\Rightarrow h^q=k^{-q}$, define $x=h^q=k^{-q}$, and let $u=|\langle x\rangle|$, which gives $x^u=e$.

Since $\langle x\rangle\le H\Rightarrow u|r\Rightarrow u|r_1\sigma$, similarly, since $\langle x\rangle\le K\Rightarrow u|s\Rightarrow u|s_1\sigma$. Therefore, we get $u|\sigma\Rightarrow \sigma=tu, x^\sigma=(x^u)^t=e$ and we have

$x^u=h^{qu}=e\Rightarrow r|qu\Rightarrow r_1\sigma|qu\Rightarrow r_1tu|qu\Rightarrow r_1t|q$

$x^u=k^{qu}=e\Rightarrow s|qu\Rightarrow s_1\sigma|qu\Rightarrow s_1tu| qu\Rightarrow s_1t|q$

Since $r_1, s_1$ are coprimes, we have $r_1ts_1|q$.

Let $q=\alpha\cdot r_1 t s_1$ and write $L=r_1\sigma s_1=u\cdot r_1t s_1$, so we get

$$\Rightarrow 1\le\alpha\le u$$

Since $q|L$, we have

$$\alpha|u$$

I stuck here, it didn't give any contradictions, can I get some hint to proceed? or a counter-example to show $|\langle hk\rangle|< L$?

Answer 1

Consider the group $\Bbb{Z}_2\times \Bbb{Z}_4$ and $H=\langle (0,1)\rangle $ and $K=\langle (1,1)\rangle $

$|H|=|K|=4$

$|\langle (0,1)+(1,1)\rangle |=|\langle (1,2)\rangle|=2 $

Answer 2

The order can be smaller.

Consider the following subgroups of $\Bbb{Z}_{30}$. The element $h=\overline{4}$ generates the unique subgroup of order $15$. The element $k=\overline{21}$ generates the unique subgroup of order $10$. The element $h+k=\overline{25}$ generates a subgroup of order $6$ only.

The problem is that in the projection to the quotient group $\Bbb{Z}_5\simeq \Bbb{Z}_{30}/\langle\overline{5}\rangle$ the images of $h$ and $k$ become each others inverses. If we pick generators more carefully this won't happen. For example $\overline{2}$ generates the same subgroup as $\overline{4}$, and $\overline{2}+\overline{21}=\overline{23}$ does generate a subgroup of order $\mathrm{lcm}(15,10)=30$.