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\begin{center} {\Large\bf Honors Algebra \\ Assignment 4 Solutions}
\end{center}
\ben
\item In this problem, assume (important!) that $G$ is an {\em Abelian}.
Set $H=\{g\in G: g^5=e\}$. (Warning: Expressions such as $x^{1/5}$ are
{\em not} well defined. Do not use them!)
\ben
\item Show $H$ is a subgroup of $G$. Point out where the assumption that
$G$ was Abelian was used.
\\ Solution. As usual, three parts.
\\ {\tt Identity:} As $e^5=e$, $e\in H$.
\\ {\tt Product:} If $x,y\in H$ then $x^5=y^5=e$ so that $(xy)^5=x^5y^5=ee=e$
so that $xy\in H$ -- where we need that the group is Abelian to say that
\[ (xy)^5=xyxyxyxyxy=xxxxxyyyyy=x^5y^5 \]
\\ {\tt Inverse:} If $xin H$ then $x^5=e$ so that $(x^{-1})^5 = (x^5)^{-1}=e^{-1}=e$
so $x^{-1}\in H$.
\item Show $H$ is a normal subgroup of $G$.
\\ $G$ is Abelian so that all subgroups of $G$ are Normal!
\item Assume further that $G$ is finite and that $H=\{e\}$. Show
that the map $\phi:G\ra G$ given by $\phi(g)=g^5$ is an automorphism.
(Definition: An automorphism is an isomorphism from a group to itself.)
\\ Proof: $\phi$ is a homomorphism as
\[ \phi(xy)=(xy)^5=x^5y^5=\phi(x)\phi(y) \mbox{ for all } x,y\in G \]
$\phi$ is injective as the kernal \[ K_{\phi}=\{g\in G: \phi(g)=e\} = H= \{e\} \]
by assumption, and we proved in class that $\phi$ is injective if and only if the
kernal is $\{e\}$. Finally, since $G$ is finite, an injective map from $G$ to itself
{\em must} be surjective. This is the pigeonhole principle. Say $G$ has $m$ elements.
We have a map $\phi$ from $m$ elements (pigeons) to $m$ elements (pigeonholes) and no
two pigeons go into the same pigeonhole (injectivity) so all pigeonholes are filled
(surjectivity). BTW -- it goes the other way -- surjectivity would imply injectivity.
But the set MUST be finite!
\een
\item
Let $G$ be any group and $g$ a fixed element of $G$. Define
$\phi: G\ra G$ by $\phi(x) = g^{-1}xg$.
\ben
\item Show that $\phi$ is a homomorphism.
\\ Solution:
\[ \phi(xy)=gxg^{-1}gyg^{-1}=g(xy)g^{-1}=\phi(x)\phi(y) \]
\item Show that $\phi$ is an injection. To do this you have to
show that the equation $\phi(x)=e$ has {\em only} the solution
$x=e$.
\\ Solution: If $g^{-1}xg=e$ then premultiply by $g$, postmultiply
by $g^{-1}$ so $x=gg^{-1}=e$.
\item Show that $\phi$ is a surjection. To do this you have to
show, given any $y\in G$, that the equation $\phi(x)=y$ has a
solution. [With these three you may conclude that $\phi$ is an
automorphism as defined above.]
\\ Solution: Set $x=gyg^{-1}$.
\item Find an isomorphism $\phi: (Z_{12},+)\ra (Z_{13}^*,\cdot)$.
(Idea: $\phi(1)$ determines $\phi$, try various $\phi(1)$ until
one works.)
\\ Solution: $\phi(1)=2$ works (others do as well) giving
\begin{center}\begin{tabular}{rrrrrrrrrrrr}
0&1&2&3&4&5&6&7&8&9&10&11\\
1&2&4&8&3&6&12&11&9&5&10&7
\end{tabular}
\end{center}
\item Let $G$ be the positive reals under multiplication and let
$H$ be numbers $2^i$ where $i\in Z$.
\ben
\item Show $H$ is a subgroup of $G$.
\\ Solution: $1=2^0\in H$. If $x,y\in H$ then $x=2^i,y=2^j$ so
$xy=2^{i+j}\in H$. If $x\in H$ then $x=2^i$ so $x^{-1}=2^{-i}\in H$.
\item Show $H$ is a Normal subgroup of $G$.
\\ Solution: $G$ is Abelian so all subgroups are Normal.
\item Give a natural representation of the factor group $G/H$.
By this I mean that each element should be uniquely describable
as $\ol{a}$ where $a$ ranges over some natural set. (So, for
example, you couldn't have $3.1$ and $12.4$ as $\ol{3.1}=\ol{12.4}$.)
Have the identity represented as $\ol{1}$.
\\ Solution: Restrict $a$ to $1\leq a < 2$. The $\ol{a}$ are
distinct as if $1\leq a < b < 2$ then $1