Let $G$ be a group with $|G|=102 ( = 2 \cdot 3 \cdot 17)$, with $|Z(G)| = 2$. Prove the following:

A proof for 1) without Sylow: Note that for any Group $H$ if $H/Z(H)$ is cyclic then $H$ is abelian.
So if $|Z(G/Z(G))|>1$ we get $[G/Z(G):Z(G/Z(G))]=1,3$ or $17$. By the above remark $G/Z(G)$ will be abelian. Since $|G/Z(G)|=3\cdot 17$ it follows that $G/Z(G)$ is cyclic and hence $G$ is abelian, a contradiction.
Since $|Z(G)|=2$ claim 3) will follow from 2).
For 2) notice that by Cauchy's theorem there is an element and therefore also a subgroup of order $17$ in $G/Z(G)$.
Edit: A self contained proof that $G/Z(G)=:H$ contains a subgroup of order $17$ (similar to the proof of Cauchy's theorem): Let $x_1,\dots,x_r$ be a system of representatives for the non-trivial conjugacy classes of $H$ (i.e. those with more than one element). Then we have $$H=\underbrace{Z(H)}_{\{e\}}\cup\bigcup_{i=1}^r[x_i]$$ where $[x_i]$ denotes the conjugacy class of $x_i$ and the union is disjoint. This gives us the class equation: $$3\cdot 17=1+\sum_{i=1}^r |[x_i]|$$ Since $|[x_i]|$ divides $|H|$ and $|[x_i]|>1$ there has to be some $i$ with $|[x_i]|=3$. It follows that the centralizer $C_H(x_i)$ has order $17$ which is our subgroup of order $17$.