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\title{Group Theory: Math30038, Sheet 8}
\date{GCS Solutions}
\author{}
\begin{document}
\maketitle

\begin{enumerate}
\item Let $P$ be a Sylow $p$-subgroup of the finite group $G$.
Suppose that $N \unlhd G$. 
\begin{enumerate}
\item {\it Show that $P \cap N \in \mbox{Syl}_p(N).$} \newline
\noindent {\bf Solution }\ $P \cap N \leq P$ so
$|P \cap N|$ is a power of $p$. Also $PN/N \simeq P/P \cap N$
so $|PN:N| = |P: P \cap N|$ but
\[ |PN:N| \cdot |N : P \cap N| = |PN:P \cap N|=
|PN:P| \cdot |P : P \cap N|\]
and therefore $|N : P \cap N| = |PN:P|$.
Now $|PN:N|$ is a divisor of $|G:P|$ and so is coprime to $p$.
Thus $P \cap N$ is a Sylow $p$-subgroup of $N$.
\item {\it Show that $PN/N \in \mbox{Syl}_p(G/N).$}  \newline
\noindent {\bf Solution }\ $PN/N$ has $p$-power order by the 2nd 
isomorphism theorem. Now $|G:PN|$ divides $|G:P|$ an so is 
coprime to $p$. Now $|G/N : PN/N| = |G:PN|$ since if $T$ is
a left transveral for $PN$ in $G$, then $\{ tN \mid t \in T\}$
is easily seen to be a left transversal for $PN/N$ in $G/N$.
\end{enumerate}
\item {\it Show that every group of order $15$ must be abelian.} \newline
\noindent {\bf Solution }\ Sylow's theorem tells us that there are
unique Sylow 5-subgroups and 3-subgroups $P$ and $Q$ respectively.
These subgroups must be normal in $G$ since conjugation
must leave each one invariant. Now for all $a \in P$ and 
$b \in Q$ we have $a^{-1}b^{-1}ab \in P \cap Q = 1$ so
$a$ and $b$ commute. Now if $a \in A$, then 
$A \leq C_G(a)$ since $A$ is cyclic (of prime order). We have
just shown that $B \leq C_G(A)$ so $\langle A, B \rangle \leq C_G(a)$.
Now $\langle A, B \rangle$ has order divisible by both 3 and 5 and
so $G = \langle A, B \rangle.$ Thus $a \in Z(G)$ so
$A \leq Z(G).$ Similarly $B \leq Z(G)$ so
$G = \langle A, B \rangle \leq Z(G)$ and therefore $G$ is abelian.
\item Show that every group of order $35$ must be abelian. \newline
\noindent {\bf Solution }\ The solution is a copy of the previous
solution.
\item {\it Show that there is no non-abelian finite simple group
of order less than 60.} \newline
\noindent {\bf Solution }\ A group of $p$-power order has a non-trivial 
centre. This will prevent $G$ from being simple unless $G = Z(G)$ is abelian
(in which case $G$ will be simple if and only if $|G| = p$ but this is not 
needed). Thus there is no non-abelian finite simple group
of prime power order. Now using the results of the next question
we may focus on groups of the following orders: 24, 40, 48, 54, 30 and 56.
\begin{enumerate}
\item[(a)] A simple group of order 24 would have 3 Sylow 2-subgroups, and so
there would (see Poincar\'e's theorem) be a non-trivial homomorphism
from $G$ to $S_3$ which will be injective by simplicity and therefore
$24$ would divide 6 which is absurd.
\item[(b)] A group $G$ of order 40  must have  a unique Sylow $5$-subgroup and 
so can not be simple. 
\item[(c)] A simple group of order 48 must have 3 sylow
$2$-subgroups, and so (see |G| = 24) we can deduce that 48 divides
6 which is absurd. 
\item[(d)] A group of order 54 must have a Sylow 3-subgroup
$H$ of index 2 in $G$. Now any subgroup of index 2 is normal so $G$
can not be simple. 
\item[(e)] A simple group of order 30 would have 6 Sylow 5-subgroups.
Since any pair of these groups intersects in the identity, it follows
that there are 24 elements of order 5. Similarly this this group must
have 10 Sylow $3$-subgroups and so 20 elements of order 3. However,
$24 + 20 > 30$ so this is absurd. 
\item[(f)] A simple group of order 56 would
have 8 Sylow $7$-subgroups, any pair of which would intersect in the 
trivial group. Thus there would be 48 elements of order 7. This leaves
8 elements remaining, which must all belong to a Sylow $2$-subgroup. 
This Sylow $2$-subgroup is unique and therefore invariant under
conjugation and thus is normal in $G$. This is absurd. 
\end{enumerate}
\item Let $p$ and $q$ be distinct prime numbers.
\begin{enumerate}
\item {\it Show that a group of order $pq$ can not be simple.}  \newline
\noindent {\bf Solution }\ Suppose w.l.o.g. that $p < q$. By Sylow's
theorem there is a unique Sylow $q$-subgroup $Q$. Now $Q$ is invariant
under conjugation by elements of $G$, and so must be a normal subgroup
of $G$.
\item {\it Show that a group of order $p^2q$ can not be simple.}\newline
\noindent {\bf Solution }\ Suppose (for contradiction) that $G$ is simple.
By Sylow's theorem the number of Sylow $q$-subgroups is $p$ or $p^2$,
and this number must be congruent to 1 modulo $q$. Therefore
$q$ divides $p-1$ or $p^2-1 = (p-1)(p+1)$. In the first event $q< p$.
In the second event $q < p$ or $(p,q) = (2,3).$

Now, if $(p,q) = (2,3)$ it follows that $|G| = 12$. In a non-simple
group of order 12 there must be 4 Sylow 3-subgroups, any pair of 
which intersect in the identity. There are therefore 8 elements of
order 3. The number of elements of order dividing 4 is therefore
at most $12-8 = 4$. Let $P$ be a Sylow $2$ subgroup of $G$, a group
of size 4, all elements of $P$ will have order dividing 4. Thus
$P$ must consist of all the elements of $G$ which are not of order $3$. 
Thus there is a unique Sylow $2$-subgroup which violates simplicity.

Thus we may assume that $q < p$. Now (see Poincar\'e's theorem)
we have a non-trivial homomorphism $G \rightarrow S_q$ which
must be injective by the simplicity of $G$. Now by Lagrange's theorem
$|G|$ divides $q!$ and so $p$ divides $q!$ which is false.
\item {\it Show that a group of order $p^2q^2$ can not be simple.}\newline
\noindent {\bf Solution }\ Assume (for contradiction) that
$G$ is simple. The number 
of Sylow $q$ subgroups $q$ and must be $p$ or $p^2$ (by simplicity)
so $q$ divides $p-1$ or divides $p^2-1 = (p-1)(p+1)$. Thus
either $q < p$ or $q = 3, q = 2$ (so $|G| = 36$). Reversing the
roles of $p$ and $q$ we obtain that either $p < q$ or $|G| = 36$.
Thus we are done unless $|G|= 36$, but then a Sylow 3-subgroup
has index 4 and so (see Poincar\'e's theorem) there is a 
non-trivial homomorphism from $G$ to $S_4$ which must be injective
by simplicity. Now by Lagrange's theorem 36 divides 24 which is absurd. 
\end{enumerate} 
\end{enumerate}  
\end{document}
