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Math 541 \\
Problem Set 4
\begin{enumerate}


\item[1.7.8.]
\begin{enumerate}
\item Let $\sigma, \tau \in S_A$ and let $a_1, \dotsc, a_k \in A$ be distinct.  Then $\sigma$ is a bijection, so $\sigma(a_1), \dotsb, \sigma(a_k)$ are distinct, so we really do have a map $G \times B \to B$ and not to some different range.  Now $\sigma \cdot (\tau \cdot \{ a_1, \dotsc, a_k \}) = \sigma \cdot \{ \tau(a_1), \dotsc, \tau(a_k) \} = \{ \sigma(\tau(a_1)), \dotsc, \sigma(\tau(a_k)) \} = (\sigma \circ \tau) \{ a_1, \dotsc, a_k \}$, and $1 \cdot \{ a_1, \dotsc, a_k \} = \{ 1(a_1), \dotsc, 1(a_k) \} = \{ a_1, \dotsc, a_k \}$, so we have a group action.
\item $(1\ 2)$ acts as follows:
\begin{align*}
\{ 1, 2 \} &\mapsto \{ 1, 2 \} &
\{ 1, 3 \} &\mapsto \{ 2, 3 \} &
\{ 1, 4 \} &\mapsto \{ 2, 4 \} \\
\{ 2, 3 \} &\mapsto \{ 1, 3 \} &
\{ 2, 4 \} &\mapsto \{ 1, 4 \} &
\{ 3, 4 \} &\mapsto \{ 3, 4 \}.
\end{align*}
$(1\ 2\ 3)$ acts as follows:
\begin{align*}
\{ 1, 2 \} &\mapsto \{ 2, 3 \} &
\{ 1, 3 \} &\mapsto \{ 1, 2 \} &
\{ 1, 4 \} &\mapsto \{ 2, 4 \} \\
\{ 2, 3 \} &\mapsto \{ 1, 3 \} &
\{ 2, 4 \} &\mapsto \{ 3, 4 \} &
\{ 3, 4 \} &\mapsto \{ 1, 4 \}.
\end{align*}
\end{enumerate}


\item[2.1.2.] Throughout, let $H$ denote the subset in question.  In each case, $H$ is not closed under products.
\begin{enumerate}
\item $(1\ 2) \circ (1\ 3) = (1\ 3\ 2)$, which is not a 2-cycle.  Even more problematic, $1 \notin H$.
\item Again, $1 \notin H$, but also $s \cdot sr = r$, which does not have order 2, hence is not a reflection.
\item Let $n = ab$, where $1 < a < n$ and $1 < b < n$.  If $x \in G$ has order $n$ then $x^a$ has order $b$, so $x^a \notin H$.
\item $1 + 1 = 2$, which is not odd.
\item $\frac{1}{\sqrt 2} + \frac{1}{\sqrt 3} \notin H$ since $\left( \frac{1}{\sqrt 2} + \frac{1}{\sqrt 3} \right)^2 = \frac{5}{6} + \sqrt\frac{2}{3}$.
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\item[2.1.8.]  If $K \subseteq H$ then $H \cup K = H$, which is a subgroup.  Similarly, if $H \subseteq K$ then $H \cup K = K$.  Inversely, suppose there exist $h \in H - K$ and $k \in K - H$.  If $hk \in H$ then $k = h^{-1} \cdot hk \in H$ since $H$ is a subgroup, but $k \notin H$, so $hk \notin H$.  Similarly, $hk \notin K$.  Thus $hk \notin H \cup K$, so $H \cup K$ is not a subgroup.


\item[2.2.3.]  Suppose $g \in C_G(B)$.  For all $a \in A$, $a \in B$, so $gag^{-1} = a$.  Thus $g \in C_G(A)$.  Now $C_G(B)$ is a subset of $C_G(A)$, and it is shown in the text that $C_G(B)$ is a subgroup of $G$, so it is nonempty and closed under products and inverses, hence is a subgroup of $C_G(A)$.


\item[2.2.6.]
\begin{enumerate}
\item Fix $h \in H$.  Then $h H h^{-1} \subset H$, since for all $x \in H$, $h x h^{-1} \in H$; and $H \subset h H h^{-1}$, since for all $x \in H$, $h^{-1} x h \in H$ and $x = h (h^{-1} x h) h^{-1}$, so $x \in h H h^{-1}$.  Thus $h H h^{-1} = H$, so $h \in N_G(H)$.

For a counterexample when $H$ is not a subgroup, consider the transpositions $a = (1\ 2)$ and $b = (2\ 3)$ in $S_3$, and take $H = \{ a, b \}$.  Then $a b a^{-1} = (1\ 2) (2\ 3) (1\ 2) = (1\ 3) \notin H$, so $a H a^{-1} \not\subset H$, so $a \notin N_G(H)$.

\item The following are equivalent: $H \subset C_G(H)$; for all $h \in H$, $h \in C_G(H)$; for all $h \in H$ and all $x \in H$, $h x h^{-1} = x$; for all $h, x \in H$, $hx = xh$; $H$ is abelian.
\end{enumerate}


\end{enumerate}
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