Question

Best proof of sylow's first theorem well describe.

Answer 1

Im studying the proof of sylows first theorem (from abstract algebra by beachy and balair). I have 2 questions about it, I will present the proof below:
(theorem: Let
G
G
be a finite group, if
p
p
is a prime such that
p
α
pα
is a divisor of
∣G∣
∣G∣
for some
α≥0
α≥0
, then
G
G
contains a subgroup of order
p
α
pα
.)
The proof is by induction on
∣G∣=n
∣G∣=n
where one states that the theorem holds for
n=1
n=1
and assume it holds for all groups of order less than
n
n
, the proof also makes use of the class equation,
∣G∣=∣Z(G)∣+∑[G:C(x)]
∣G∣=∣Z(G)∣+∑[G:C(x)]
where the sum ranges over one entry from each nontrivial conjugacy class. It is split up in two cases:
Case (1)
Case (1)
: For each
x∉Z(G)
x∉Z(G)
,
p
p
is a divisor of
[G:C(x)]
[G:C(x)]
Case (2)
Case (2)
: For some
x∉Z(G)
x∉Z(G)
,
p
p
is not a divisor of
[G:C(x)]
[G:C(x)]
Proof for case (1):
Proof for case (1):
In this case the class equation shows that
p
p
must be a divisor
∣Z(G)∣
∣Z(G)∣
and so
Z(G)
Z(G)
contains an element
a
a
of order
p
p
(by Cauchy´s theorem). Then
⟨a⟩
⟨a⟩
is a normal subgroup of
G
G
since
a∈Z(G)
a∈Z(G)
, and so by the induction hypothesis,
G /⟨a⟩
G /⟨a⟩
contains a subgroup of order
p
α−1
pα−1
, since
p
α−1
pα−1
is a divisor of
∣G /⟨a⟩∣
∣G /⟨a⟩∣
. The inverse image in
G
G
of this subgroup has order
p
α
pα
since each coset of
⟨a⟩
⟨a⟩
has
p
p
elements.
My first question is, what if
G
G
is abelian then it doesnt exist an element
x∉Z(G)
x∉Z(G)
, although it seems like the proof for case 1 will be valid whenever
G
G
is abelian, is this the authors intension?
My second question is, I dont understand the last part of the proof for case 1, what does the following sentence mean? "The inverse image in
G
G
of this subgroup has order
p
α
pα
since each coset of
⟨a⟩
⟨a⟩
has
p
p
elements."