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QUESTION

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\item[(a)]
Define the following terms:

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\item[(i)]
homomorphism,

\item[(ii)]
kernel,

\item[(iii)]
isomorphism,

\item[(iv)]
normal subgroup,

\item[(v)]
quotient group.

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\item[(b)]
Show that the kernel of a homomorphism is a normal subgroup (you
may assume that it is a subgroup), and state and prove the First
Isomorphism Theorem.. Illustrate the theorem by using an example
of a surjective homomorphism from $S_4$ to $\mathbf{Z}_2$.

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ANSWER


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\item[(a)]

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\item[(i)]
A homomorphism is a function $f:G\rightarrow H$ between groups $G$
and $H$, such that $f(gk)=f(g)f(k)$ for every $g,k\in G$.

\item[(ii)]
The kernel of a homomorphism is the set ker$(f)=\{g\in
G|f(g)=e_H\}$

\item[(iii)]
An isomorphism is a bijective homomorphism.

\item[(iv)]
A normal subgroup $H\triangleleft G$ is a subgroup such that
$g^{0-1}Hg=H\ \forall g\in G$

\item[(v)]
The quotient $\frac{G}{N}$ is the group of left cosets $\{gN|g\in
G\}$ with $gNg'N=gg'N$.

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\item[(b)]
Let $K=$ker$F$ and $g\in G$. Then $g^{-1}Kg=\{g^{-1}kg|f(k)=e_H\}$
so for any element $g^{-1}kg\in g^{-1}Kg$ we have
$f(g^{-1}kg)=f(g^{-1})f(k)f(g)=f(g^{-1}e_Hf(g)=f(g)^{-1}e_Hf(g)=e_H$

The First Isomorphism Theorem

Let $f:G\Rightarrow H$ be a surjective homomorphism. Then
$\overline{f}:\begin{array}{c}\frac{G}{\textrm{ker}f}\Rightarrow
H\\g\textrm{ker}f\mapsto f(g)\end{array}$ is an isomorphism.

Proof

Let $K=$ker$f$. $\overline{f}$ is well defined since if $gK=g'K$
then $\overline{f}(gK)=f(g)$ and $\overline{f}(g'K)=f(g;)$, but
$g\in g'K$ so $g=g'k$ for some $k\in\textrm{ker}f\Rightarrow
f(g)=f(g'k)=f(g')f(j)=f(g')e_H=f(g')$ as required.

$\overline{f}$ is a homomorphism since
$\overline{f}(g'KgK)=\overline{f}(g'gg^{-1}KgK)=\overline{f}(g'gKK)
=\overline{f}(g'gK)=g'g=\overline{f}(g'K)\overline{f}(gK)$

$\overline{f}$ is surjective since $f$ was (for any $h\in H\exists
g\in G$ with $f(g)=h$ so $\overline{f}(gK)=h$)

$\overline{f}$ is injective since
$\overline{f}(gK=e_H\Leftrightarrow f(g)=e_H\Leftrightarrow g\in
K\Leftrightarrow gK=K$.

Let sgn:$S_n\rightarrow\mathbf{Z}_2$ denote the sign homomorphism
with the kernel $A_n$ so by the theorem $\frac{S_n}{A_N}$ is
isomorphic to $\mathbf{Z}_2$. It's elements are $A_n$ and
$(12)A_n$ and its multiplication table is

\begin{tabular}{c|cc}
&$A_n$&$(12)A_n$\\ \hline $A_n$&$A_n$&$(12)A_n$\\
$(12)A_n$&$(12)A_N$&$A_n$
\end{tabular}

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