# Sylow's Theorem

• November 12th 2011, 06:57 PM
dwsmith
Sylow's Theorem
Let $|G|=p^{\alpha}m, \ p\nmid m$. Then $Syl_p(G)\neq\O$ that is $n_p\geq 1$ where $n_p$ number of Sylow p subgroups.

Let G be a minimal counterexample.

$p\mid |Z(G)| \ \ \text{or} \ \ p\nmid |Z(G)|$

Why are me looking at p dividing or not dividing the center of G?
• November 12th 2011, 07:41 PM
Deveno
Re: Sylow's Theorem
Quote:

Originally Posted by dwsmith
Let $|G|=p^{\alpha}m, \ p\nmid m$. Then $Sly_p(G)\neq\O$ that is $n_p\geq 1$ where $n_p=|Sly_p(G)|$.

Let G be a minimal counterexample.

$p\mid |Z(G)| \ \ \text{or} \ \ p\nmid |Z(G)|$

Why are we looking at p dividing or not dividing the center of G?

what is your working definition of $Syl_p(G)$? also, is $n_p$ supposed to number of subgroups $Syl_p(G)$, or the cardinality of such a subgroup?

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).
• November 12th 2011, 08:15 PM
dwsmith
Re: Sylow's Theorem
I will just type up the whole proof but my question is still in regards to the center.

proof:
Let G be a minimal counterexample. Either $p||Z(G)| \ \text{or} \ p\nmid |Z(G)|$.

Case 1: $p||Z(G)|$
The center is abelian. By Cauchy's Theorem (Let G be a finite abelian group and $p||Z(G)|$, then G has an element of order p), there is an element $g\in Z(G)$ of order p.
Set $=H$. $H\trianglelefteq G$ since $g\in Z(G)$.
$|G|=|G/H||H|=|G/H|p$ since the order of H is p.
$|G|=p^{\alpha}m=|G/H|p$
So $|G/H|<|G|$.
Since G is a minimal counterexample, G\H has a subgroup of order $p^{\alpha-1}$ (WHY), say $K.
By the 4th Isomorphism Theorem, there exist $N\leq G$ such that $H\leq N$ and $N/H=K$.
Then $|N/H||H|=|N|$ which is $p^{\alpha-1}p=p^{\alpha}$.
Therefore, $N\in Syl_p(G)$

Case 2: $p\nmid |Z(G)|$
By the Class Equation, $p^{\alpha}m=|G|=|Z(G)|+\sum_{i=1}^r|G:C_G(g_i)|$.
So $|Z(G)|=|G|-\sum_{i=1}^r|G:C_G(g_i)|$.
p must not divide $|G:C_G(g_i)|$ for some i (WHY). By Lagrange's Theorem, $|G:C_G(g_i)||C_G(g_i)|=|G|$. Since p divides G and by Euclid's Lemma, $p||C_G(g_i)|$; moreover, $p^{\alpha}||C_G(g_i)|$.
So $|C_G(g_i)|=p^{\alpha}m'$ where $m' otherwise $G=C_G(g_i)$ which contradicts the Class Equation.
By induction, there exist $P\leq C_G(g_i)$ such that $|P|=p^{\alpha}$ so $P\in Syl_p(G)$
• November 12th 2011, 09:54 PM
Deveno
Re: Sylow's Theorem
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 $p^{\alpha-1}$. 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 $p^{\alpha-1}$ (this subgroup is N in the proof).

N/<g> is isomorphic to the p-sylow subgroup of G/<g>, and |N| = |N/<g>||<g>| = $(p^{\alpha-1})(p) = p^\alpha$,

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.