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\noindent
\parbox{2.25in}{\bf Math 4010-101}
\hfill {\Large \bf Homework \#4} \hfill
\parbox{2.25in}{\bf \hfill Due: Tue., Jan. $21^\mathrm{st}$, 2014}

\vspace{0.2in}

\begin{enumerate}
\item Suppose $a,b \in G$. We write $[a,b]=aba^{-1}b^{-1}$. This is called the {\it commutator} of $a$ and $b$. Define:
$$ G' = \langle [a,b] \;|\; a,b \in G \rangle = \langle aba^{-1}b^{-1} \;|\; a,b \in G \rangle $$
So $G'$ is the subgroup \underline{generated} by the commutators.\footnote{One might ask, ``Is $\{ aba^{-1}b^{-1} \;|\; a,b \in G \}$ a subgroup?'' In other words, can we avoid using ``{\it generated by}'' and just use the set of commutators? It turns out that the answer is ``No.'' But the first example of a group whose commutators do not form a subgroup is a group of order 96.} $G'$ is called the {\it derived} or {\it commutator} subgroup of $G$.

\begin{enumerate}[(a)]
\item Prove that $G' \lhd G$ (we already know it's a subgroup, so just show it's normal). What does $G' = \{ 1\}$ mean?
\item Let $N \lhd G$. Prove that $\diagquotient{G}{N}$ is Abelian if and only if $G' \subset N$.\\ {}[This means that $G/G'$ is the largest Abelian quotient of $G$.] 
\end{enumerate}

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\item Prove that $S_4$ does not have a subgroup isomorphic to $Q$ (the group of Quaternions).\\ {}[{\it Hint:} Remember that subgroups of $S_4$ are either all even or half even/half odd.]

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\item Let $H \leq K \leq G$ (subgroups).

\begin{enumerate}[(a)]
\item Assume $G$ is finite. Prove that $[G:H] = [G:K] \; [K:H]$.

\item {}[{\bf Grad Problem}] Do part (a) again. But now without assuming $G$ is finite.
\end{enumerate}

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\item Suppose that $H,K \lhd G$ and that $G=HK$ (which, since the subgroups are normal, implies that $G=KH$ as well). Prove that $\diagquotient{G}{H \cap K} \cong \diagquotient{G}{H} \times \diagquotient{G}{K}$.

\end{enumerate}

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