Next: Isomorphisms Up: MT2002 Algebra Previous: Cosets and Lagrange's Theorem Contents

Homomorphisms of groups

Here we study mappings between groups which respect the operations.

$\begin{de} Let $G$\ and $H$\ be groups with operations $\ast$\ and $\bullet$\ re... ...morphism} if for all $x,y\in G$\ we have $f(x\ast y)=f(x)\bullet f(y)$. \end{de}$

If the operation in both groups is denoted as multiplication, then the above rule becomes . (The image of the product is the product of images.)

$\begin{ex} The mapping $f\st {\mathbb{Z}}\longrightarrow {\mathbb{Z}}_n$defined by $f(x)=x\pmod{n}$\ is a homomorphism. \end{ex}$

$\begin{ex} For any two groups $G$\ and $H$\ the mapping $f\st G\longrightarrow H... ...identity of $H$) is a homomorphism. Indeed $f(xy)=e_H=e_He_H=f(x)f(y)$. \end{ex}$

$\begin{ex} For any group $G$\ the identity mapping $G\longrightarrow G$is a homomorphism. \end{ex}$

$\begin{ex} The mapping $f\st GL(n,{\mathbb{R}})\longrightarrow {\mathbb{R}}\setm... ...b{R}}\setminus\{0\}$, because $\vert AB\vert=\vert A\vert\vert B\vert$. \end{ex}$

$\begin{ex} The mapping $f\st S_n\longrightarrow {\mathbb{Z}}_2$defined by \begin... ..., $\tau$ even} \end{array}\right\}= f(\sigma)+f(\tau). \end{displaymath}\end{ex}$

Next we derive some basic properties of homomorphisms.

$\begin{thm} Let $G$\ and $H$\ be two groups with identities $e_G$\ and $e_H$resp... ...item[\rm (ii)] $f(a)^{-1}=f(a^{-1})$\ for every $a\in G$. \end{itemize}\end{thm}$

$\begin{proof} (i) Take arbitrary $x\in G$. Since $xe_G=x$, it follows that $f(x)... ...d the uniqueness of inverses, it follows that $f(a^{-1})=f(a)^{-1}$. \end{proof}$

We also note that there are certain natural subgroups related to homomorphisms.

$\begin{de} Let $f\st G\longrightarrow H$\ be a homomorphism of groups. The set \... ...G)$\ is also denoted by $\im(f)$\ and is called the {\em image} of $f$. \end{de}$

$\begin{ex} % latex2html id marker 3341Let $f\st {\mathbb{Z}}\longrightarrow {\... ... $n$. The image of $f$\ is whole ${\mathbb{Z}}_n$\ (i.e. $f$\ is onto). \end{ex}$

$\begin{exc} % latex2html id marker 3351Prove that for the homomorphism $f$\ fr... ...(f)=SL(n,{\mathbb{R}})$, and that $\im(f)={\mathbb{R}}\setminus\{0\}$. \end{exc}$

$\begin{exc} % latex2html id marker 3356Prove that for the homomorphism $f$\ from Example \ref{ex43} we have $\ker(f)=A_n$\ and $\im(f)={\mathbb{Z}}_2$. \end{exc}$

$\begin{thm} Let $f\st G\longrightarrow H$\ be a homomorphism of groups. \begin{i... ...f(K)$\ of any subgroup $K$\ of $G$\ is a subgroup of $H$. \end{itemize}\end{thm}$

$\begin{proof} (i) Let $a,b\in\ker(f)$, so that $f(a)=f(b)=e_H$. Then we have \be... ...1}=(f(x))^{-1}=f(x^{-1})\in f(K). \end{eqnarray*}Hence $f(K)\leq H$. \end{proof}$

$\begin{exc} For a homomorphism $f\st G\longrightarrow H$\ of groups and a subset... ...H$\ is a subgroup of $G$. Also, prove that $\ker(f)=f^{-1}(\{ e_H\})$. \end{exc}$

$\begin{exc} Prove that a homomorphism $f\st G\longrightarrow H$\ of groups is one-one if and only if $\ker(f)=\{e_G\}$. \end{exc}$

Edmund F Robertson

11 September 2006