# Prove that intersection of two non-trivial subgroups of Z is again non-trivial

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.

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 $$H_1\cap H_2$$.