algebra problem sets

algebra--nf--1.tex → abstract-algebra--nf--1.tex

@@ -83,9 +83,9 @@ \section*{Problem  Set 1}
 First note that the elements of $\Z/36\Z$ are $\{1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35\}$.
 
 There are $12$ elements, and we can compute the order of an element $x$ by doing calculations like
-\begin{minted}{python3}
-  [(i, (x ** i) % 36) for i in range(1, 12+1)]
-\end{minted}
+% \begin{minted}{python3}
+%   [(i, (x ** i) % 36) for i in range(1, 12+1)]
+% \end{minted}
 and observing when we first see a $1$.
 
 
@@ -226,4 +226,4 @@ \section*{Problem  Set 1}
   It's true for $i = 0$, so let $i \neq 0$. By the Division Algorithm there exist unique $q, r$ such
   that $i = qn + r$ where $0 \leq r < n$.
   Therefore $x^i = x^{qn}x^r = (x^n)^qx^r = x^r \in \{1, x, x^2, \ldots, x^{n-1}\}$.
-\end{proof}
+\end{proof}

abstract-algebra--nf--2.tex

@@ -0,0 +1,300 @@
+\section*{Problem  Set 2}
+
+\begin{mdframed}
+  \includegraphics[width=400pt]{img/algebra--nf--2-d8d7.png}
+\end{mdframed}
+
+\blue{{\it Note: This exercise was not among those set, but is a sort of lemma for exercise 5.}}
+
+{\it {\bf Intuition:} The only one which is its own inverse is the one ``half way around​'' the cycle, which only exists if $n$ is even. }
+
+
+\begin{proof}~\\
+  {\bf (a)} Suppose there exists $i \in \{1, 2, \ldots, n - 1\}$ such that $x^i = x^{-1}$.
+  Then $x^{2i} = e$. Therefore \green{$n | 2i$ and, since $n$ is odd, $n | i$ which
+    contradicts $0 < i < n$} (my original argument was more convoluted).
+
+\begin{comment}
+  {\bf (b)}\\
+  $\implies$~\\
+  Suppose $i = k = n/2$. Then $(x^i)^2 = x^{2i} = x^n = e$, therefore $x^i = x^{-i}$.
+
+
+  $\impliedby$~\\
+  Suppose $x^i = x^{-i}$ for some $i \in \{1, 2, \ldots, n - 1\}$. Then
+  $(x^i)^2 = x^{2i} = e$. Since $|x| = n$, we have $2i \geq n$. Suppose $2i > n$. Then $2i - n < n$
+  since $2i < 2n - 1$. But $x^{2i - n} = x^{2i}x^{-n} = e \cdot e = e$ which is a contradiction
+  since $|x| = n$.
+\end{comment}
+\end{proof}
+
+\begin{mdframed}
+In these exercises, $D_{2n}$ has the presentation $D_{2n} = \langle r, s ~|~ r^n = s^2 = 1, rs = sr^\1 \rangle$.
+\end{mdframed}
+
+\begin{mdframed}
+\includegraphics[width=400pt]{img/algebra--nf--2-21ce.png}
+\end{mdframed}
+
+
+\begin{proof}
+  Let $n$ be odd and let $x$ be a non-identity element of $D_{2n}$.
+
+  We will exhibit, for any given non-identity element of $D_{2n}$, an element with which it does
+  not commute. There are two cases:
+
+  {\bf Case 1}:  $x = r^i$ for some $1 \leq i < n$ \\
+  Take $y = s$. Then $xy = r^is$ and $yx = sr^i = r^{-i}s$ via a dihedral relation. But from
+  exercise 33a above, for odd $n$, we have $r^i \neq r^{-i}$ for all $1 \leq i < n$, hence $xy \neq yx$.
+
+
+  {\bf Case 2:}  $x = r^is$ for some $0 \leq i < n$ \\
+  Take $y = r$. Then $xy = r^isr = r^ir^\1s = r^{i - 1}s$ and
+  \begin{align*}
+    xy &= r^isr = r^ir^\1s = r^{i - 1}s
+  \end{align*}
+  and
+  \begin{align*}
+    yx &= rr^is = r^{i+1}s.
+  \end{align*}
+  But $r^{i-1} \neq r^{i+1}$ so $xy \neq yx$.
+\end{proof}
+
+
+
+\newpage
+\begin{mdframed}
+\includegraphics[width=400pt]{img/algebra--nf--2-9bae.png}\\
+\includegraphics[width=400pt]{img/algebra--nf--2-e89d.png}
+\end{mdframed}
+
+\begin{proof}
+  First note that every element of $D_{2n}$ which is not a power of $r$ can be written
+  as $x = r^is$ for some $0 \leq i < n$, and that such an element is not the identity.
+
+  We see that $x^2 = r^isr^is = sr^{-i}r^is = s^2 = 1$, therefore every element of $D_{2n}$ which is
+  not a power of $r$ has order $2$.
+
+  Furthermore note that $s\cdot sr = r$ hence $D_{2n}$ is generated by $s$ and $sr$, since we can
+  compute the known generators $s$ and $r$ from them.
+\end{proof}
+
+
+\begin{mdframed}
+\includegraphics[width=400pt]{img/algebra--nf--2-bb2c.png}
+\end{mdframed}
+
+
+\begin{proof}
+  Let $a = s$ and $b = sr$ and note that $ab = s^2r$.
+
+  $\implies$\\
+  For the forward implication, suppose $a^2 = b^2 = (ab)^n = 1$. We want to show that the standard
+  relations $r^n = s^2 = 1$ and $rs = sr^\1$ hold.
+
+  $s^2 = 1$ is immediate, since we have $a = s$ by definition and $a^2 = 1$ by supposition.
+
+  This gives us $ab = s^2r = 1r = r$. Therefore $r^n = 1$, since $(ab)^n = 1$ by supposition.
+
+  Finally, we have $b = sr$ and $b^2 = 1$, therefore $srsr = 1$. Now, multiplying on the left
+  by $s^\1$ and on the right by $r^\1$, we have $rs = s^\1r^\1 = sr^\1$ as required, since
+  $s^2 = 1$ implies $s^\1 = s$.
+
+
+
+  $\impliedby$\\
+  For the reverse implication, suppose $r^n = s^2 = 1$ and $rs = sr^\1$. We want to show that $a^2 = b^2 = (ab)^n = 1$.
+
+  $a^2 = 1$ is immediate since we have $a = s$ by definition and $s^2 = 1$ by supposition.
+
+  Next, we have $ab = s^2r = r$ since $s^2 = 1$ by supposition. Therefore $(ab)^n = 1$ since
+  $r^n = 1$ by supposition.
+
+  Finally, we have $rs = sr^\1$ by supposition. Multiplying on the left by $s$ and on the right
+  by $r$, gives $srsr = s^2r^\1r$, hence $b^2 = 1$ since $b = sr$ by definition and $s^2 = 1$ by
+  supposition.
+
+
+  We've shown that the relations for $r$ and $s$ in the standard presentation for $D_{2n}$ hold if
+  and only if the relations $a^2 = b^2 = (ab)^n = 1$ hold. Therefore the latter relations also give a
+  presentation for $D_{2n}$.
+\end{proof}
+
+
+
+
+\newpage
+\begin{mdframed}
+\includegraphics[width=400pt]{img/algebra--nf--2-0db6.png}
+\end{mdframed}
+
+
+\begin{example*}
+\begin{verbatim}
+
+sigma^0       1  2  3  4  5  6         (1)(2)(3)(4)(5)(6)
+sigma^1       6  1  2  3  4  5         (1  2  3  4  5  6)
+sigma^2       5  6  1  2  3  4         (1  3  5)(2  4  6)
+sigma^3       4  5  6  1  2  3         (1  4)(2  5)(3  6)
+sigma^4       3  4  5  6  1  2         (1  5  3)(2  6  4)
+sigma^5       2  3  4  5  6  1         (1  6  5  4  3  2)
+sigma^6       1  2  3  4  5  6         (1)(2)(3)(4)(5)(6)
+\end{verbatim}
+\end{example*}
+
+\begin{proof}
+  Under repeated application of $\sigma^i$, the positions visited by the element $1$ form the
+  sequence $(ki + 1 \mod m)$ for $k=0, 1, 2, \ldots$.
+
+  Note that an element of this sequence is equal to $1$ if and only if $ki$ is a multiple of $m$.
+  Therefore, the first time the sequence returns to $1$ occurs when $ki = \lcm(i, m)$ and we see
+  that in the cycle decomposition of $\sigma^i$, the element $1$ belongs to a cycle of
+  length $\lcm(i, m) / i$.
+
+  Therefore $\sigma^i$ is an $m$-cycle if and only if $\lcm(i, m) = im$, which is true if and only
+  if $i$ and $m$ are relatively prime.
+
+  (To see the last claim, note that from a fundamental theorem regarding $\lcm$ and $\gcd$ we
+  have $\lcm(a, b) = ab / \gcd(a, b)$. Therefore $\lcm(i, m) = im$ is true if and only
+  if $\gcd(i, m) = 1$, i.e. they are relatively prime.)
+\end{proof}
+
+
+
+~\\~\\
+\begin{mdframed}
+\includegraphics[width=400pt]{img/algebra--nf--2-3beb.png}
+\end{mdframed}
+
+\begin{proof}
+  Let $x \in S_n$ with cycle decomposition $x = \sigma_1\sigma_2\cdots\sigma_k$ and
+  let $N = \lcm(|\sigma_1|, |\sigma_2|, \ldots, |\sigma_k|)$. We must show that $N$ is the smallest positive integer
+  such that $x^N = 1$.
+
+  Since the cycles in the decomposition commute and since $N$ is a multiple of the order of every
+  cycle in the decomposition, we have $x^N = (\sigma_1\sigma_2\cdots\sigma_k)^N = \sigma_1^N\sigma_2^N\cdots\sigma_k^N = 1^N1^N\cdots1^N = 1$.
+
+  Finally, suppose that $x^M = 1$ for some $M < N$, so that we have
+  $\sigma_1^M\sigma_2^M\cdots\sigma_k^M = 1$. But since the cycles in the decomposition are disjoint, this implies (*)
+  that $\sigma_i^M = 1$ for every $i = 1, 2, \ldots, k$. Therefore $M$ is a multiple of
+  $|\sigma_i|$. But this contradicts the fact that $N$ is the $\lcm$, therefore no such $M < N$ exists.
+
+  To see the claim (*), suppose that $\sigma_i^M \neq 1$ for some $i$. Then, rearranging the order of the
+  commuting cycles if necessary, we have $\prod_{j \neq i}\sigma_j^M = (\sigma_i^M)^\1$. But this is impossible: the
+  cycles $\sigma_j$ are disjoint, therefore their powers $\sigma_j^M$ are disjoint also, and we cannot form
+  the inverse of a non-identity element $\sigma_i^M$ from a product of elements that are disjoint
+  with $\sigma_i^M$.
+\end{proof}
+
+
+
+~\\~\\
+\begin{mdframed}
+\includegraphics[width=400pt]{img/algebra--nf--2-b38d.png}
+\end{mdframed}
+
+\begin{proof}
+  For $n \geq 4$, a permutation in $S_n$ which is the product of two disjoint $2$-cycles is defined by
+  an unordered specification of two disjoint subsets of $\{1, 2, \ldots, n\}$ of size $2$. The number of
+  ways to do so is equal to
+  \begin{align*}
+    \frac{1}{2}{n \choose 2}{n - 2 \choose 2} = \frac{1}{2}\frac{n(n-1)}{2}\frac{(n-2)(n-3)}{2} = \frac{n(n-1)(n-2)(n-3)}{8},
+  \end{align*}
+  where the factor of $\frac{1}{2}$ is the double-counting correction that converts a count of
+  ordered pairs to a count of unordered pairs.
+\end{proof}
+
+
+
+
+~\\~\\
+6\\
+\begin{mdframed}
+\includegraphics[width=400pt]{img/algebra--nf--2-9aa2.png}
+\end{mdframed}
+
+Note: in this proof, $(\cdots)$ is used to denote a permutation in ``one-line​'' notation; not cycle notation.
+
+
+\begin{proof}
+  Let $\sigma \in S_n$ and for $i, j = 1, 2, \ldots, n$, let $\tau_{i,j}$ be the transposition of elements
+  $i$ and $j$, where $\tau_{i,i} = e$. We construct $\sigma$ as a product of transpositions as follows:
+
+  \begin{enumerate}
+  \item Set $x = (1, 2, \ldots, n)$
+  \item For $i = 1, 2, \ldots, n$
+    \begin{enumerate}
+    \item Let $j$ be the index position of $\sigma(i)$ in $x$.
+    \item Apply the transposition $\tau_{i,j}$ to $x$.
+    \end{enumerate}
+  \end{enumerate}
+  Note that:
+  \begin{itemize}
+  \item Instruction 2a is well-defined because $\sigma(i)$ always occurs at exactly one location in
+    $x$. This is true because $\sigma(i) \in \{1, 2, \ldots, n\}$, and $x$ is initialized to a permutation
+    of $\{1, 2, \ldots, n\}$, and thereafter is altered only by applying transpositions.
+  \item On execution of instruction 2b, $x_i = \sigma(i)$. Thereafter, the value of $x_i$ never changes,
+    since in the subsequent iterations of the loop, instruction 2a always sets $j$ equal to the
+    index position of $\sigma(i')$ for some $i' > i$.
+  \end{itemize}
+
+  Therefore, when this procedure terminates, $x = (\sigma(1), \sigma(2), \ldots, \sigma(n))$, i.e.
+  $x$ is $\sigma$ written in ``one-line​'' notation. Since we used only transpositions in the algorithm,
+  and $\sigma$ was arbitrary, $S_n$ is generated by its transpositions.
+\end{proof}
+
+\begin{mdframed}
+\includegraphics[width=400pt]{img/algebra--nf--2-d8ce.png}
+\end{mdframed}
+
+
+The identity is even since it has zero inversions.
+
+A transposition is odd since it has one inversion.
+
+
+\begin{mdframed}
+\includegraphics[width=400pt]{img/algebra--nf--2-179c.png}
+\end{mdframed}
+
+\begin{proof}
+  Let $\sigma \in S_n$ be any permutation, let $\tau \in S_n$ be a transposition, and let $I_\rho$ be the set of
+  inversions of any permutation $\rho \in S_n$.
+
+  First, suppose that the cycle decomposition of $\sigma$ does not contain $\tau$. Then
+  $\tau\sigma$ contains an inversion that $\sigma$ does not, so that $|I_{\tau\sigma}| = |I_\sigma| + 1$.
+
+  Alternatively, suppose that the cycle decomposition of $\sigma$ contains $\tau$, and write
+  $\sigma = \tau\sigma^-$, where $\sigma^-$ is the product of all cycles other than $\tau$ in the cycle decomposition
+  of $\sigma$. Note that exactly one of the inversions of $\sigma$ is due to $\tau$, so
+  that $|I_\sigma| = |I_\sigma^-| + 1$. We have $\tau\sigma = \tau^2\sigma^- = \sigma^-$, so
+  that $|I_{\tau\sigma}| = |I_{\sigma^-}| = |I_\sigma| - 1$ (this is non-negative, since the cycle decomposition
+  of $\sigma$ contains $\tau$, therefore $|I_\sigma| \ge 1$.)
+
+  In both cases, $|I_{\tau\sigma}| = |I_\sigma| \pm 1$, proving the claim.
+\end{proof}
+
+
+\begin{mdframed}
+\includegraphics[width=400pt]{img/algebra--nf--2-05a9.png}
+\end{mdframed}
+
+\begin{proof}
+  First note that it follows from result (c) above that the product of an even number of
+  transpositions is an even permutation, and the product of an odd number of transpositions is an
+  odd permutation. To see this, note that we can rewrite any product $\tau_1\tau_2\cdots\tau_K$
+  as $\tau_1(\tau_2(\cdots(\tau_K(1))))$ and, since $1$ is an even permutation, it follows from
+  result (c) that this product is even if and only if $K$ is even.
+
+  Therefore, since $S_n$ is generated by its transpositions, an even permutation can be written as
+  a product of an even number of transpositions (but not an odd number) and an odd permutation can
+  be written as a product of an odd number of transpositions (but not an even number).
+
+  The desired result follows: if $\sigma_1$ and $\sigma_2$ are both even (resp. odd), then write them both as
+  products of even (resp. odd) numbers of transpositions, and observe that the total number of
+  transpositions in the product of the two is even, and therefore that that product is an even
+  permutation. Alternatively if one of $\sigma_1$ and $\sigma_2$ is odd and the other even, then write one as
+  a product of an odd number of transpositions and the other as the product of an even number of
+  transpositions, and observe that the total number of transpositions in the product of the two is
+  odd, and therefore that that product is an odd permutation.
+\end{proof}

algebra--nf--all.tex → abstract-algebra--nf--all.tex

@@ -1,6 +1,8 @@
 \documentclass{article}
 \usepackage{mathematics}
 \usepackage[outputdir=.build]{minted} \setminted{fontsize=\footnotesize}
+
+
 % \title{Abstract Algebra - Dummit \& Foote}
 % \author{Dan Davison}
 
@@ -11,5 +13,6 @@
 % \tableofcontents
 % \mainmatter
 
-\include{algebra--nf--1}
+\include{abstract-algebra--nf--1}
+\include{abstract-algebra--nf--2}
 \end{document}