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QUESTION

\begin{description}

\item[(a)]
Define the following terms

\begin{description}

\item[(i)]
direct product,

\item[(ii)]
isomorphism,

\item[(iii)]
normal subgroup.

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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).

\item[(c)]
Let $G$ be a group with identity element $e$ and let $H$ and $K$
be subgroups of $G$ with $H\cap K=\{e\}$. Show that if $hk=kh$ for
any $h\in H$ and any $k\in K$ then the function $f:H\times
K\longrightarrow G$ given $f(h,k)=hk$ is an injective
homomorphism. Show that if $G$ is a group in which every element
has order 2 then $G$ is abelian, and deduce that any two
non-identity elements of $G$ generate a subgroup isomorphic to the
Klein 4-group.

Give an example to show that an abelian group can contain two
elements of order 3 without containing a subgroup isomorphic to
$Z_3\times Z_3$.

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ANSWER


\begin{description}

\item[(a)]

\begin{description}

\item[(i)]
$(G,*),(H,.)$ are groups.

$\{(g,h)|g\in G,h\in H\}=G\times H$ with
$(g_1,h_1)(g_2,h_2)=(g_1*g_2,h_1.h_2)$ is the direct product.

\item[(ii)]
An isomorphism is a bijective function $f:G\longrightarrow H$ with
$f(g*k)=f(g).f(k)\forall g,k\in G$.

\item[(iii)]
A subgroup $H<G$ is normal if $g^{-1}Hg=H\forall g\in G$.

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\item[(b)]
\begin{eqnarray*}
f(g^{-1}kg)&=&f(g^{-1}f(k)f(g)\forall g\in G, k\in
\textrm{kernel}\\ &=&f(g^{-1})e_Hf(g)\\ &=&f(g^{-1})f(g)\\
&=&f(g^{-1}g)=f(e_G)=e_H
\end{eqnarray*}

\item[(c)]
$f(h,k)=e\Leftrightarrow hk=e\Leftrightarrow h=k^{-1}$. But
$h=k^{-1}\Rightarrow h\in H\cap K=\{e\}$ so $h=e$.

Similarly $k=e$ and $\textrm{Ker}(f)=\{(e,e)\}$ and $f$ is
injective.

\begin{eqnarray*}
f((h_1,k_1)(h_2,k_2))&=&f(h_1h_2,k_1k_2)\\ &=&h_1h_2k_1k_2\\
&=&h_1k_1h_2k_2\\ &=&f(h_1,k_1)f(h_2,k_2)
\end{eqnarray*}

If every element in $G$ has order 2 then $(gh)^2=e\forall g,h\in
G$ and $g=g^{-1},\ h=h^{-1}$ so
$e=(gh)^2=ghgh=ghg^{-1}h^{-1}\Rightarrow gh=hg \forall g,h\in G$.

Now $\left<g,h\right>=\left<g\right>\times\left<h\right>$ since
the map $f:\left<g\times\right>\longrightarrow G$ is an
isomorphism onto its image.

$C_3$ contains 2 elements of order 2 but is not isomorphic to
$c_3\times C_3$

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