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\newcommand{\ZZ}{\mathbb{Z}}
\DeclareMathOperator{\im}{im}
\renewcommand{\phi}{\varphi}
\DeclareMathOperator{\Aut}{Aut}
\DeclareMathOperator{\Inn}{Inn}

\begin{document}

\noindent
Math 541 \\
Problem Set 7
\begin{enumerate}


\item[3.5.12.]  Write $\ZZ/2\ZZ$ additively, so $\epsilon(\sigma)$ is 0 if $\sigma$ is even and 1 if $\sigma$ is odd, and $\epsilon(\sigma \tau) = \epsilon(\sigma) + \epsilon(\tau)$.  Consider $S_n$ as a subgroup of $S_{n+2}$.  Let $\tau = (n-1\ n) \in C_G(S_n)$.  Define a map $\phi: S_n \to S_{n+2}$ by $\phi(\sigma) = \tau^{\epsilon(\sigma)} \sigma$.  Then $\phi$ is a homomorphism:
\[ \phi(\sigma_1) \phi(\sigma_2) = \tau^{\epsilon(\sigma_1)} \sigma_1 \tau^{\epsilon(\sigma_2)} \sigma_2 = \tau^{\epsilon(\sigma_1) + \epsilon(\sigma_2)} \sigma_1 \sigma_2 = \tau^{\epsilon(\sigma_1 \sigma_2)} \sigma_1 \sigma_2 = \phi(\sigma_1 \sigma_2). \]
Thus $\im \phi$ is a subgroup.  In fact $\im \phi \subseteq A_{n+2}$:
\[ \epsilon(\tau^{\epsilon(\sigma)} \sigma) = \epsilon(\tau^{\epsilon(\sigma)}) + \epsilon(\sigma) = \epsilon(\sigma) \cdot \epsilon(\tau) + \epsilon(\sigma) = \epsilon(\sigma) \cdot 1 + \epsilon(\sigma) = 0. \]
$\phi$ is clearly injective, hence is an isomorphism onto its image.


\item[4.3.2.]
\begin{enumerate}
\item The elements of $D_8$ are $1, r, r^2, r^3, s, sr, sr^2, sr^3$.  If we conjugate these by $r$ we get $1, r, r^2, r^3, sr^2$, $sr^3, s, sr$.  If we conjugate by $s$ we get $1, r^3, r^2, r, s, sr^3, s^2, sr$.  Thus the conjugacy classes are $\{1\}$, $\{ r^2 \}$, $\{ r, r^3 \}$, $\{ s, sr^2 \}$, and $\{ sr, sr^3 \}$.  The class equation is $8 = 2 + 2 + 2 + 2$.

\item The elements of $Q_8$ are $1, -1, i, -i, j, -j, k, -k$.  If we conjugate by $i$ we get $1, -1, i, -i, -j, j, -k, k$.  If we conjugate by $j$ we get $1, -1, -i, i, j, -j, -k, k$.  Since $k = ij$, conjugating by it won't give us more information.  Thus the conjugacy classes are $\{1\}$, $\{-1\}$, $\{i,-i\}$, $\{j,-j\}$, and $\{k,-k\}$.  The class equation is $8 = 2 + 2 + 2 + 2$.

\item The elements of $A_4$ are the identity, the three products of disjoint transpositions $(1\ 2)(3\ 4)$, $(1\ 3)(2\ 4)$, and $(1\ 4)(2\ 3)$, and eight 3-cycles: for each $n \in \{ 1, 2, 3, 4 \}$, there are two 3-cycles that leave $n$ fixed.

The products of transpositions are all conjugate to one another: $(1\ 3)(2\ 4) = (1\ 3\ 2) \circ (1\ 2)(3\ 4) \circ (1\ 2\ 3)$ and $(1\ 4)(2\ 3) = (2\ 4\ 3) \circ (1\ 2)(3\ 4) \circ (2\ 3\ 4)$.  They have order 2, hence are not conjugate to the 3-cycles, which have order 3.

The order of a conjugacy class divides the order of the group, so the eight 3-cycles do not form a single conjugacy class.  But if $\sigma$ is a 3-cycle fixing $n$ and $\tau \in A_4$ then $\tau \circ \sigma \circ \tau^{-1}$ is a 3-cycle fixing $\tau(n)$, so each 3-cycle is conjugate to at least three others, so in fact the 3-cycles split into two conjugacy classes of four elements each.

Thus the conjugacy classes of $A_4$ are $\{1\}$, $\{ (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3) \}$, $\{ (1\ 2\ 3), (2\ 4\ 3), (1\ 3\ 4), (1\ 4\ 2) \}$, and $\{ (1\ 3\ 2), (2\ 3\ 4), (1\ 4\ 3), (1\ 2\ 4) \}$.  The class equation is $12 = 1 + 3 + 4 + 4$.
\end{enumerate}


\item[4.3.4.]  To show that $N_G(g S g^{-1}) = g N_G(S) g^{-1}$, observe that the following are equivalent:
\begin{gather*}
x \in N_G(g S g^{-1}) \\
x g S g^{-1} x^{-1} = g S g^{-1} \\
g^{-1} x g S g^{-1} x^{-1} g = S \\
g^{-1} x g \in N_G(S) \\
x \in g N_G(S) g^{-1}.
\end{gather*}
To show that $C_G(g S g^{-1}) = g C_G(S) g^{-1}$, observe that the following are equivalent:
\begin{gather*}
x \in C_G(g S g^{-1}) \\
x g s g^{-1} x^{-1} = g s g^{-1} \quad \forall s \in S \\
g^{-1} x g s g^{-1} x^{-1} g = s \quad \forall s \in S \\
g^{-1} x g \in C_G(S) \\
x \in g C_G(S) g^{-1}.
\end{gather*}


\item[4.4.1.] For all $x \in G$ we have
\[ \sigma(\phi_g(\sigma^{-1}(x))) = \sigma(g \sigma^{-1}(x) g^{-1}) = \sigma(g) x \sigma(g)^{-1} = \phi_{\sigma(g)}(x), \]
as desired.  Thus $\sigma \Inn(G) \sigma^{-1} \subseteq \Inn(G)$, so by Theorem 3.6(5), $\Inn(G) \trianglelefteq \Aut(G)$.  (That $\Inn(G) \le \Aut(G)$ is shown on page 134).


\item[4.4.13.] Observe that the conjugacy class of an element $g \in G$ has order 1 if and only if $g \in Z(G)$.  The order of a conjugacy class divides the order of the group, so a conjugacy class may have order 1, 7, 29, or 203.  Since $H$ is normal, it is a union of conjugacy classes, so either it consists of 7 conjugacy classes of order 1, hence is contained in $Z(G)$, or it consists of a single conjugacy class of order 7, but this is impossible since $1 \in H$.

Now choose $x \notin H$ and let $K = \langle H, x \rangle$.  Since $H \subseteq Z(G)$, $K$ is abelian, as we showed two weeks ago in Exercise 2.4.3.  Now $|K| > 7$, $|K|$ divides 203, and 7 divides $|K|$ since $H \le K$, so $|K| = 203$, so $K = G$.  Thus $G$ is abelian.


\end{enumerate}
\end{document}