Consider two non-trivial subgroups $H_1,H_2$ of $\mathbb Z$. $H_1=m\mathbb Z, H_2=n\mathbb Z$ . We know that the intersection $H_1 \bigcap H_2$ is a subgroup, and therefore can be written as $t\mathbb Z$ for some non-negative integer $t$. Prove that $t = lcm(m,n)$. This proves that the intersection of two non-trivial subgroups of $\mathbb Z$ is again non-trivial.
This is what I tried so far:
I assumed that the Lagrange theorem applied for the number of subgroups as well. Since the intersection of the two subgroups must include both of the subgroups, the number of subgroups $m,n$ must divide $t$. In order to fulfill this condition, $t$ must be $lcm(m,n)$. I’m just in doubt because I don’t think this is how I’m supposed to solve this question.
Best Answer
You can simply show the equality of sets. Let $H_1=m\Bbb Z$ and $H_2=n\Bbb Z$. Then $H_1=\{km:k\in\Bbb Z\}$ and $H_2=\{kn:k\in\Bbb Z\}$. We claim that $H_1\cap H_2=\{k\ell:k\in\Bbb Z\}$, where $\ell=\operatorname{lcm}(m,n)$. Clearly $k\ell$ is a multiple of $m$ and $n$ for any $k\in\Bbb Z$, so $\{k\ell:k\in\Bbb Z\}\subseteq H_1\cap H_2$. Conversely, if $x\in H_1\cap H_2$, then $x$ is divisible by $m$ and $n$, which implies $x$ is divisible by $\ell=\operatorname{lcm}(m,n)$. This shows $H_1\cap H_2\subseteq \{k\ell:k\in\Bbb Z\}$.
If you are only trying to prove that the intersection is nontrivial, you can just note that $mn$ is a nonzero element of $H_1\cap H_2$.