Let G be a group of order p
am where p is a prime not dividing m. Show the following
1. Sylow p-subgroups of G exist; i.e. Sylp(G) 6= ∅.
2. If P ∈ Sylp(G) and Q is any p-subgroup of G, then there exists g
∈ G such that Q 6
gP g−1
; i.e. Q is contained in some conjugate of P. In particular, any
two Sylow p-
subgroups of G are conjugate in G.
3. np ≡...
(a) Let G be a finite abelian group and p prime with p | | G |.
Show that there is only one p - Sylow subgroup of G. b) Find all p
- Sylow subgroups of (Z2500, +)
1. Consider the group Zp for a prime p with multiplication
multiplication mod p). Show that (p − 1)2 = 1 (mod p)
2. Is the above true for any number (not necessarily prime)?
3. Show that the equation a 2 − 1 = 0, has only two solutions
mod p.
4. Consider (p − 1)!. Show that (p − 1)! = −1 (mod p) Remark:
Think about what are the values of inverses of 1, 2, . . ....
Prove that an abelian group G of order 2000 is the direct
product PxQ where P is the Sylow-2 subgroup of G, and Q the Sylow-5
subgroup of G. (So order of P=16 and order or Q=125).
Throughout this question, let G be a finite group, let p be a
prime, and suppose that H ≤ G is such that [G : H] = p.
Let G act on the set of left cosets of H in G by left
multiplication (i.e., g · aH = (ga)H). Let K be the set of elements
of G that fix every coset under this action; that is,
K = {g ∈ G : (∀a ∈ G) g · aH...
Let gcd(a, p) = 1 with p a prime. Show that if a has at least
one square root, then a has exactly 2 roots. [hint: look at
generators or use x^2 = y^2 (mod p) and use the fact that ab = 0
(mod p) the one of a or b must be 0(why?) ]