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\title{Advanced Group Theory: Math0038, Sheet 8}
\author{GCS: Solutions}
\date{}
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\maketitle

{\it There next lectures for this course are
on Friday November 19th and Monday November 22nd.
The course web site is available via
\newline \begin{center} 
{\tt http://www.bath.ac.uk/$\sim$masgcs/}
\end{center}
There is information there about Group Theoretic
facilities and links.}

\begin{enumerate}
\item {\em Hint: Groups of order $p^n, pq$ ($p$ and $q$ distinct primes)
cannot be non-abelian simple groups (why?).}
Show that there is no non-abelian finite simple
group of order less than 60.
\newline{\bf Solution }
{\em The trivial group is not simple by definition. 
If $G$ is a group of order $p^n$ with $p$ prime and $n \geq 1,$
then $Z(G) \not = 1,$ but $Z(G) \unlhd G$ so the simplicity of $G$
forces $G = Z(G)$ to be abelian. Another useful case is where
$|G| = pr$ where $p$ is a prime number and $\sqrt n < p.$
Now Sylow's theorem forces $|\mbox{Syl}_p(G)| = 1$ and simplicity
forces $\mbox{Syl}_p(G) = \{G\}.$ Thus $G$ has prime order and so is abelian.
The integers less than 60 which are not disposed of as candidate
orders for non-abelian simple groups are as follows:
\[ 12, 18, 24, 30, 36, 40, 45, 48, 50, 54, 56\]
Now the integers 18, 50 and 54 force a Sylow subgroup to have index 2, 
and therefore to be normal. Now deploy Poincar\'e's theorem 
on some of the others. In groups of order 12, 24 and 48 a Sylow 2-subgroup
has index 3 but $12, 24$ and $48$ do not divide $3!.$
In a group of order 36 a Sylow 3-subgroup has index 4 but 
36 does not divide 4!.
In a group of order 45 a Sylow $3$-subgroup has index 5 but 45 does not
divide $5!$. 
The remaining candidate orders are
\[ 30, 40 \mbox{ and } 56.\]
A simple group of order 30 would have to contain 6 Sylow 5-subgroups,
and therefore 24 elements of order 5. It would also have to contain 
10 Sylow 3-subgroups and so 20 elements of order 3.
Now $24 + 20 > 30$ so this is absurd. In a group of order $40$
the number of Sylow 5-subgroups must be 1 by Sylow's theorem, so it
cannot be simple. In a simple group of order 56 there must be 8
Sylow 7-subgroups and so 48 elements of order 7. The remaining
$56-48 = 8$ elements of the group must comprise the unique Sylow 2-subgroup,
which must therefore be normal which is absurd.} 

\item Show that there is no non-abelian finite simple group
of order greater than 60 but less that 168.
\newline{\bf Solution }
{\em We use the same two criteria as in the solution to the
previous question to eliminate some orders. The remaining cases are:
\[ 63, 70, 72, 75, 80, 84, 90, 96, 98, 100, 
105, 108, 112, 120,\]\[ 
126, 135, 140, 144, 147, 150, 154, 160, 162, 165.\]
Now Sylow subgroups of index 2 arise when the order
is 98 and 162. Poincare'e theorem will come into
play for the orders 75 because 75 does not divide 3!, 80 because 
80 does not divide 5!, 96 because 96 does not
divide 3!, 100 because 100 does not divide 4!,
108 because 108 does not divide 4!, 135 because
135 does not divide 5!, 147 because 147 does not divide 
3!, 150  because 150 does not divide 6!,
160  because 160 does not divide 5!. The surviving
candidate orders are
\[ 63, 70, 72, 84, 90, 105, 112, 120, 126, 140, 144, 154, 165.\]
In $G$ a simple group of order 63 there must be 7 Sylow $3$-subgroups.
Choose $P, Q$ distinct Sylow $3$-subgroups. Now $|PQ| \leq 63$
so $|P||Q|/|P \cap Q| \leq 63.$ Thus $|T|= 3.$ Now
$T$ is a proper subgroup of each of the finite $3$-groups
$P$ and $Q$, and so must be properly contained in its normalizer
in each of them. However, $T$ is a maximal subgroup of each of $P$ and $Q$,
so $P, Q \leq N_G(T).$ Thus $\langle P, Q \rangle \leq N_G(T),$
but $P$ is a maximal subgroup of $G$ and $Q$ is not a subgroup of $P$
so $\langle P, Q \rangle = G.$ Thus $T \unlhd G$ and we have a contradiction.

That argument is too good to waste. By replacing $3$ by a prime number
$p$ and $7$ by the prime number $q \not = p,$ it shows that
there is no simple group of order $p^2q$ whenever $p^2 > q$. On the
other hand, if $p^2 < q$ then $q > \sqrt{|G|},$ so $G$ is not simple.
Thus no simple group can have order $p^2q.$ 

Now suppose that $G$ is a simple group of order $72 = 2^33^2.$
Choose $P, Q$ distinct Sylow 3-subgroups. Now $|PQ| \leq 72$ 
so $T = P \cap Q$ has order 3, and is normalized by both
$P$ and $Q$. Since $T$ is not normal in $G,$ the only possible
order for $S = N_G(T)$ is 24, so $|G:L| = 3.$ Now Poincar\'e's theorem
forces $|G| = 72$ to divide $3! = 6,$ which is absurd.

Next suppose that $G$ is a simple group of order $84 = 2^2 \cdot 3 \cdot 7.$
Sylow's theorem tells us that there is a unique Sylow 7-subgroup which
must be normal and this is absurd.

Now suppose that $G$ is a simple group of order $90 = 2 \cdot 3^2 \cdot 5.$
There must be 6 Sylow $5$-subgroups, and 10 Sylow 3-subgroups.
There are 24 elements of order 5. If each pair of distinct Sylow 3-subgroups
intersects in the trivial group then there would be 80 elements 
of order 3 or 9. Now $24 + 80 > 90$ so there are $P, Q \in \mbox{Syl}_3(G)$
with $T = P \cap Q$ of order 3. Let $S = N_G(T),$ so
$P, Q \leq S.$ Thus $|S| = 18, 45$ or $90.$ Now $T$ is not normal in
$G$ by simplicity, so $|S| \not = 90.$ If $|S| = 45,$ then
$|G:S| = 2,$ so $S \unlhd G$ which is absurd so
$|S| = 18.$ Now Poincar\'e's theorem applies and we deduce that
$90$ divides $5! = 120$ which is absurd.

Next suppose that $G$ is a simple group of order $105 = 3 \cdot 5 \cdot 7.$
By Sylow's theorem there must 15 subgroups of order 7 and so
90 elements of order 7. Sylow's theorem also tells us that there
are 21 subgroups of order 5 and therefore 84 elements of order
5 in $G$. Now $90 + 84 > 105$ so this is absurd.

Next suppose that $G$ is a simple group of order $112 = 2^4 \cdot 7.$
There must be 7 Sylow 2-subgroups and and 8 Sylow 7-subgroups.
Let $P, Q$ be distinct Sylow 2-subgroups with intersection $T$
of maximal order. Since $256/112 > 2$ it follows that $|T| = 4$ or $8$.
If $T$ has order 8, then both $P$ and $Q$ normalize $T$ so
$T$ is normal in $\langle P, Q \rangle = G.$ This is absurd so 
$|T| = 4.$ Let $S = N_G(T)$. Suppose that $P \leq S,$ then 
because $S \cap Q$ is not contained
in the maximal subgroup $P,$ $T \unlhd \langle P, S \cap Q \rangle = G$
which is absurd. Let $\widehat P = P \cap S$ and $\widehat Q
= Q \cap S.$ These groups $\widehat P, \widehat Q$ both have order
$8$ and normalise $T = \widehat P \cap \widehat Q.$
Now $R = \langle  \widehat P, \widehat Q\rangle$ normalizes $T$, and
so cannot be $G$ and has order strictly larger than $8.$
Thus $|R| = 16$ or $56$. Now $T$ is not normal in $P$ but
$T$ is normal in $R$ so $P \not = R.$ Also $\widehat P$ has order
8 and is a subgroup of both $P$ and $R.$ Since $P, R
\in \mbox{Syl}_2(G)$ this violates the maximality of the order of the intersection
of $P \cap Q.$ Finally we conclude that $|R| = 56$ so $|G:R| = 2$
and therefore $R \unlhd G$ which is absurd.

Next suppose that $G$ is a simple group of order $120 = 2^3 \cdot 3 \cdot 5.$
There must be 6 Sylow-5 subgroups, so if $L$ is the normalizer
of a subgroup of order 5, then $|G:L| = 6.$ By a Poincar\'e argument
there is a monomorphism of $G$ into $S_6.$ Let the image of this
isomorphism be $\widehat G.$ Now $A_6 \cap \widehat G \unlhd \widehat G$
and $|\widehat G: A_6 \cap \widehat G| = |\widehat G A_6: A_6| \leq 2.$
The simplicity of $\widehat G$ forces $\widehat G \leq A_6$
so $A_6$ has a subgroup of index 3. However, $A_6$ is a simple group
so Poincar\'e's argument yields that $360$ divides $3! =6$ which is
absurd.

Next suppose that $G$ is a simple group of order $126 = 2 \cdot 3^2 \cdot 7.$
By Sylow's theorem there must be a unique subgroup of order 7 which
is absurd.

Next suppose that $G$ is a simple group of order $140 = 2^2 \cdot 5 \cdot 7.$
By Sylow's theorem there must be a unique subgroup of order 7 which
is absurd.

Next suppose that $G$ is a simple group of order $144 = 2^4 \cdot 3^2.$
Consider the Sylow $3$-subgroups. 
By Sylow's theorem there must be 4 or 16 of them. However,
if there were just 4, then the normalizer $L$ of a Sylow $3$-subgroup
would have index $4$ in $G,$ and Poincar\'e's argument shows
that $144$ divides $4! = 24$ which is absurd. Therefore
there are 16 Sylow $3$-subgroups. If each pair of distinct
Sylow $3$-subgroups intersect in the trivial group, then $G$ has
128 elements whose order is 3 or 9. The remaining 16 elements
must comprise the unique Sylow $2$-subgroup which violates simplicity.
Therefore there are distinct Sylow 3-subgroups $P,Q$ such
that $P \cap Q = T$ has order 3. Now $P, Q \leq N_G(T),$ so
$|N_G(T)| > 9$ and 9 divides $|N_G(T)|.$ Now $|N_G(T)| \not = 18$
since any group fo order $18$ has a unique Sylow 3-subgroup
by Sylow's theorem. Thus $|N_G(T)| = 36, 72$ or $144$. The last
option does not arise since it would entail $T \unlhd G.$
However, Poincar\'e's argument will not allow
$|N_G(T)| = 36$ since $144$ divides neither $2!$ nor $4!$.

Next suppose that $G$ is a simple group of order $154 = 2 \cdot 7 \cdot 11.$
By Sylow's theorem there must be a unique subgroup of order 11 which
is absurd.

Next suppose that $G$ is a simple group of order $165 = 3 \cdot 5 \cdot 11.$
By Sylow's theorem there must be a unique subgroup of order 11 which
is absurd.}

\item Suppose that $G$ is a finite group with $N \unlhd G$.
Choose $Q \in \mbox{Syl}_p(N).$ Show that $N_G(Q)N = G.$
{\em This proof is called the Frattini argument after an Italian
group theorist.}
\newline{\bf Solution }
{\em Suppose that $g \in G,$ so conjugation by $g$ induces an automorphism
of the normal subgroup $N.$ Thus $Q^g \in \mbox{Syl}_p(N).$ It follows
from Sylow's theorem that $Q$ and $Q^g$ are conjugate in $N.$ Thus
there is $n \in N$ such that $Q^g = Q^n$ so $Q^{gn^{-1}} = Q.$
Therefore $gn^{-1} = x \in N_G(N).$ Now $g = xn \in N_G(Q)N$.
Now $g \in G$ was arbitrary so $G \leq N_G(Q)N \leq G$ and therefore
$G = N_G(Q)N.$}
\item Suppose that $G$ is a finite group and that
$P \in \mbox{Syl}_p(G).$ Suppose that $N_G(P) \leq T \leq G.$
Prove that $N_G(T) = T.$
\newline{\bf Solution }
{\em $T \unlhd  N_G(T)$ and $P \in \mbox{Syl}_p(T)$ so the Frattini
argument applies. We deduce that $N_G(T) = N_G(P)T = T$ since
$N_G(P) \leq T.$ We are done.}
\end{enumerate}
{\em Hint: Use the result of Question 3.}
{\bf Please hand in solutions at the lecture on Monday
November 22nd, or to the folder on the door of 1W3.32
before 18:15 on that day.}
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