group of order 28 is not simple I have a proof from notes but I don't quite understand the bold part:
Abelian case: $a \in G / \{1\}$. If $\langle a\rangle \neq G$, then we are done. If $\langle a\rangle = G$, then $\langle a^4\rangle $ is a proper normal subgroup of $G$. General case: WLOG we can assume $G \neq Z(G)$. $\langle 1\rangle \neq Z(G)$ which is a proper normal subgroup of $G$. Done. Otherwise $|Z(G)|= 1$. 
$$
28 = 1 + \sum_{**}\frac{|G|}{|C_G(x)|}
$$
There must be some $a\in G$ such that 7 does not divide
$$
\frac{|G|}{|C_G(a)|}
$$
It follows that
$\frac{|G|}{|C_G(a)|} = 2 $ or $4$ $\Rightarrow [G:C_G(a)] = 2$ or $4$ $\Rightarrow 28 \mid2!$ or $28\mid4!$.
Therefore group of order 28 is not simple.
Why are they true?

Reading the class equation modulo 7 gives the existence of one $x$ such that $\frac{|G|}{|C_G(x)|}$ is NOT divisible by 7. Hence 7 divides $|C_G(x)|$. Now the factors of the numerator $|G|$ are 1, 2, 4, 7 $\cdots$. Since $\frac{|G|}{|C_G(x)|}$ cannot be 1 and cannot divide 7, the only possibilities are 2 and 4. 
