Let $H, K$ be two subgroups of $G$. If $|H| = 12$ and $|K|=17$ then $H \cap K = \{e\}$ My reasoning:
Since $|K| = 17$ and $17$ is prime, then any subgroup of $K$ is cyclic. Also, the order of any subgroup must divide the order of the group. But since the subgroups of $K$ must have an order that divides $|G|$, the only possible order is $1$. Then, the group $K$ is generated by any element (except the identity). Since the order of the generator of a cyclic group is the same as the order of the group, the order of any element in $K$ except $e$ is $17$.
Now, in the other group, the order of each element may vary, since the order of a element $a \in H$ can't be higher than the order of the group, we have no elements with order $17$, then the only thing in common between $K$ ans $H$ is the identity.
Is my reasoning correct? Is there an easier way to solve this? I'm kinda assuming the last part 

"the order of a element $a \in H$ can't be higher than the order of
  the group"

Is there any way to justify this?
 A: It is a much more general phenomena:
Claim: If $|H|=n$ and $|K|=m$ and $\gcd(m,n)=1$ then $H\cap K=\{e\}$.
Proof: Let $g\in H\cap K$. Then the order of $g$, $|g|$ divides both $n$ and $m$ (why?) and hence $|g|$ divides $1$
So, $g=e$.
A: Here is another proof of the general result:

If $|H|=n$ and $|K|=m$ and $\gcd(m,n)=1$ then $H\cap K=\{e\}$. 

Since $\gcd(m,n)=1$, we can write $1=mu+nv$ for some integers $u$ and $v$.
Let $g\in H\cap K$. Then $g=g^1=g^{mu+nv}=g^{mu}g^{nv}=(g^m)^u(g^n)^v=ee=e$, because $g^m=e=g^n$.
A: Yes, it is great. I would say $|H\cap K|$ divides $12$ and $17$ by Lagrange and the fact  $H\cap K\subseteq H$ and $H\cap K\subseteq K$  (for this you need to know the intersection of two subgrous of $G$ is a subgroup of $G$). From here $|H\cap K|$ divides the greatest common factor of $12$ and $17$ which is $1$. Since every subgroup has the identity the subgroup is $\{e\}$
A: I’m only going to answer the second question since the first has been answered multiple times. In general, let $g\in G$, then the group generated by $g$, denoted $\langle g\rangle$ is a subgroup of $G$, hence its order divides the order of $G$. Also,
$o(\langle g\rangle)=o(g)$, hence the order of the element must be less or equal to the order of the group (because it divides it).
A: Yes your reasoning's correct.
Assume an element $t$ of the group $G$ has an order $o_t$, and the order of the group was $o_G$. Then the set $\{e, t, t^1, \cdots t^{o_t-1}\}$ would form a cyclic group. Now this cyclic group would be a subgroup of $G$ (why?), and hence would have an equal to or less order than $G$. Hence $o_t<o_G$.
