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\noindent
Math 541 \\
Problem Set 3
\begin{enumerate}


\item[1.3.2.] $\sigma = (1\ 13\ 5\ 10) (3\ 15\ 8) (4\ 14\ 11\ 7\ 12\ 9)$. \\
$\tau = (1\ 14) (2\ 9\ 15\ 13\ 4) (3\ 10) (5\ 12\ 7) (8\ 11)$.


\item[1.3.15.] First we show that an $m$-cycle has order $m$.  If $\sigma = (a_1\ a_2\ \dots\ a_m)$ is a $m$-cycle, then $\sigma(a_1) = a_2$, $\sigma^2(a_1) = a_3$, $\sigma^3(a_1) = a_4$, and similarly $\sigma^k(a_1)$ is different from $a_1$ for all $k < m$, but $\sigma^m$ is the identity.

Now let $\sigma_1, \dotsc, \sigma_k$ be disjoint cycles of lengths $m_1, \dotsc, m_k$, let $\sigma = \sigma_1 \dotsb \sigma_m$, and let $m$ be the least common multiple of $m_1, \dotsc, m_k$.  Since disjoint cycles commute, $\sigma^m = {\sigma_1}^m \dotsb {\sigma_k}^m = 1$, so $|\sigma| \le m$.  On the other hand, let $p = |\sigma|$.  Then $1 = \sigma^p = {\sigma_1}^p \dotsb {\sigma_k}^p$, and ${\sigma_1}^p, \dotsc, {\sigma_k}^p$ are still disjoint cycles, so ${\sigma_i}^p = 1$ for all $i = 1, \dotsc, k$.  Thus $p$ is a multiple of each $m_i$, so $m \le p = |\sigma|$.


\item[1.4.8.] Every field contains a 0 and a 1 which are not equal, so let
\[
A = \begin{pmatrix}
1 & 1 \\
0 & 1 \\
  &   & 1 \\
  &   &   & \ddots \\
  &   &   &        & 1
\end{pmatrix}
\qquad
B = \begin{pmatrix}
1 & 0 \\
1 & 1 \\
  &   & 1 \\
  &   &   & \ddots \\
  &   &   &        & 1
\end{pmatrix}.
\]
Then
\[
A B = \begin{pmatrix}
2 & 1 \\
1 & 1 \\
  &   & 1 \\
  &   &   & \ddots \\
  &   &   &        & 1
\end{pmatrix}
\qquad
B A = \begin{pmatrix}
1 & 1 \\
1 & 2 \\
  &   & 1 \\
  &   &   & \ddots \\
  &   &   &        & 1
\end{pmatrix}
\]
where $2 = 1+1 \ne 1$.  Thus $AB \ne BA$, so $GL_n(F)$ is not abelian.


\item[1.6.1.]
\begin{enumerate}
\item $\phi(x^n) = \phi(\overbrace{x \cdot x \dotsb x}^{n\text{ times}}) = \overbrace{\phi(x) \phi(x) \dotsb \phi(x)}^{n\text{ times}} = \phi(x)^n$.
\item First, $\phi(1) = \phi(1 \cdot 1) = \phi(1) \phi(1)$, so $1 = \phi(1)$, so the claim is true for $n = 0$.  Next, $\phi(x) \phi(x^{-1}) = \phi(x x^{-1}) = \phi(1) = 1$, so $\phi(x^{-1}) = \phi(x)^{-1}$.  If $n > 0$ then
\[ \phi(x^{-n}) = \phi(\overbrace{x^{-1} x^{-1} \dotsb x^{-1}}^{n\text{ times}}) = \overbrace{\phi(x^{-1}) \phi(x^{-1}) \dotsb \phi(x^{-1})}^{n\text{ times}} = \overbrace{\phi(x)^{-1} \phi(x)^{-1} \dotsb \phi(x)^{-1}}^{n\text{ times}} = \phi(x)^{-n}. \]
\end{enumerate}


\item[1.6.9.] An element of $S_4$ is one of the following: the identity, a transposition, a product of disjoint transpositions, a 3-cycle, or a 4-cycle.  Thus an element of $S_4$ has order at most 4.  But $D_{24}$ has an element of order 24, namely $r$.


\end{enumerate}
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