Let . Then that is where number of Sylow p subgroups.
Let G be a minimal counterexample.
Why are me looking at p dividing or not dividing the center of G?
without having a bit more context, i can only hazard a guess, but my thought is, you are looking at an inductive proof, and will eventually consider G/Z(G).
I will just type up the whole proof but my question is still in regards to the center.
Let G be a minimal counterexample. Either .
The center is abelian. By Cauchy's Theorem (Let G be a finite abelian group and , then G has an element of order p), there is an element of order p.
Set . since .
since the order of H is p.
Since G is a minimal counterexample, G\H has a subgroup of order (WHY), say .
By the 4th Isomorphism Theorem, there exist such that and .
Then which is .
By the Class Equation, .
p must not divide for some i (WHY). By Lagrange's Theorem, . Since p divides G and by Euclid's Lemma, ; moreover, .
So where otherwise which contradicts the Class Equation.
By induction, there exist such that so
ok, now we have some context.
the reason why we look at |Z(G)|, is in order to use the class equation.
specifically, if p does not divide |Z(G)|, then the sum of the centralizer indexes must not be divisible by p, either. because if the sum was, then
since p divides |G|, and the sum of centralizer indexes, it would divide the difference, which is |Z(G)|, a contradiction.
in turn, if the sum of centralizer indexes is not divisible by p, at least one centralizer index is not divisible by p.
but since |G| = [G:C(g)]|C(g)| for any centralizer C(g), all the powers of p in the factorization of |G| are in |C(g)|.
since C(g) is strictly less than G (since otherwise g is in Z(G)), it has a p-sylow subgroup, which in turn is a subgroup of G.
however, if p divides Z(G), we can't use the class equation, because it might be that p divides the index of all our centralizers.
so then we use cauchy's theorem for abelian groups to produce a normal subgroup of G, <g>, of order p, and consider G/<g>.
since |G/<g>| < |G|, we can assume G/<g> has a p-sylow subgroup (since G is, by assumption, the minimal counter-example).
this subgroup has order . we then use the 1-1 correspondence of subgroups of G/<g> and subgroups of G
containing <g> (guaranteed by the 4th isomorphism theorem, which is a consequence of the 1st isomorphism theorem),
to produce a subgroup of G containing <g> which has order (this subgroup is N in the proof).
N/<g> is isomorphic to the p-sylow subgroup of G/<g>, and |N| = |N/<g>||<g>| = ,
which is the desired p-sylow subgroup of G.
in either case, we get a contradiction: by assuming G is a minimal counter-example, we prove G is NOT a counter-example.
therefore, there is no minimal counter-example, so every finite group G has a p-sylow subgroup.