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QUESTION


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\item[(a)]
Define the direct product of two groups $(G,e_g,*)$ and $H,e_H,.)$
and prove that the direct product of $(G,e_g,*)$ and $(H,e_H,.)$
is isomorphic to the direct product of $(H,e_H.)$ and $(G,e_g,*)$

\item[(b)]
State the internal direct products theorem. Use it to prove that:

\begin{description}

\item[(i)]
the group of symmetries of a regular hexagon is isomorphic to
$S_3\times Z_2$ (you may assume the classification of groups of
order 6).

\item[(ii)]
The permutations (123) and (45) generate a subgroup of $S_5$
isomorphic to the cyclic group $Z_6$.

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ANSWER


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\item[(a)]
The direct product has $G\times H=\{(g,h)|g\in G,h\in H\}$ for its
elements, $(e_G,e_H)$ for its identity element and group operation
$(g_1,h_1)@(g_2,h_2)=(g_1*g_2,h_1.h_2)$. The function
$\phi((g,h))=(h,g)$ defines a map $G\times H\longleftarrow H\times
G$ which is bijective. We will denote multiplication in $H\times
G$ by juxtaposition.

\item[(b)]
The internal direct product theorem states that if $H$ and $K$ are
subgroups of a group $G$ such that $H\cap K=\{e\}$ and every
element $h\in H$ commutes with every element $k\in K$ then the
subgroup they generate, $\left<H\cup K\right>$, is isomorphic to
the direct product $H\times K$.

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\item[(i)]
The subgroup $H$ generated by the rotation of $\pi$ around the
center is central of order 2. The subgroup $K$ generated by
reflections in lines joining opposite vertices is a non-abelian
group of order 6 so is isomorphic to $S_3$. Since the angle
between any two of these lines is $\frac{2\pi}{3}$ this subgroup
does not contain the rotation of $\pi$ so $H\cap K=\{e\}$, and by
centrality of $H$, $hk=kh$ for any $h\in H$, and any $k\in K$, So
$\left<h\cup K\right>$ is isomorphic to $S_3\times z_2$ as
required.

\item[(ii)]
The cycles (123) and (45) are disjoint and commute so they
generate commuting cyclic subgroups $H$ and $K$ of orders 2,3
respectively. Clearly $H\cap K=\{e\}$ so they generate a group
isomorphic to $Z_2\times Z_3$ which is isomrphic to $Z_6$ since
hcf(2,3)=1.

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