Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last updated: 23 September 2014

Lecture 18: Groups and group homomorphisms

A group is a set G with a function G×G G (g1,g2) g1g2 such that

(a) If g1,g2,g3G then g1(g2g3)=(g1g2)g3,
(b) There exists 1G such that if gG then 1·g=g and g·1=g.
(c) If gG then there exists g-1G such that g·g-1=1and g-1·g=1.

An abelian group is a set A with a function A×A A (a1,a2) a1+a2 such that

(a) If a1,a2,a3A then a1+(a2+a3)=(a1+a2)+a3,
(b) There exists 0A such that if aA then 0+a=a and a+0=a.
(c) If aA then there exists -aA such that a+(-a)=0and (-a)+a=0,
(d) If a1,a2A then a1+a2=a2+a1.

Every abelian group is a group. GL2() = { 2×2invertible matrices with entries in } = { (abcd) |a,b,c,d andad-bc0 } . GL2() is a group with product matrix multiplication. GL2() is not an abelian group.

Let G be a group. The order of G is Card(G), the number of elements in G.

The order of an element gG is the smallest k>0 such that gk=1. If there does not exist k>0 such that gk=1 then the order of gG is .

A subgroup of G is a subset HG such that

(a) If h1,h2H then h1h2H.
(b) 1H,
(c) If hH then h-1H.

Group homomorphisms are for comparing groups.

Let G and K be groups. A group homomorphism from K to G is a function f:KG such that if k1,k2K then f(k1k2)=f(k1)f(k2).

An isomorphism from K to G is a group homomorphism f:KG such that there exists a group homomorphism f-1:GK such that ff-1=idG and f-1f=idK.

Let f:KG be a group homomorphism. The kernel of f is the set kerf= {gG|f(g)=1} The image of f is the set imf= {f(g)|gG}.

Let f:KG be a group homomorphism. Then f:KG is an isomorphism if and only if f:KG is bijective.


Assume f:KG is an isomorphism from K to G.
To show: f:KG is bijective.
Since f is an isomorphism, there exists an inverse function to f, g:GK such that gf=idG and fg=idK.
Thus, by theorem Theorem Let f:KG be a function. An inverse function to f exists if and only if f is bijective. which is proved fully in Lecture notes, f is bijective.
Assume f:KG is a group homomorphism and f:KG is bijective.
To show: f:KG is an isomorphism.
To show:
(a) There exists a function g:GK such that gf=idGand fg=idK.
(b) g:GK is a group homomorphism.
(a) follows from Theorem.
(b) To show: If x1,x2G then g(x1x2)=g(x1)g(x2).
Assume x1,x2G.
To show: g(x1x2)=g(x1)g(x2).
Since f is bijective, f is injective, which means if k1,k2K and f(k1)=f(k2) then k1=k2. To show: f(g(x1x2))=f(g(x1)g(x2)). f(g(x1x2))= x1x2, since fg=idG, and f(g(x1)g(x2)) = f(g(x1)) f(g(x2)), sincefis a homomorphism = x1·x2,since fg=idG. So f(g(x1)g(x2))=f(g(x1x2)).
So g(x1)g(x2)=g(x1x2).
So g is a homomorphism.

Notes and References

These are a typed copy of Lecture 18 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on September 2, 2011.

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