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QUESTION


\begin{description}

\item[(a)]
State the three axioms A1-A3 for a group $(G,e,*)$.

\item[(b)]
Show that every finite group is isomorphic to a group of
permutations of a finite set.

\item[(c)]

\begin{description}

\item[(i)]
Carefully prove the following statement using only the axioms for
a group, and saying which axiom you use at each stage of the
argument: Given a group $(G,e,*)$ and elements $g,h,k\in G$, if
$g*h=g*k$ then $h=k$.

\item[(ii)]
Use your result from the previous part to prove the following
statement: Given a group $(g,e,*)$ and an element $g\in G$ there
is a unique element $k\in G$ such that $g*k=e$.

\item[(iii)]
Given a group $(G,e,*)$ and elements $g,k\in G$ show that if
$g*k=e$ than $k*g=e$.

\end{description}

\item[(d)]
Let $(G,e,*)$ be a group. Show that the binary operation $\cdot$
on $G$, defined by $a\cdot b=b*a$, is a group operation on $G$.
Show also that the map $\phi:G\longleftarrow G$ defined by
$g\mapsto g^{-1}$ is an isomorphism from the group $(G,e,*)$ to
the group $(G,e,\cdot)$.

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ANSWER


\begin{description}

\item[(a)]

\begin{description}

\item[A1)]
For any three element $g,h,k\in G, (g*h)*k=g*(h*k)$.

\item[A2)]
There is an element $e\in G$ such that for any element $g\in G$,
we have $e*g=g=g*e$.

\item[A3)]
For any element $g\in G$ there is an element $g^{-1}$ (called the
inverse of $g$) such that $g*g^{-1}=g^{-1}*g=e$.

\end{description}

\item[(b)]
For each element $g\in G$ define a function $f_g:G\Longrightarrow
G$ by $f_g(k)=g*k$. This is a bijection since its inverse is given
by the function $f_{g^{-1}}$. The function $\phi:G\Longrightarrow
S_G$ given by $g\mapsto f_g$ is then an isomorphism since
$\phi(g*h)$ is the function $f_{g*h}$ defined by $k\mapsto
(g*h)*k$, which by associativity is the same as $f_g\circ f_h(k)$,
hence $\phi(g*h)=\phi(g*h)=\phi(g)\circ \phi(h)$ as required.

\item[(c)]

\begin{description}

\item[(i)]
$g*h=g*k\Rightarrow
g^{-1}*(g*h)=g^{-1}*(g*h)\Rightarrow_{A1}(g^{-1}*g)*h=(g^{-1}*g)*k
\Rightarrow_{A1}e*h=e*k\Rightarrow_{A2}h=k$.

\item[(ii)]
By A3 there is at least one element, denoted $g^{-1}$ such that
$g*g^{-1}=e$. Now suppose there is another denoted $k$. Then
$g*g^{-1}=g*k\Rightarrow g^{-1}=k$ by the previous result.

\item[(iii)]
$g*k=e\Rightarrow
k*(g*k)=k*e=_{A2}k\Rightarrow_{A1}(k*g)*k=k\Rightarrow((k*g)*k)*k^{-1}=k*k^{-1}=_{A3}e$.
Now by A1
$((k*g)*k)*k^{-1}=(k*g)*(k*k^{-1})=_{A3}(K*g)*e=_{A2}k*g$.

Hence $k*g=e$ as required.

\end{description}

\item[(d)]

\begin{description}

\item[A1)]
For any elements $a,b,c\in G,
a.(b.c)=(b.c)*a=(c*b)*a=c*(b*a)=(b*a).c=(a.b).c$.

\item[A2)]
For any $a\in G,\ a.e=e*a=a=a*e=e.a$.

\item[A3)]
For any $a\in G$ let $a^{-1}$ denote the inverse of $a$ for the
binary operation $*$. Then
$a.a^{-1}=a^{-1}*a=e=a*a^{-1}=a^{-1}.a$.

\end{description}
For any $a,b\in
G.\phi(a*b)=(a*b)^{-1}=b^{-1}*a^{-1}=\phi(b)*\phi(a)=\phi(a),\phi(b)$,
so $\phi$ is a homomorphism. Sine $\phi\circ\phi(a)=a$ for any
$a\in G,\ \phi$ is invertible and therefore an isomorphism.

\end{description}




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