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QUESTION

\begin{description}

\item[(a)]
Define the terms \textit{homomorphism} and \textit{isomorphism}.

\item[(b)]
In each of the following cases decides whether or not the given
pair of groups are isomorphic. If they are give an explicit
isomorphism between them, and if they are not give brief reasons
how you can be sure of this.

\begin{description}

\item[(i)]
$Z_3\times Z_5$ and $Z_{15}$.

\item[(ii)]
$Z_3\times Z_6$ and $Z_{18}$.

\item[(iii)]
$D_5$ and $Z_{10}$.

\end{description}

\item[(c)]
For each positive integer $n$ less than or equal to 8 give a list
of groups of order $n$ so that no two groups in your list are
isomorphic, but every group of order $n$ is isomorphic to one of
the groups on your list. For each order where there is more than
one isomorphism class of group of that order, indicate how you
distinguish between the different isomorphism classes.

\item[(d)]
Decide which of the groups in your list from part (iii) is
isomorphic to the group given by the following Cayley table,
giving brief reasons for your answer:


\begin{tabular}{c|cccccccc}
.&$e$&$g$&$c$&$b$&$h$&$d$&$f$&$k$\\ \hline
$e$&$e$&$g$&$c$&$b$&$h$&$d$&$f$&$k$\\
$g$&$g$&$c$&$b$&$e$&$k$&$h$&$d$&$f$\\
$c$&$c$&$b$&$e$&$g$&$f$&$k$&$h$&$d$\\
$b$&$b$&$e$&$g$&$c$&$d$&$f$&$k$&$h$\\
$h$&$h$&$d$&$f$&$k$&$c$&$b$&$e$&$g$\\
$d$&$d$&$f$&$k$&$h$&$g$&$c$&$b$&$e$\\
$f$&$f$&$k$&$h$&$d$&$e$&$g$&$c$&$b$\\
$k$&$k$&$h$&$d$&$f$&$b$&$e$&$g$&$c$
\end{tabular}


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ANSWER


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\item[(a)]
Given groups $(G,e_G,*)$ and $(H,e_H,.)$ a homomorphism is a
function $\phi:G\Longrightarrow H$ such that for every $g,k\in G,
\phi(g*h)=\phi(g),\phi(h)$. An isomorphism is a bijective
homomorphism.

\item[(b)]

\begin{description}

\item[(i)]
The function $Z_3\times Z_3\Longrightarrow Z_{15}$ defined by
$(1,1)\mapsto1$ is an isomorphism since hcf(3,5)=1.

\item[(ii)]
$Z_3\times z_6$ is not isomorphic to $Z_{18}$ since every element
of $Z_3\times Z_6$ has order dividing lcm(3,6)=6.

\item[(iii)]
$D_5$ is not abelian so it cannot be isomorphic to $Z_{10}$.

\end{description}

\item[(c)]

\begin{tabular}{c|c|l}
Group&order&distinguishing features\\ \hline $\{e\}$&1\\ $Z_2$&2\\
$Z_3$&3\\ $Z_4$&4&contains an element of order 4,\\ &&and two
elements of order 2\\ $z_2\times Z_2$&2&3 elements of order 2\\
$Z)_5$&5\\ $Z_6$&6&abelian\\ $D_3$&6&non-abelian\\ $Z_7$&7\\
$Z_8$&8&abelian with one element of order 2\\ $Z_4\times
Z_2$&8&abelian with three elements of order 2\\ $Z_2\times
Z_2\times Z_2$&8&abelian with seven elements of order 2\\
$D_4$&8&non-abelian with five elements of order 2\\
$\mathcal{Q}$&8&non-abelian with one element of order 2\\
\end{tabular}

\item[(d)]
$hg=d\neq k=gh$ so the group is non-abelian. It has one element of
order 2 so it is isomorphic to $\mathcal{Q}$.

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