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QUESTION


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\item[(a)]
Say what it means for a subset $H$ of a group $(G,e,*)$ to be a
subgroup.

\item[(b)]
Giving brief reasons for your answer, say how many elements of
$S_4$ lie in a subgroup generated by the two elements (12) and
(234).

\item[(c)]
Show that a non-empty subset $H$ of a group $(G,e,*)$ is a
subgroup if $g*h^{-1}\in H$ for all $g,h\in H$.

\item[(d)]
Say what it means for a subgroup to be normal. Show that the
kernel of a homomorphism is always a normal subgroup, first using
the result stated in part (c) to show that it is a subgroup.

\item[(e)]
State Lagrange's theorem and use it to show that any group of
prime order must be cyclic.

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ANSWER


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\item[(a)]
A subgroup  of a group $(G,e,*)$ is a subset $H\subseteq G$
satisfying the following conditions.

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\item[S1)]
If $h,k\in H$, then $h*k\in H$.

\item[S2)]
The identity element of $e\in G$ is also an element of $H$.

\item[S3)]
If $h\in H$ then $h^{-1}\in H$.

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\item[(b)]
They all do, since conjugating (12) by the powers of (234) yields
the transpositions (12),(13),(14) which generate all of $S_4$.

\item[(c)]
Since $H$ is non empty we can choose an element $h\in H$. Putting
$g=h$ we see that $e=g*h^{-1}=h*h^{-1}$ is an element of $H$ by
the hypothesis, so $H$ satisfies axiom S2. Now given any element
$h\in H$ we can put $g=e$ to get $h^{-1}=e*h^{-1}$ as an element
of $H$, so $H$ satisfies axiom S3. Finally given any two elements
$g,h\in h$ we see that $g,h^{-1}\in H$ so $g*(h^{-1})^{-1}\in H$,
or $g*h\in H$. So $H$ also satisfies axiom S1.

\item[(d)]
A subgroup $H<G$ is said to be normal if for any element $c\in G,
gH=Hg$. Let $\phi:G\longleftarrow H$ be a homomorphism with kernel
$K=\{g\in G|\phi(g)=e_h\}$. Clearly $e_G\in K$, so $K$ is
non-empty. Now for any $g,h\in K$ we have
$\phi(gh^{-1})=\phi(g)\phi(h)^{-1}=e_H$ so the kernel is a
subgroup. Furthermore if $k\in K,g\in G$ then
$\phi(g*k)=\phi(g)=\phi(k*g)$ so both $gK$ and $Kg$ are the
pre-image of $\phi(g)$ in $G$ hence $gK=Kg$.

\item[(e)]
Lagrange's theorem: If $H$ is a subgroup of a finite group $G$
then $|H|$ divides $|G|$. If $G$ has a prime order $p$ then any
subgroup of $G$ must have order 1 or $p$. Take a non-trivial
element $g\in G$ and consider the subgroup it generates. It has at
least 2 elements, so it must have $p$ elements, and
$G=\left<g\right>$ as required.

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