Any group of order $85$ is cyclic. 
Any group of order $85$ is cyclic.

My attempt:
Let $|G|=85=5\times17$
Let $H_1$ be the sylow-$5$ and $H_2$ be sylow-$17$ subgroup of $G.$
Then $H_1\cap H_2=\{e\}$
So $|H_1\times H_2|=85$
and hence $G\simeq H_1\times H_2$
Since $H_1\times H_2$ is cyclic so is $G.$

I'm not sure about my proof. Please tell me whether it's correct! Apart from voting please leave comment.
 A: "Proof by Lagrange-Cauchy-Frobenius-Burnside Theorems"
(*) If $H,K$ are subgroups of a finite group, then $|HK|=\frac{|H|.|K|}{|H\cap K|}$.
1) Suppose $G$ is abelian. Then by Cauchy theorem for finite abelian groups, $G$ has elements $x,y$ of order $17$ and $5$, respectively. As $x,y$ comute and have relatively prome order, the order of $xy$ must be $85$, and $G$ is cyclic. (Cauchy theorem for finite abelian groups can be very easily proved by * ).
2) Suppose $G$ is non-abelian. Then $Z(G)<G$, hence center has order either $1,5$ or $17$ by Lagrange theorem. The later two are impossible, otherwise $G/Z(G)$ will be cyclic, so $Z(G)=1$. The Frobenius class equation for $G$ becomes 
$|G|=|Z(G)|+|C_G(x_1)|+|C_G(x_2)| + \cdots |C_G(x_r)|$ (where $C_G(a)=$ conjugacy class of $a$ in $G$). Let $|C_G(x_i)|=n_i$. 
Hence, $85=1+n_1+n_2+\cdots n_r$.
2.1) For $a\in G$, $a\neq 1$, by Lagranges theorem, $a$ has order $5$, $17$ or $85$; if it is $85$, then $G$ is cyclic, so consider first two possibilities. If $a$ has order $5$, then the centralizer of $a$ has order $\geq 5$, and it divides $|G|$, it should be $5$ since $Z(G)=1$. Therefore, $|C_G(a)|=85/5=17$. Similarly, if $a$ has order $17$, then $|C_G(a)|=5$. Therefore, in the class equation, $n_i$ is one of $5$ and $17$. In particular, $n_i\geq 5$, and from class equation, we deduce that, $r\leq 16$. Hence $G$ has at most $17$ conjugacy classes.
2.2) By a theorem of Burnside (probably 1902), in a group of odd order, the number of conjugacy classes is congruent to $|G|$ mod $16$. Hence, $(r+1)\equiv 85$ mod $16$, and $r+1\leq 17$ (by 2.1). The only solution of this equation is $r=4$. Hence $85=1+n_1+n_2+n_3+n_4$, with $n_i\in \{5,17\}$. This is impossible, since RHS is at most $69$. 
Hence $G$ is abelian, and by (1), it is cyclic.

Comment The purpose of this solution is to show how very basic theorems (shown with bold case letters) can be beautifully used to understand structure of finite groups. Especially, I have not used general Cauchy theorem, but only for finite abelian groups, which can be quickly proved by *. But, the theorem of Burnside has non-trivial proof. It is included in very less books of group theory, and this is an opportunity to show it to the people, who never met it.
A: One more step is needed. Recall that one of a definition of direct product $G = H \times K$ is $H, K \trianglelefteq G$, $G = \langle H, K \rangle$, and $H \cap K = 1$. Hence, you have to show that $H_1, H_2 \trianglelefteq G$. (This can be easily shown by Sylow's theorem.)
Otherwise your proof fails. Such an example is $S_3$. Though Sylow subgroups $C_2$, $C_3$ of $S_3$ have trivial intersection $C_2 \cap C_3 = 1$ and $\lvert C_3C_2 \rvert = 6$ (which implies $S_3 = C_3C_2$), $S_3$ is not the direct product of the two. (Observe that  $C_2 \not \trianglelefteq S_3$ in this case.)
A: I'll give a shot at a solution without using the Sylow theorems -- please tell me if anything I say is false.
We begin by considering $|Z(G)|$. If this is 85 we are done. We claim that it is neither $5$ nor $17$. Indeed, if it is $5$, we have that the action of $G$ on itself by conjugation has kernel containing $Z(G)$; hence taking the quotient, we find that $G/Z(G) \simeq \mathbb Z / ( 17 \mathbb Z)$ acts on $G$ by conjugation. Hence non-trivial conjugacy classes must be size $17$, which is incompatible with the class equation since $5 + 17 n \neq 85$. Similarly if $|Z(G)| = 17$ then $17 + 5n \neq 85$ which shows that this case is impossible as well.
Hence the only case left is $|Z(G)|  =1$, which we will also show is impossible. Since the size of a conjugacy class must divide the order of $G$, the class equation tells us
$1 + 5n + 17m = 85$,
which has the unique solution $n=10, m=2$. 
Now, by Cauchy's theorem there exists a cyclic subgroup $H \leq G$ of order $17$, i.e. index $5$. We look at the action of $G$ on the cosets $G/H$ which gives a homomorphism $G \rightarrow S_5$. We find that the kernel of this action is contained in $H$ and has index at most $5$ and hence is $H$, i.e. $H$ is a normal subgroup.
However, a normal subgroup must contain the identity and be a union of conjugacy classes, and since $1 + 5n' + 17 m' \neq 17$ for any choice of $0\leq n' \leq n, 0 \leq m' \leq m$, this is impossible as well.
So $G = Z(G)$ is Abelian of order 15, which means it must be cyclic.
A: It is worth noting that underlying point here is that $5$ does not divide $17-1=16$. Using Sylow's theorems, if a group has order $pq$ for $p<q$ then the Sylow $q$-subgroup $H_q$ is normal. Combined what what you said, this means we have a semi-direct product.
Note that $H_p$ and $H_q$ are cyclic. Now, it is "well-known" (It is! Honest!) that if $C_q$ is the cyclic group of order $q$ then $\operatorname{Aut}(C_q)\cong C_{q-1}$. Therefore, the action of $H_p$ on $H_q$ is non-trivial if and only if $p$ divides $q-1$. Here, $5$ does not divide $17-1=16$ and so the action is trivial. Therefore, we have a direct product.
If you were to keep plugging away at the (non-direct) semi-direct product case, you will find that there is precisely one isomorphism class (for fixed $p$, $q$), basically because $p$ is prime. Therefore, we can conclude with the following result,
Lemma: Suppose $G$ has order $pq$, $p<q$. If $p$ does not divide $q-1$ then $G$ is cyclic. If $p$ divides $q-1$ then there are precisely two isomorphism classes of order $pq$,  that is, either $G\cong C_{pq}$ or $G$ is the semidirect product $G\cong\langle a, b; a^p, b^q, a^{-1}ba=b^p\rangle$.
For example, $S_3$ has order $6=2\cdot 3$, and $2$ divides $3-1=2$. This is why $S_3$ is not simply $C_2\times C_3\cong C_6$.
