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QUESTION

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\item[(a)]
Define the following terms:

\begin{description}

\item[(i)]
Subgroup.

\item[(ii)]
Right coset.

\item[(iii)]
The order of a group.

\end{description}

\item[(b)]
State Lagrange's Theorem. Let $G$ be a group and $H$ a subgroup of
$G$. Show that the relation $d\sim h\Leftrightarrow gh^{-1}\in H$
is an equivalence relation and describe its equivalence classes.
Show that the equivalence classes all have the same number of
elements. Use this to prove Lagrange's Theorem. Define the order
of a group element and explain how and why Lagrange's Theorem
constrains the order of an element in a finite group.

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ANSWER


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\item[(a)]

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\item[(i)]
A subgroup $H<G$ is a subset $H\subseteq G$ which:

\begin{enumerate}

\item
is closed under group multiplication $h,k\in  H\Rightarrow hk\in
H$

\item
is closed under taking inverses:$h\in H\Rightarrow h^{-1}\in H$

\item
contains the identity:$e\in H$.

\end{enumerate}

\item[(ii)]
The right coset $H_g=\left\{hg|h\in H\right\}$.

\item[(iii)]
The order of $G,\ |G|$= number of elements in $G$.

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\item[(b)]
Lagrange's Theorem

If $G$ is a finite group and $H$ is a subgroup of $G$ then $|H|$
divides $|G|$.

$g\sim h\Leftrightarrow gh^{-1}\in H$ is

\begin{description}

\item[(i)]
reflexive since $gg^{-1}=e\in H$

\item[(ii)]
symmetric since $gk^{-1}\in H\Leftrightarrow(gk^{-1})^{-1}\in
H\Leftrightarrow kg^{-1}\in H\Leftrightarrow k\sim g$

\item[(iii)]
transitive since if $f\sim g\sim k$ then $fg^{-1}\in H$ and
$gk^{-1}\in H$ so $f^{-1}ggk^{-1}\in H$ i.e. $fk^{-1}\in H$ and
$f\sim k$

The class $[g]=\{k\in G|k\sim g\}=\{k|kg^{-1}\in H\}=\{k|k\in
Hg\}=$right coset $H_g$

Given two right cosets $H_g$ and $H_k$ define $\phi:H_g\rightarrow
H_k$ by $\phi(hg)=hk$.

This is injective since $\phi(hg)=\phi(hg')\Leftrightarrow
hk'\Leftrightarrow k=k'$.

It is surjective since the pre-image of $hk$ is $hg$.

The equivalence classes therefore partition $G$ into equal sized
subsets, all with the same number of elements as $H=[e]$, so
$|G|=n|H|$ where $n$ is the number of distinct equivalence
classes.

The order of a group element $g$ is the least positive integer $n$
such that $g^n=e$, or $\infty$ if none such exists. It is also
equal to the number of elements in the cyclic subgroup
$\left<g\right>$ which must therefore divide $|G|$.

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