\documentclass[a4paper,12pt]{article}

\begin{document}

\parindent=0pt

QUESTION


Define the following terms:

\begin{description}

\item[(i)]
homomorphism

\item[(ii)]
isomorphism

\end{description}

Let $G$ be a group. Show that the function $f:G\Rightarrow G$
given by $f(g)=g^2$ is a homomorphism if and only if $G$ is
abelian. Show that if $G$ is finite, abelian and has no elements
of order 2 then $f$ is an isomorphism.

Define Euler's $\phi$-function, state Euler's $\phi$ formula and
use it to find the number of elements $m\in \mathbf{Z}_8$ such
that $\left<m\right>=\mathbf{Z}_8$.



ANSWER


\begin{description}

\item[(i)]
A homomorphism is a function $f:G\rightarrow H$ between two groups
$(G,e_G,*)$ and $(H,e_H,.)$ such that for any elements $g_1,g_2\in
G,\ f(g_1*g_2)=f(g_1).f(g_2)$.

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

\end{description}

Suppose first that $G$ is abelian, so for any $g_1,g_2\in G,\
g_1*g_2=g_2*g_1$. Then
$f(g_1*g_2)=(g_1*g_2)^2=g_1*g_2*g_1*g_2=g_1*(g_1*g_2)*g_2=g_1^2*g_2^2=g(g_1)*f(g_2)$
so $f$ is a homomorphism.

Now suppose $h$ is a homomorphism so for any $g_1,g_2\in G\
f(g_1*g_2)=f(g_1)*f(g_2)$.

Then $(g_1*g_2)^2=g_1^2g_2^2\forall g_1,g_2\in G$ so

\begin{eqnarray*}
&&g_1*g_2*g_1*g_2=g_1*g_1*g_2*g_2\\
&\Rightarrow&g_1^{-1}*g_1*g_2*g_1*g_2*g_2^{-1}=g_1^{-1}*g_1*g_1*g_2*g_2*g_2^{-1}\\
&\Rightarrow&g_2*g_1=g_1*g_2\forall g_1,g_2\in G
\end{eqnarray*}

If $G$ is abelian then $f$ is a homomorphism. If $G$ has no
elements of order 2 then $f(g)=e\Leftrightarrow
g^2=e\Leftrightarrow g=e$ so ker$f=\{e\}$. Hence $f$ is injective.
Since $G$ is finite any injective function $f:G\rightarrow G$ is
also surjective so $f$ is an isomorphism.

$\phi(n)$=number of integers less than $n$ and coprime to it.

Euler's formula $\phi(n)=\sum_{d|n}\phi(d)$

$\phi(8)$=number of generators of $Z_8$= number of integers
coprime to 8 and less than 8= number of odd integers less then 8=
4 so there are 4 generators.




\end{document}