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Background: 

I: There are three basic methods for coming up with normal subgroups of finite groups.

(1) You can apply the Sylow theorems. For example, the Sylow theorems tell us
immediately that a group of order 111 must have a normal subgroup
of order 37. 

(2) Normal subgroups also arise as isotropy subgroups of group actions.
For example, if a Sylow 11-subgroup of a group $G$ of order $1452=11^2*12$ is not normal, 
Sylow tells us that there are 12 Sylow 11-subgroups, and that $G$ acts on them transitively by conjugation. 
This action induces a non-trivial homomorphism $G\to S_{12}$, and since 1452 does not divide $12!$, 
the kernel of the homomorphism (i.e., the isotropy subgroup of the action) is neither $G$ nor $(1_G)$, 
so $G$ has a non-trivial normal subgroup. 

(3) Finally, there are counting arguments. For these it is useful to keep in mind that
the union $U_p$ of the conjugates of the Sylow $p$-subgroups is disjoint (apart from the identity)
from the union $U_q$ of the conjugates of the Sylow $q$-subgroups if $p\neq q$,
since $a\in U_p$ has order a positive power of $p$ while $b\in U_q$ has order a positive power of $q$.
Note also that the distinct subgroups of prime order are disjoint, apart from the identity,
since any non-identity element of a subgroup of prime order generates the subgroup.
We can apply these ideas to show that
a group $G$ of order $132=11*12$ cannot be simple. Let $n_p$ be the number of Sylow $p$-subgroups
in $G$. If $G$ were simple, Sylow tells us that $n_{11}=12$, that $n_3\geq 4$, and that $n_2\geq 3$. 
Thus there must be $12(11-1)=120$ non-identity elements in the union of the Sylow 11-subgroups, and
at least $4(3-1)=8$ non-identity elements in the union of the Sylow 3-subgroups. There are also clearly
at least $4+1$ elements in the union of the Sylow 2-subgroups. Since all of these elements must be different,
$G$ must have at least $120+8+5=133$ elements. This is impossible since $|G|=132$.

All three of these techniques come up in Problem [1] (at least, if you do it the way I did it!).

II: The following background will be useful for proving Wedderburn's little theorem
(Problem [5] below,
that a finite division ring is a field). A {\it left module\/} $M$ over a ring $R$ (also known as a left $R$-module)
is an abelian group with an operation $\theta: R\times M\to M$ 
where we denote $\theta((r,m)$ by $rm$ and we think of the operation as a scalar multiplication,
which we require to satisfy the following: 
$a(bm)=(ab)m$ for all $a,b\in R$ and all $m\in M$;
$(a+b)m=am+bm$ for all $a,b\in R$ and all $m\in M$;
and $a(m+n)=am+an$ for all $a\in R$ and $m,n\in M$. 
Lang also requires $1_Rm=m$ for all $m\in M$, while Hungerford
calls a module satisfying this condition a {\it unitary} $R$-module.
We'll go by Lang's definition. When $R$ is not commutative, 
one can also define a right $R$-module where the multiplication is on the right.
When $R$ is commutative, every left $R$-module automatically has
a right $R$-module structure and vice-versa, so one does not talk about left or right
$R$-modules when $R$ is a commutative ring, only about $R$-modules.
The usual facts apply to modules; for example, $-m=(-1_R)m$ and $0_Rm=0_M$.
Here are some examples of modules. If $R$ is a field, an $R$-module is called a {\it vector space}.
If $R$ is a division ring, a left $R$-module is called a {\it left vector space}.
When $k\subseteq K$ are fields, then $K$ is a $k$-vector space and so
has a basis over $k$ such that $K$ is isomorphic (as a $k$-module) to a direct sum
of copies of $k$; the number of copies is called the dimension of $K$ over $k$.
Similarly, when $R\subseteq S$ are division rings,
$S$ is a left $R$-vector space (also a right $R$-vector space, but never mind)
and by the same proof as in the commutative case $S$ has a left basis over $R$, so
$S$ is isomorphic (as a left $R$-module) to a direct sum
of copies of $R$; the number of copies is called the dimension 
(or rank) of $S$ over $R$.

\newpage
\begin{center}{\bf M901, Assignment 6: Due Friday, November 11, 2011}\end{center}

{\it Instructions}: Do any three problems.

\begin{itemize}

\item[(1)] Show that a group $G$ of order $7^2*2^3$ is solvable.
\vskip\baselineskip

\item[(2)] Let $F$ be a finite field. 
\begin{itemize}
\item[(a)] Show that ${\bf Q}[x]$ has irreducible polynomials for every degree $d\geq 1$.
\vskip\baselineskip
\item[(b)] Show that $F[x]$ has infinitely many irreducible polynomials
(mimic Euclid's proof that ${\bf Z}$ has infinitely many primes); conclude
for every integer $d$ that $F[x]$ has an irreducible polynomial $f$ of degree $\deg(f)\geq d$.
\end{itemize}
\vskip\baselineskip

\item[(3)] A subgroup $G$ of a group $H$ is said to be a {\it characteristic\/} subgroup if
for every automorphism $f$ of $H$ we have $f(G)=G$. 
\vskip\baselineskip
\begin{itemize}
\item[(a)] Show that a characteristic subgroup is always normal.
\vskip\baselineskip
\item[(b)] Show that the derived subgroups $H^{(i)}$ of a group $H$ are characteristic subgroups
(and hence normal).
\end{itemize}
\vskip\baselineskip

\item[(4)] Let $D$ be a division ring.
For each $x\in D$, define the centralizer of $x$ in $D$ to be 
$$C_D(x)=\{y\in D:yx=xy\}.$$ 
Define the center $Z(D)$ of $D$ to be 
$\cap _{x\in D}C_D(x)$.
\begin{itemize}
\vskip\baselineskip
\item[(a)] Show that $C_D(x)$ is a division ring.
\vskip\baselineskip
\item[(b)] Show that $Z(D)$ is a field.
\end{itemize}
\vskip\baselineskip

\item[(5)] Let $D$ be a division ring with finitely many elements. Let $x\in D$.
\vskip\baselineskip
\begin{itemize}
\item[(a)] Show that $|Z(D)|=q$ where $q=p^r$ for some prime $p$ and some
integer $r\geq 1$.
\vskip\baselineskip
\item[(b)] Show that $|D|=q^s$ for some integer $s\geq 1$.
\vskip\baselineskip
\item[(c)] Show that $|C_D(x)|=q^t$ for some integer $t\geq 1$ that divides $s$.
\vskip\baselineskip
\item[(d)] Use the class equation applied to the group of non-zero elements of $D$ to show that
$$q^s-1=(q-1)+\sum_i \frac{q^s-1}{q^{t_i}-1}$$
for some positive integers $t_i$, each of which divides $s$.
\vskip\baselineskip
\item[(f)] Look up and apply Zsigmondy's Theorem; conclude that either $s=1$, or $s=2$, or $q=2$ and $s=6$.
\vskip\baselineskip
\item[(g)] Show that $s=2$ is impossible. [Hint: if $s=2$, pick a basis for $D$ over $Z(D)$ and show 
explicitly that $D$ is commutative, implying $s=1$.]
\vskip\baselineskip
\item[(h)] Show that $s=6$ with $q=2$ is impossible. [Hint: plug it into the displayed equation above.]
\vskip\baselineskip
\item[(i)] Conclude that $s=1$, so $D=Z(D)$, hence $D$ is a field. 
\end{itemize}

\end{itemize}
\end{document}
