Math 6310 • Algebra

\begin{enumerate}
\item[(a)]
Let \(H\) be a maximal subgroup of the finite group \(G\) and let \(M\) be a minimal normal subgroup of
\(G\). Assume that \(M\) is abelian but \(H\) does not contain \(M\). Prove that \(G=HM\),
\(H\cap M=\{1\}\) (the identity subgroup), and \(|G:H|\) (the index of \(H\) in \(G\)) is a power of a
prime.
\item[(b)]
Use the result in (a) to show that that if \(G\) is a finite solvable group and \(H\) is a maximal
subgroup of \(G\), then \(|G:H|\) is a power of a prime.
\end{enumerate}