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\begin{center}
{\large\textbf{Dartmouth College}\\}
Mathematics 101\\\medskip Homework 3 \small{(due Wednesday, Oct 17)}
\end{center}
\begin{enumerate}

\item Let $G$ be a finite group, and for positive integers $n$ define:
  \begin{align*}
    \delta(n) &= \#\{x \in G \mid |x| = n\}, \textrm{\ and}\\
    \varepsilon(n) &= \#\{x \in G \mid x^n = e\}.
  \end{align*}
  \begin{enumerate}
  \item Show that $\varepsilon(n) = \sum_{m\mid n}
    \delta(m)$.
  \item Show that $\varepsilon(n) \le n$ implies $\delta(n) = 0$ or
    $\phi(n)$; here $\phi$ is the Euler
    $\phi$-function.

  \item If $G$ is cyclic and $m \mid |G|$, show that $\delta(m) =
    \phi(m)$ and $\varepsilon(m) = m$.
  \item Show that $G$ is cyclic if and only if $\varepsilon(n) \le n$
    for all $n\ge 1$.   \textit{Hint:} It may be useful to show that 
    $\varepsilon(n) = \varepsilon(d )$, where $d = \gcd (n, |G|).$

  \end{enumerate}

\item Let $G$ be a group and let $G'$ be the subgroup of $G$ generated
  by the set \newline $\{xyx^{-1}y^{-1} \mid x,y \in G\}$.  $G'$ is
  called the commutator subgroup of $G$.
  \begin{enumerate}
  \item Show that if $H$ is a subgroup of $G$, then $H \supseteq G'$
    if and only if $H \triangleleft G$ and $G/H$ is abelian.  In
    particular, $G'\triangleleft G$ and $G/G'$ is abelian.

  \item Show that if $\varphi: G \to H$ is a homomorphism of groups,
    and $H$ is abelian, then $\varphi$ factors through $G/G'$, that is
    there is a map $\varphi_* : G/G' \to H$ with $\varphi = \varphi_*
    \circ \pi$ with $\pi: G \to G/G'$ the standard projection.
  \end{enumerate}


\item Solvable groups.

  \begin{enumerate}
  \item Let $G$ be a group, and $H$ a normal subgroup.  Show that $G$
    is solvable if and only if $H$ and $G/H$ are solvable.

  \item Let $G$ be a group, $H \normal G$ with $G/H$ finite and
    abelian.  Show that there is a sequence of subgroups $H_i$ with $H
    = H_0 \subseteq H_1 \subseteq \cdots \subseteq H_n = G$ with $H_i
    \normal H_{i+1}$ and $H_{i+1}/H_i$ cyclic for $0 \le i < n$.

  \item For a group $G$, define a sequence of subgroups $G^{(k)}$, by
    $G^{(1)} = G'$ (the commutator subgroup), and $G^{(k+1)} =
    [G^{(k)}]'$, the commutator of $G^{(k)}$.  Show that $G$ is
    solvable if and only if $G^{(k)} = \{e\}$ for some $k \ge 1$.
  \end{enumerate}

 






\end{enumerate}


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