Cyclic Groups and Order

Abstract Algebra

The group U(n)

The group of units of the multiplicative monoid (Z_n, dot) is the set of invertible elements, i.e. the units, in (Z_n, dot).

Order of a group

The order of a group G is denoted |G| and is the cardinality of the set G.

Order of an element

The order of an element g in G is denoted |g| and equals the smallest positive integer n such that gⁿ=e.

Order of infinite group

The |(Z,+)|= infinity, and in fact the order of any infinite group is infinity.

Finite and Infinite Theorem

Let g be in G

(1) <g>={e,g,g²,…,g^n-1} is finite with |g|=n. Then g^k=e ←→ n|k and g^k=g^m ←→ k (triple equals) m (mod n).

(2) <g>={…,g^-2,g^-1,e,g,g²,…,g^n-1} is infinite with |g|= infinity. Then g^k=e ←→ k=0 and g^k=g^m ←→ k=m

Identity e order

The identity e is the only element of order 1 in a group G.

Corollary

|g|=|<g>| for all g in G

Inverse order

|g|=|g^-1| for all g in G

Order of an element in U(n)

There is no formula :( but a corollary to Lagrange’s Theorem will show us that for g in G, |g| divides |G|

r-cycle

r=(k₁k₂…k_r) is an r-cycle in S_n → |r|=r

Disjoint Cycles

r=omega₁omega₂…omega_r where omega_i are disjoint cycle → |r|=lcm(|omega₁|,|omega₂|,…,|omega_r|)

Is every cyclic group abelian?

Every cyclic group is abelian, but the converse does not hold.

Subgroup of a cyclic group

Every subgroup of a cyclic group is cyclic

gcd

Let G=<g> be a cyclic group, where |g|=n. Then, G=<g^k> if and only if gcd(k,n)=1.

The Fundamental Theorem of Finite Cyclic Groups

Let G=<g> be a cyclic group of order n

(1) H<G → H=<g^d> for some d|n. Hence |H| divides n.

(2) k|n → <g^n/k> is the unique subgroup of order k.