exam Flashcards
(108 cards)
inverse element of x
unique y∈G : xy=e=yx
y=x^(-1)
show that xy=e
inverse element properties
(a^(-1))^(-1) = a
(a1…an)^(-1) = an^(-1)…a1^(-1)
(a^n)^(-1) = a^(-n) = (a^(-1))^n
x^0=
e
a and b “congruent modulo N”
N|a-b
a≡bmodN
equivalence relation
[reflexivity]
a≡amodN
[symmetry]
a≡bmodN ⟺ b≡amodN
[transitivity]
a≡bmodN and b≡cmodN
⟹ a≡cmodN
residue class of a modulo N
amodN := {b∈Z| b≡amodN}
amodN=bmodN
⟺a≡bmodN
a=qN+r
⟹a≡rmodN
⟹amodN=rmodN
amodN
= a_
= {a+Nk|k∈Z}
a is a “unit modulo N”
∃b_ ∈ Z/NZ :
a_ * b_ = 1_
gcd(a,N) = 1
(Z/NZ)^X
subset of Z/NZ containing all units modulo N
Euler’s totient function
ρ(N) = #(Z/NZ)^X
Euler’s theorem
for all a_∈ (Z/NZ)^X,
(a_)^ρ(N) = 1_
Fermat’s little theorem
if p is prime,
(amodp)^p = amodp
subgroup H of G
H<=G
H is subset of G and group law and unit element are the same.
subgroup criterion
[H1]
e∈H
[H2]
x,y∈H ⟹ x*y ∈H
[H3]
x ∈H ⟹ x^(-1) ∈H
Lagrange
H is a subgroup of finite G
⟹
#H | #G
ord(x)
- minimum m>0 such that x^m=e
- no such m exists ⟹ ord(x)=∞
ord(x^(-1))
- ord(x) = ord(x^(-1))
- ord(x)<∞
⟹
<x>={x,x^2,...,x^ord(x)=e}
-
</x>
ord(x)<∞
<x> = {x , x^2, ..., x^(ord(x)) = e}
</x>
<x></x>
= ord(x)
G<∞
⟹ ord(x) < ∞ ⟹ ord(x)|#G
x^n=e
⟹ ord(x) | n
G “cyclic”
if G=<g> for some g∈G
g is then the "generator" of G</g>