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{\Large {\bf Homework 8}} \hfill MATH 301/601 \\
{Solutions to Graded Problems\\[-2mm]
\line(1,0){470}

\medskip





\medskip
\begin{Exercise}[Section 6.5 \#18]
If \( [G:H] = 2 \), prove that \( gH = Hg \).
\end{Exercise}

\begin{proof}
As the index of \( H \) in \( G \) is two, we know that \( H \) has two left cosets; moreover, we know that these two left cosets are disjoint and that their union is all of \( G \) (as they partition \( G \)).
Therefore, the two left costs of \( H \) in \( G \) are \( H \) and \( G \smallsetminus H \). 
The same holds for right cosets, so the two right cosets of \( H \) in \( G \) are also \( H \) and \( G \ssm H \). 

Now, let \( g \in G \). 
If \( g \in H \), then \( g H = H \) and \( Hg = H \), so \( gH = Hg \).
Otherwise, \( g \in G \ssm H \), implying \( gH = G \ssm H \) and \( Hg = G \ssm H \), so \( gH = Hg \).
Thus, \( gH = Hg \). 
\end{proof}


\setcounter{Exercise}5
\medskip
\begin{ExChallenge}
Let \( G \) be a group acting on a set \( X \). 
Let \( x \in X \). 
\begin{enumerate}[(a)]
\item Let \( g, h \in G \).
Prove that \( gx = hx \) if and only if \( h^{-1}g \in \mathrm{Stab}_G(x) \). 

\begin{Solution}
If \( gx = hx \), then \( h^{-1}(gx) = h^{-1}(hx) = (h^{-1}h)x \), and so \( (h^{-1}g)x = x \), implying \( h^{-1}g \in \mathrm{Stab}_G(x) \).
For the converse, if \( h^{-1}g \in \mathrm{Stab}_G(x) \), then \[ x =  (h^{-1}g) x = h^{-1}(gx), \] implying \( h x = g x \). 
\end{Solution}

\item Let \( \mathcal L \) be the set of left cosets of \( \mathrm{Stab}_G(x) \) in \( G \).
Let \( \psi \co \mathcal L \to \mathcal O_x \) be given by \( \psi(g \mathrm{Stab}_G(x)) = gx \).
	\begin{enumerate}[(i)]
		\item Prove that \( \psi \) is a well-defined.
		
		\begin{Solution}
			Suppose \( g\mathrm{Stab}_G(x) = h\mathrm{Stab}_G(x) \).
			Then there exists \( s \in \mathrm{Stab}_G(x) \) such that \( gs = h \), and hence
			\[ \psi(g \mathrm{Stab}_G(x)) = g x = g(sx)=(gs)x = hx = \psi(h \mathrm{Stab}_G(x)), \]
			implying \( \psi \) is well-defined. 
		\end{Solution}
		\item Prove that \( \psi \) is bijective.
		
		\begin{Solution}
			The surjectivity of \( \psi \) follows immediately from the definition.
			Suppose that \( \psi(g\mathrm{Stab}_G(x)) = \psi(h \mathrm{Stab}_G(x)) \).
			Then \( g x = h x \), which implies that \( h^{-1}g \in \mathrm{Stab}_G(x) \). 
			Therefore, \( g \mathrm{Stab}_G(x) = h \mathrm{Stab}_G(x) \), and \( \psi \) is injective.
		\end{Solution}
	\end{enumerate} 
\item The previous part implies that \( |\mathcal O_x| = [G: \mathrm{Stab}_G(x)] \).
Apply Lagrange's theorem to obtain \( |G| = |\mathcal O_x|\cdot |\mathrm{Stab}_G(x)| \). 

\begin{Solution}
By Lagrange's theorem, \( |G| = | \mathrm{Stab}_G(x) | \cdot [G: \mathrm{Stab}_G(x)] \), and hence \( |G| = | \mathrm{Stab}_G(x)| \cdot | \mathcal O_x | \) by part (b). 
\end{Solution}
\item Now suppose \( G \) is a finite group, and let \( G \) act on itself by conjugation, that is, the action is given by  \( g\cdot a = gag^{-1} \).
Apply the orbit--stabilizer theorem to show that, for \( a \in G \), the cardinality of the set \( \{ gag^{-1} : g \in G\} \) (that is, the conjugacy class of \( a \)) divides \( |G| \). 

\begin{Solution}
The orbit of \( a \) under the conjugation action is exactly the conjugacy class of \( a \). 
By the Orbit-Stabilizer theorem, \( |G| \) is a multiple of \( |\mathcal O_a| \). 
\end{Solution}
\end{enumerate}
\end{ExChallenge}



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