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

*

*$|Z(G/Z(G))| = 1$

*$G/Z(G)$ has a subgroup of order 17

*$G$ has a subgroup of order 34


I am preparing for my Algebra exam and I have been stuck on this question for a bit.
1.
So the only thing I know is that $|Z(G/Z(G))| = 1$, which means that $Z(G) \in G/Z(G)$ is the only element for which $A \cdot Z(G) = Z(G) \cdot A$ for every $A \in G/Z(G)$. But I don't know how to prove that there are no other subgroups for which that holds. At first I thought it might be a general property, but apparently it isn't (according to this example: $Z(G/ Z(G)) = 1$ Prove or disprove.).
So I know that it has to do something with the order of the groups, but I don't know how to use the information I have to get the conclusion. So far what I know is that with Lagrange I get $|G/Z(G)| = 102/2 = 51 = 3 \cdot 17$. I see that both are prime but I don't know how this is supposed to help me any further.
2.
I thought of maybe using the fact that $|Cl(A)|$, for every $A \in G/Z(G)$, has to be a divisor of $|G/Z(G)|$, but then I have that $|Cl(A)|$ for every $A$ has to be either 3 or 17, but how to prove that there exists one of 17 I don't know. Or maybe I can use the Class equation but I'm not sure.
I unfortunately don't know how to get any further with this question.
3.
I suppose that this will follow from (1) and (2), since if $G/Z(G)$ has a group of order 17, since $|Z(G)| = 2$, it follows with Lagrange that $G$ has a subgroup of order 34.

Can someone please help me by giving hints that will help me solve this problem? This is a rather introductory course, I've seen other people solve this with 'Sylow theory' but we haven't seen that yet in this course.
 A: 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 $17$.
A: Hints:

*

*We have that $\;\left|G/Z(G)\right|=\frac{102}2=51\;$. Now, how many groups of order $\;51\;$ are there? This question is pretty symple if you know semidirect product, but if you don't it is also pretty simple knowing only Sylow Theorems

2)+ 3) are immediate consequences of (1) and Sylow Thms. and the given info (what happens when you take the product of two subgroups $\;HK\;$ ,with $\;K\lhd G\;$ ?)
