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\begin{document}

\maketitle

\begin{center}
{{\Large\bf ALGEBRA I}\\
\bigskip

{\Large\bf HOMEWORK ASSIGNMENT VII}\\
\bigskip
{\bf (Turn in underlined problems.)}}
\end{center}

\vspace{.5in}

\noindent
{\bf 
\begin{tabbing}
\underline{\large\bf CHAPTER}\makebox[2in]{}\=\underline{\large\bf PROBLEMS}\\

\bigskip
\noindent
3. \> 3.2, \underline{3.3}, 3.4, 3.9, 
\underline{3.25}, \underline{3.26}, 3.33

\end{tabbing}

}
\vspace{.5in}

\noindent
\underline{\large\bf ADDITIONAL PROBLEMS}
\vspace{.25in}

{\bf 

We want to prove that any chief factor $N_i/N_{i+1}$
of a finite group is isomorphic to a power of a simple group. By
forming the quotient $G/N_{i+1}$, it is easy to see
that it suffices to prove only that the {\it first} 
(or {\it bottom}) chief
factor in a chief series is isomorphic 
to a power of a simple group.
That is, we want to prove that if $N$
is a minimal normal subgroup of $G$, then $N\cong S^k$ for
some simple group $S$. We will prove this by 
induction on the length of a composition series for $G$.
The basis for induction is the case where $G = N = S$ is a simple
group. This case is trivial, so the exercises concentrate 
on proving the inductive step.

Let $N$ be a minimal normal subgroup of a finite group $G$.
Since we are not in the basis case, $N\neq G$. There is
a composition series for $G$ that involves $N$, so
there is a composition series for $N$
that is shorter than any composition series for $G$.
By induction, if $K$ is a minimal normal subgroup of $N$,
then $K\cong S^m$ for some simple group $S$. 
If we show that $N\cong K^n$ for some $n$, then
$N\cong (S^m)^n\cong S^{mn}$ and we are done.
So let's show that $N\cong K^n$ for some $n$!

\bigskip

1. Show that the conjugates of $K$ in $G$
are isomorphic, pairwise disjoint subgroups
that are all minimal normal subgroups of $N$.

\bigskip

\underline{2.} Show that the conjugates of $K$
are subgroups of $G$ whose join equals $N$.

\bigskip

3. If $\{K, K', K'', \ldots, \}$
are the conjugates of $K$ in $G$, show that it is possible 
to pick a sequence of these groups: $K_1, K_2, \ldots, K_{\ell}$
with the following properties:
\begin{enumerate}
\item[(1)] $(K_1\vee K_2\vee\cdots\vee K_i)\cap K_{i+1} = \{1\}$
for $i=1,\ldots,{\ell}-1$; and
\item[(2)] $K_1\vee K_2\vee\cdots\vee K_{\ell} = N$.
\end{enumerate}

\bigskip

\underline{4.} Use the above results to
show that $N\cong K_1\times \cdots\times K_{\ell}\cong 
K^{\ell}$.

\end{document}

