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\title{Group Theory: Math30038, Sheet 3}
\author{GCS}
\date{Revised 31-x-2003}
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\maketitle

{\it The course web site is available via
{\tt http://www.bath.ac.uk/$\sim$masgcs/}}

\begin{enumerate}
\item Let $\phi$ denote the Euler $\phi$-function.
Prove that for every integer $n$ we have
\[ \sum \phi(d) = n\]
where the sum is taken over all natural numbers $d$ which
divide $n$. \newline
\noindent {\bf Solution: } In a cyclic group of order $n$, there
are exactly $\phi(d)$ elements of order $d$ where $d$ is any divisor
of $n$.
\item Suppose that $G$ is a finite abelian group. Suppose
that $p$ is a prime number which divides $|G|$. Prove that
there is an element $g \in G$ such that $o(g) = p$.
{\em Hint: multiply together all the cyclic subgroups of $G$.}\newline
\noindent {\bf Solution: } Follow the hint. If each element of $G$
has order coprime to $p$, then it follows from the formula for 
the size of a product of subgroups that $G$ has size coprime with $p$.
This is not the case so there is an element $x \in G$ of order
$pm$. Now $y = x^m$ has order $p$.

\item Suppose that $G$ is a group and that $K$, $L$ are both
normal subgroups with the property that $K \cap L = 1$ (i.e.
$K$ and $L$ intersect to form the trivial subgroup consisting 
of the identity element). Prove that every element of $K$
commutes with every element of $L$. {\em Hint: consider elements
of the form $k^{-1}l^{-1}kl$ where $k \in K$ and $l \in L$.} \newline
\noindent {\bf Solution: } $k^{-1}L^k = L$ and $l^{-1}Kl= K$ by normality
so $k^{-1}l^{-1}k \in L$ and $l^{-1}kl \in K$. Therefore
$k^{-1}l^{-1}kl \in K \cap L = \{ 1 \}.$ Thus $kl = lk$ wherever
$k \in K$ and $l \in L$.

\item Suppose that $G$ is a group and that $H$ is a subgroup
of $G$ of finite index. Suppose that $K$ is also a subgroup of $G$.
Prove that $\vert K : H \cap K \vert \leq \vert G : H \vert.$
What can you say if this inequality is an equality? \newline
\noindent {\bf Solution: } Let $T$ be a right transversal 
for $H \cap K$ in $K$, so if $t, t'\in T$ are distinct,
then $t, t' \in K$
but $tt'^{-1} \not \in H \cap K$. Therefore $tt'^{-1} \not \in H$.
It follows that $Ht,$ $Ht'$ are distinct cosets. The inequality
is established. We claim that equality holds if and only if 
$HK = G$. We prove this as follows. If $HK = G$ then it is 
possible to choose a right transversal $S$ for $H$ in $G$
consisting of elements of $K$. Now if $s, s'$ are distinct elements
of $S$ then $ss'^{-1}  \not \in H$ and so
$ss'^{-1}  \not \in H \cap K$. Therefore $|G:H| \leq |K: H \cap K|.$
However, the reverse inequality was established earlier 
Thus $|G:H| = |K: H \cap K|.$  Next suppose insteab that 
$|G:H| = |K: H \cap K|$. The elements $u$ of
right transversal $U$ for $H \cap K$ in $K$ give rise to distinct
right cosets $Hu$ (why?), but by assumption these are all the right
cosets of $H$ in $G$, thus $HU = G$ and therefore $HK = G$.

\item Let $G$ be a group. Suppose that $H \leq G$ and that
$\vert G : H \vert = 2$. Prove that $H \unlhd G$. Can one derive
the same conclusion when 2 is replaced by 3?  \newline
\noindent {\bf Solution: } Suppose that $g \in G$. If $g \in H$, then
$gH = H = Hg$. On the other hand if $g \not \in G$, then
$gH \not = H$, but there are only two left cosets of $H$ in $G$
so $gH = G - H$ (i.e. the set of elemenst of $G$ which are not
elements of $H$). Similarly $Hg = G - H$ so $gH = Hg$. Thus $H$
is a normal subgroup of $G$. The argument will not work when $2$
is replaced by 3, 
for we may let $G = S_3$, and put $H = \langle (1,2) \rangle.$
Now $H$ has order $2$ and therefore index 3 in $G$. Moreover
$(2,3)H = \{ (2,3), (1,2,3) \}$ but $H(2,3) = \{ (2,3), (1,3,2)\}.$ 
\end{enumerate}
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