Prove that statements are equivalent Let $m,n \in \mathbb{N}$. Prove that following statements are equivalent:
(1) $m, n$ - mutually prime numbers;
(2) for every group $G$, any subgroup $A \subseteq G$ of order $m$ and any subgroup $B \subseteq G$ of order $n$ the following is true: $A \cap B =\{e\}$.

It is important to notice that $G$ is not necessarily a finite group.
I have following ideas on my mind:
$1 \rightarrow 2:$ if $\gcd(m, n)=1$ then as of Lagrange's theorem given that $Q=A\cap B$, $|Q|$ should be a divisor of $|A|$ and $|B|$, which is true only for $|Q|=1$, given that and the fact that every subgroup should contain unitary element we have: $A\cap B=Q=\{e\}$.
$2 \rightarrow 1:$ let's prove that by contradiction: if $\gcd(m, n)\neq 1$, then lets take group  $G=(\mathbb{Z}_{m\cdot n}, +)$, $A=(\mathbb{Z}_{m}, +)$ and $B=(\mathbb{Z}_{n}, +)$. Then $A\cap B=(\mathbb{Z}_{gcd(m, n)}, +)$, so if $\gcd(m, n)\neq 1$, then $A \cap B \neq \{e\}$, what lead us to the contradiction itself. We may say that, as in the statements there are words: 'every' and 'any'.
Are my ideas correct? I'm asking as my teacher said, that mentioned solution wouldn't work for infinite G and some other specific cases. Can you help me with fixing it? And may anyone explain me, why $A\cap B=(\mathbb{Z}_{\textbf{gcd(m, n)}}, +)$, we were given that with no proof, however I don't want to use it with no ideas why does it work.
Your "$1\to 2$" implication is fine.

The "$2\to 1$" implication is not precise enough for my taste. First, what exactly is $\mathbb{Z}_k$? For the sake of this post I will assume that $\mathbb{Z}_k=\{0,1,\ldots,k-1\}$ with "modulo $k$" addition.

Now you consider $\mathbb{Z}_{mn}$, that is fine. But neither $\mathbb{Z}_m$ nor $\mathbb{Z}_n$ is a subgroup of $\mathbb{Z}_{mn}$. The $\mathbb{Z}_{mn}$ group contains cyclic subgroups of order $m$ and $n$, but these are only isomorphic to $\mathbb{Z}_m$ and $\mathbb{Z}_n$, not literally equal. For example $\mathbb{Z}_6=\{0,1,2,3,4,5\}$ and $\mathbb{Z}_2=\{0,1\}$ but $\{0,1\}$ is not even a subgroup of $\mathbb{Z}_6$. The cyclic subgroup of order $2$ in $\mathbb{Z}_6$ is actually $\langle 3\rangle=\{0,3\}$. It is indeed, isomorphic to $\mathbb{Z}_2$, but not literally equal.

Now take again $G=\mathbb{Z}_{mn}$. The subgroups you are looking for are $A=\langle n\rangle$ and $B=\langle m\rangle$, which are of order $m$ and $n$ respectively. So how to show that $A\cap B\neq \{0\}$? This follows from the fact that if $G$ is a cyclic group and $d$ is a divisor of $|G|$ then $G$ contains a unique cyclic subgroup $H\subseteq G$ of order $d$. See this: Each finite cyclic group of order $n$ contains unique subgroup of order $d$ where $d\mid n$. And so both $A$ and $B$ contain cyclic subgroups of order $\gcd(n,m)\neq 1$. By the uniqueness this is one and the same subgroup. And so it is a subgroup of $A\cap B$, which thus is nontrivial.
For $\ 2\rightarrow 1$, we again use Lagrange's theorem.  

$\ | A\cap B |$ divides $\ |A|$ and $\ |B|$ as the $\ A\cap B $ is subgroup of both $\ A$ and $\ B$. Now if $\ gcd(m,n)\ne 1$ then $\ |A\cap B| \ne 1\implies  A\cap B \ne \{e\}$. Contrapositively you can say if $\ A\cap B =\{e\}$ then $\ gcd(m,n)= 1$