Need help determining what this center is isomorphic to I am looking at this released exam, problem 2c. It states:

Let $G$ be a non-abelian group of order $8$ and $Z$ be the center of $G$. To which group is $Z$ isomorphic?

It gives a hint to recall a theorem from what I believe is this released past midterm. I think he is referring to 4a, and 4b:

I'm still not exactly sure how to use them. As for b, $G$ is not of the form $pq$ it is $2^3$. And the first part, I'm not sure how I would fit that in.
I can "solve" it, because in the next question it is mentioned that $G/Z$ is isomorphic to Klein-4 (although I have not proven it). Hence $Z$ must have order $2$, as $G$ has order $8$ therefore $Z(G)$ divides $G$ and $8/2=4$. From this, I know there is only one possible group of order $2$, implying $Z \cong \mathbb{Z}_2$.
This is begging the question, as I don't know how to prove the one after this. And proving the original question would make the next question trivial.
Any hint in the right direction would be appreciated.

Let $\frac{G}{Z(G)}$ be cyclic group. So, $\frac{G}{Z(G)}=<gZ(G)>$ for some $g\in G$. It's mean for every $x\in G$, we can find $z\in Z(G)$ such that $x=zg^n$. Now let: $a,b\in G$. So, $a=g^kz_1$ and $b=g^mz_2$ which $z_1,z_2 \in Z(G)$ and $m,k\in \mathbb Z$. Thus: $$ab=g^kz_1g^mz_2=g^kg^mz_1z_2=g^{m+k}z_2z_1=g^mg^kz_1z_2=g^mz_2g^kz_1=ba$$  Which means $G$ is an abelian group. So, $Z(G)=G$ contradiction. So, $\frac{G}{Z(G)}$ when $G$ is non-abelian group, can not be cyclic.
For the prt $(b)$, because of $G$ is an non-abelian group, we have the following cases:
$(1)$  $\left|\frac{G}{Z(G)}\right|=p$
$(2)$ $\left|\frac{G}{Z(G)}\right|=q$
$(3)$ $\left|\frac{G}{Z(G)}\right|=pq$.
The cases $(1),(2)$ can not happen. Since in this cases, $\frac{G}{Z(G)}$ shoud be cyclic and $G$ should be abelian group. contradict. So,the only possible  case is $(3)$  and it's mean $Z(G)$ is trivial.
