TITLE: group of order 28 is not simple
QUESTION [8 upvotes]: 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?

REPLY [3 votes]: @Mark - or you could skip the argument based on the class formula and apply Sylow's theorems: the 7-Sylow subgroup must be normal, since the number of such subgroups must divide 28 and must be congruent to 1 (mod 7). Probably at this point in your notes the Sylow theorems haven't crossed your path ...
Probably more basic: by Cauchy's theorem, $G$ has a subgroup of order 7, say $H$. Now $G$ acts on the four left cosets of $H$, by left multiplication. The kernel of this action is the normal subgroup $core_G(H)$, and is contained in $H$, hence it equals $H$ or must be trivial (its order divides the prime 7). But $|G/core_G(H)|$ must divide $4!$ (this is hopefully one of the theorems you learned already). Hence $H = core_G(H)$, so $H$ is normal, whence $G$ not simple.