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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 $f(xy)=f(x)f(y)$. (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