Groups Flashcards
Injective
one-to-one
(if f(x)=f(y) implies x=y for all x,y in A)
Surjective
all used or onto
(for any b in B, there exists a in A such that f(a)=b
Bijective
one-to-one and all used
(injective and surjective)
Group (G,*)
A set G together with a binary operation * such that:
1) is associative, meaning for a,b,c in G we have
(ab)c=a(bc).
2) there exists an identity element e such that for all a in G we have
ae=a=ea.
3) each element has an inverse, meaning for any g in G there exists g^(-1) in G such that
gg^(-1)=g^(-1)g.
Note: Check that it is a binary operation (for closure).
Sym(x)
Symmetric group on n letter denoted Sn
Permutations
Elements of Sn
(bijection or relabeling of those elements)
Function Composition
Product in S3
(applying one function to the results of another)
Cycle Notation
A permutation consisting of a single cycle. Example:
(1)(2)(3)
(12)(3)
(123)
a|b
notation for a divides b.
a|b if there exists x in the integers such that ax=b
a≅b mod m
If a and b are integers, we say that a is congruent to b modulo m if m|(a-b)
a is the same as b (up to) skip counting by m (not actually the same but the same answer; ex: mod 12 –> 1am and 1pm)
modulo
Two numbers are congruent modulo a given number if they give the same remainder when divided by that number.
Ex. 19 and 64 are congruent modulo 5
Theorem: Congruence as an Equivalence Relation
Let m be a positive integer. Congruence modulo m satisfies the following properties:
1) Reflexivity
2) Symmetry
3) Transitivity
Partitioning Integers
Define the congruence class of n mod m by:
[n]m:={a:a≅n mod m}
give one that does live there: n
ex. m=2:
[0]2 ={…,-6,-4,-2,0,2,4,6,…}
[6]2={…,-6,-4,-2,0,2,4,6,…}
[3]2={…,-5,-3,-1,1,3,5,…}
Least nonnegative residue
M is a positive integer. The division algorithm gives:
a=bm+r with 0 < or = r < or = m-1
We call the remainder (r) the least nonnegative residue of a modulo m
Theorem: Modular Arithmetic
Let a,b,c,d,m be integers with m>0, a≅b mod m and c≅d mod m
a+c ≅ b+d mod m [a+c]m=[b+d]m
a-c ≅ b-d mod m [a-c]m=[b-d]m
ac ≅ bd mod m [ac]m=[bd]m
Z/mZ:={[0]m,…,[m-1]m} where:
[a]m+[c]m= [a+c]m
[a]m-[c]m= [a-c]m
[a]m([c]m)= [ac]m