Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 23 September 2014

Lecture 18: Groups and group homomorphisms

A group is a set $G$ with a function $$\begin{array}{ccc}G\times G& ⟶& G\\ (g_1,g_2)& ⟼& g_1{g}_2\end{array}$$ such that

(a)	If $g_1,g_2,g_3\in G$ then $g_1\left(g_2{g}_3\right)=\left(g_1{g}_2\right)g_3,$
(b)	There exists $1\in G$ such that if $g\in G$ then $1·g=g$ and $g·1=g\text{.}$
(c)	If $g\in G$ then there exists $g^{-1}\in G$ such that $$g·g^{-1}=1\text{and}{g}^{-1}·g=1\text{.}$$

An abelian group is a set $A$ with a function $$\begin{array}{ccc}A\times A& ⟶& A\\ (a_1,a_2)& ⟼\\ a_1+a_2\end{array}$$ such that

(a)	If $a_1,a_2,a_3\in A$ then $a_1+(a_2+a_3)=(a_1+a_2)+a_3,$
(b)	There exists $0\in A$ such that if $a\in A$ then $0+a=a$ and $a+0=a\text{.}$
(c)	If $a\in A$ then there exists $-a\in A$ such that $$a+(-a)=0\text{and}(-a)+a=0,$$
(d)	If $a_1,a_2\in A$ then $a_1+a_2=a_2+a_1\text{.}$

Every abelian group is a group. $$\begin{array}{ccc}{\displaystyle {GL}_2\left(ℝ\right)}& =& {\displaystyle \{2\times 2\hspace{0.17em}\text{invertible matrices with entries in}\hspace{0.17em}ℝ\}}\\ & =& {\displaystyle \left\{\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}a,b,c,d\in ℝ\hspace{0.17em}\text{and}\hspace{0.17em}ad-bc\ne 0\}\text{.}}\end{array}$$ ${GL}_2\left(ℝ\right)$ is a group with product matrix multiplication. ${GL}_2\left(ℝ\right)$ is not an abelian group.

Let $G$ be a group. The order of $G$ is $\text{Card}\left(G\right),$ the number of elements in $G\text{.}$

The order of an element $g\in G$ is the smallest $k\in {\mathbb{Z}}_{>0}$ such that $g^k=1\text{.}$ If there does not exist $k\in {\mathbb{Z}}_{>0}$ such that $g^k=1$ then the order of $g\in G$ is $\infty \text{.}$

A subgroup of $G$ is a subset $H\subseteq G$ such that

(a)	If $h_1,h_2\in H$ then $h_1{h}_2\in H\text{.}$
(b)	$1\in H,$
(c)	If $h\in H$ then $h^{-1}\in H\text{.}$

Group homomorphisms are for comparing groups.

Let $G$ and $K$ be groups. A group homomorphism from $K$ to $G$ is a function $f:K\to G$ such that if $k_1,k_2\in K$ then $f\left(k_1{k}_2\right)=f\left(k_1\right)f\left(k_2\right)\text{.}$

An isomorphism from $K$ to $G$ is a group homomorphism $f:K\to G$ such that there exists a group homomorphism $f-1:G\to K$ such that $f\circ f-1={\text{id}}_G$ and $f-1\circ f={\text{id}}_K\text{.}$

Let $f:K\to G$ be a group homomorphism. The kernel of $f$ is the set $$\text{ker}\hspace{0.17em}f=\{g\in G\hspace{0.17em}|\hspace{0.17em}f\left(g\right)=1\}$$ The image of $f$ is the set $$\text{im}\hspace{0.17em}f=\left\{f\left(g\right)\hspace{0.17em}\right|\hspace{0.17em}g\in G\}\text{.}$$

Let $f:K\to G$ be a group homomorphism. Then $f:K\to G$ is an isomorphism if and only if $f:K\to G$ is bijective.

		Proof.
	$\Rightarrow$ Assume $f:K\to G$ is an isomorphism from $K$ to $G\text{.}$ To show: $f:K\to G$ is bijective. Since $f$ is an isomorphism, there exists an inverse function to $f,$ $g:G\to K$ such that $g\circ f={\text{id}}_G$ and $f\circ g={\text{id}}_K\text{.}$ Thus, by theorem Theorem Let $f:K\to G$ 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. $\Leftarrow$ Assume $f:K\to G$ is a group homomorphism and $f:K\to G$ is bijective. To show: $f:K\to G$ is an isomorphism. To show: (a) There exists a function $g:G\to K$ such that $$g\circ f={\text{id}}_G\text{and}f\circ g={\text{id}}_K\text{.}$$ (b) $g:G\to K$ is a group homomorphism. (a) follows from Theorem. (b) To show: If $x_1,x_2\in G$ then $g\left(x_1{x}_2\right)=g\left(x_1\right)g\left(x_2\right)\text{.}$ Assume $x_1,x_2\in G\text{.}$ To show: $g\left(x_1{x}_2\right)=g\left(x_1\right)g\left(x_2\right)\text{.}$ Since $f$ is bijective, $f$ is injective, which means if $k_1,k_2\in K$ and $f\left(k_1\right)=f\left(k_2\right)$ then $k_1=k_2\text{.}$ To show: $f\left(g\left(x_1{x}_2\right)\right)=f\left(g\left(x_1\right)g\left(x_2\right)\right)\text{.}$ $$f\left(g\left(x_1{x}_2\right)\right)=x_1{x}_2,$$ since $f\circ g={\text{id}}_G,$ and $$\begin{array}{ccc}{\displaystyle f\left(g\left(x_1\right)g\left(x_2\right)\right)}& =& {\displaystyle f\left(g\left(x_1\right)\right)f\left(g\left(x_2\right)\right),\text{since}\hspace{0.17em}f\hspace{0.17em}\text{is a homomorphism}}\\ & =& {\displaystyle x_1·x_2,\text{since}\hspace{0.17em}f\circ g={\text{id}}_G\text{.}}\end{array}$$ So $f\left(g\left(x_1\right)g\left(x_2\right)\right)=f\left(g\left(x_1{x}_2\right)\right)\text{.}$ So $g\left(x_1\right)g\left(x_2\right)=g\left(x_1{x}_2\right)\text{.}$ So $g$ is a homomorphism. $\square$

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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