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\noindent{\Large{\textsc{Modern Algebra \hfill Assignment 15.2  \\ Spring 2023 \hfill Due April 28}}}

\large

\vspace{1cm}


\begin{exercise}
Let $G$ be a finite group and let $H \le G$.  We have seen that $G$ acts on the coset space $G/H$ by left translation:
\[
g(xH) = (gx)H
\]
for all $g \in G$ and $xH \in G/H$.  Because this is an action, the map $\lambda:G \to \operatorname{Perm}(G/H)$ given by $g \mapsto \lambda_g$, where $\lambda_g(xH) = g(xH) = (gx)H$, is a homomorphism.

\begin{itemize}

\item[\bf a.]  If $K = \ker \lambda$, show that $K \le H$.  

\item[\bf b.]  Conclude that $[G:H]$ divides $[G:K]$.

\end{itemize}

\end{exercise}

\begin{exercise}
Continuing the preceding exercise, suppose $p$ is the smallest prime dividing $|G|$, and that $[G:H] = p$.  Then $\operatorname{Perm}(G/H) \cong S_p$ and the First Isomorphism Theorem applied to $\lambda$ yields an embedding $G/K \hookrightarrow S_p$.  

\begin{itemize}

\item[\bf a.]  Explain why $[G:K]$ divides $p!$.

\item[\bf b.]  Use the preceding exercise to write $[G:K] = mp$ for some $m \in \N$.  Use part {\bf a} to show that $m$ divides $(p-1)!$.  

\item[\bf c.]  If a prime $q$ divides $m$, show that $q<p$ and $q$ divides $|G|$.  Conclude that $m=1$, so that $[G:K]=p$.

\item[\bf d.]  Show that $[H:K]=1$.  Conclude that $H \triangleleft G$.

\end{itemize}

\end{exercise}

\noindent{\bf Remark.}  Exercise 2 proves that if $G$ is a finite group, $H \le G$ and $[G:H]$ is the smallest prime dividing $|G|$, then $H \triangleleft G$.  We proved a weaker version of this result in an earlier exercise, in the case that $|G|$ is even and $[G:H] = 2$, by directly comparing the left and right coset spaces $G/H$ and $H \setminus G$.

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