Cyclic subgroups of a finite group

Question

Prove or disprove: Let $G$ be a group of order 595, then G has a cyclic subgroup of order 35.

What I think is that the statement is true. If $|G|=595=5\times 7\times 17$, then since 5 and 7 divides $|G|$, there exists elements $x,y\in G$ such that $|x|=5, |y|=7$, so $A=\langle x\rangle, B=\langle y\rangle$. By computations we see that number of Sylows 5-subgroups is 1, so there is a unique Sylows 5-subgroup of order 5, which will be $A$, since all Sylows 5-subgroup are conjugate. So, $A$ is normal subgroup of $G$, and $B$ is a subgroup of $G$, Thus, $AB$ is a subgroup of $G$, and

$$|AB|=\frac{|A||B|}{|A\cap B|}=\frac{5\times 7}{1}=35$$ because 5 and 7 are relatively prime.

S0, $|AB|=35$, and $AB\simeq \mathbb Z_{5}\times \mathbb Z_{7} \simeq \mathbb Z_{35}$ since 5 and 7 are relatively prime.

Note: A well known reasult: Any group of order 35 is cyclic, by the Fundamental Theorem of Finite Abelian Groups.

Is my argument correct in general (there are many details, using Sylows Theorem, Cauch Theorem,..)!?