Group $G$ such that $[G : Z(G)] = 4$ Let $G$ be a group, let $Z(G)$ be the center of $G$, and suppose that $[G : Z(G)] = 4$. 
(a) Prove that $x^2 \in Z(G)$ for every $x \in G$. 
(b) Prove that $Z(G)$ contains an element of order two. 
Here are my thoughts so far: 
Recall that $Z(G)$ is a normal subgroup of $G$. Thus, $G/Z(G)$ has a group structure. By the fact that $[G : Z(G)] = 4$, we have that $|G/Z(G)| = 4$. This gives that $G/Z(G)$ is either isomorphic to $\mathbb{Z}/4\mathbb{Z}$ or $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. But, $G/Z(G)$ cannot be isomorphic to $\mathbb{Z}/4\mathbb{Z}$; this would mean that $G/Z(G)$ is cyclic $\Rightarrow$ $G$ abelian $\Rightarrow$ $G = Z(G)$, which contradicts the index of $Z(G)$ in $G$ being equal to $4$. 
Thus, so far I know that $G/Z(G)$ is isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. I also know that $G/Z(G) \cong Inn(G)$, the inner automorphism group of $G$. How can I use these facts to  prove the desired facts ? 
Here's one attempt to use that $G/Z(G) \cong Inn(G)$ for part (b). If this is the case, we have that $Inn(G) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ $\Rightarrow$ $Inn(G)$ necessarily contains an element of order $2$. Thus, the map $\phi: G \longrightarrow G : h \longmapsto ghg^{-1}$ for some $g \in G$, where $\phi$ is not the identity map, has order $2$ $\Rightarrow$ $g^2hg^{-2} = h$ $\Rightarrow$ $g^2h = hg^{2}$. It looks like I have a commutativity relation here -- how can I relate this to $Z(G)$ having an element of order $2$ ? 
For part (a), is the right idea to look at the order of the cosets $G/Z(G)$ ? I suppose all cosets that aren't the identity are of order $2$, since $G/Z(G) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Is this the right route to take? 
Thanks for all of your help. (= 
 A: For the second part of the question we proceed as follows.Since $G/Z(G) \cong C_2 \times C_2$, $G$ is not abelian. Let $x \in G-Z(G)$, then observe that for the centralizer: $Z(G) \subsetneq C_G(x) \subsetneq G$, so $|G:C_G(x)|=2=|C_G(x):Z(G)|$. Hence every conjugacy class has cardinality at most $2$.
In case $G$ is finite, put $l=|G|-|Z(G)|$, then the class formula yields
$$|G|=\underbrace{1 + 1 + \cdots +1}_{|Z(G)| \text{ times}}+\underbrace{2+2+ \cdots +2}_{l\text{ times}}$$ whence $4|Z(G)|=|Z(G)|+2l$, so $3|Z(G)|=2l$. It follows that $2$ divides $|Z(G)|$ and owing to Cauchy's Theorem, $Z(G)$ has an element of order 2. In fact, from the above calculation, one can derive that the number of conjugacy classes of $G$ is $k(G)=2\frac{1}{2}|Z(G)|=\frac{5}{8}|G|$.
Let us consider the general case. We already observed that every conjugacy class has order at most $2$. We will show that this implies that $|G'|=2$. Since $G/Z(G)$ is abelian and thus $G' \subseteq Z(G)$, this shows that $Z(G)$ has an element of order $2$.

Now, for each $x \in G$ there exists a unique $f(x) \in G$ such that $\{gxg^{-1}x^{-1}:=[g,x]: g \in G\}=\{1,f(x)\}$, so that the conjugacy class of $x$ is $\{x,f(x)x\}$. Hence this defines a map $f: G \rightarrow G'$ and $G'=\langle f(G) \rangle$. We will show that the image of $f$ has cardinality $2$ and consists of elements of order at most $2$. Note that $f(z)=1$ for each $z \in Z(G)$.

Firstly, if $x$ and $y$ do not commute, $f(x)$ nor $f(y)$ is the identity element: since $yxy^{-1} \neq x, yxy^{-1}=f(x)x$ and likewise, $xyx^{-1}\neq y$ gives $xyx^{-1}=f(y)y$.
Next, observe that if $x,y$ do not commute, $xy$ does not commute with $x$ or $y$. It follows that $f(x)=f(y)$:
$$f(x)=[xy,x]=[x,xy]^{-1}=f(xy)^{-1}=[y,xy]^{-1}=[xy,y]=f(y)$$
and in addition,
$$f(x)=[y,x]=[x,y]^{-1}=f(y)^{-1}.$$
From this it follows that $f(x)=f(y)$ has order $2$. We already saw that $f$ takes the value $1$ on $Z(G)$. To finish the proof, we will show that $f$ is also constant on $G-Z(G)$ and takes the value of an element of order $2$. Let $x,y \in G-Z(G)$, hence $C_G(x)$ and $C_G(y)$ are both proper subgroups of index $2$ and it it well known that their union cannot be the whole group: $C_G(x) \cup C_G(y) \subsetneq G$. So we can find a $g \in G$ that neither commutes with $x$ or $y$, implying $f(x)=f(g)=f(y)$, and this element is of order $2$. So outside $Z(G)$, $f$ is constant and takes the value of an element of order $2$.$\square$
