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\pagestyle{fancy}
\lhead{MATH 521--01}
\chead{Problem Set \#3\answer{ solutions}}
\rhead{}
\lfoot{}
\cfoot{\answer{Page \thepage\ of \pageref{LastPage}}}
\rfoot{due Thursday, October 2, 2014}

\begin{document}
\begin{enumerate}
\item \question{Prove that $(1234)$ cannot be written as the product
    of 3-cycles.}
  
  \answer{Each 3-cycle is an even permutation, noting that
    $(abc)=(ab)(bc)$ is a way of writing $(abc)$ as a product of two
    (i.e., an even number of) swaps. Thus any product of 3-cycles is
    also an even permutation. But $(1234)=(12)(23)(34)$, which
    demonstrates that it is odd and not even.}
  
\item \question{Find an element of $S_9$ with order 20, and prove that
    no element of $S_8$ has order 20.}
  
  \answer{$(1234)(56789)$ is such an element of $S_9$, since it is a
    product of a $4$-cycle and a $5$-cycle and $\lcm(4,5)=20$.

    The possible orders in $S_8$ are those equal to the least common
    multiples of the elements of a partition of the number 8. There
    aren't that many of them (22 in total), and it's easy to check
    them all:
    \begin{align*}
      8&=8&\lcm(8)&=8\\
      &=7+1&\lcm(7,1)&=7\\
      &=6+2&\lcm(6,2)&=6\\
      &=6+1+1&\lcm(6,1,1)&=6\\
      &=5+3&\lcm(5,3)&=15\\
      &=5+2+1&\lcm(5,2,1)&=10\\
      &=5+1+1+1&\lcm(5,1,1,1)&=5\\
      &=4+4&\lcm(4,4)&=4\\
      &=4+3+1&\lcm(4,3,1)&=12\\
      &=4+2+2&\lcm(4,2,2)&=4\\
      &=4+2+1+1&\lcm(4,2,1,1)&=4\\
      &=4+1+1+1+1&\lcm(4,1,1,1,1)&=4\\
      &=3+3+2&\lcm(3,3,2)&=6\\
      &=3+3+1+1&\lcm(3,3,1,1)&=3\\
      &=3+2+2+1&\lcm(3,2,2,1)&=6\\
      &=3+2+1+1+1&\lcm(3,2,1,1,1)&=6\\
      &=3+1+1+1+1+1&\lcm(3,1,1,1,1,1)&=3\\
      &=2+2+2+2&\lcm(2,2,2,2)&=2\\
      &=2+2+2+1+1&\lcm(2,2,2,1,1)&=2\\
      &=2+2+1+1+1+1&\lcm(2,2,1,1,1,1)&=2\\
      &=2+1+1+1+1+1+1&\lcm(2,1,1,1,1,1,1)&=2\\
      &=1+1+1+1+1+1+1+1&\lcm(1,1,1,1,1,1,1,1)&=1\\
    \end{align*}
    and we note that 20 does not appear on the right side of this
    list: the elements of $S_8$ have orders 1, 2, 3, 4, 5, 6, 7, 8, 10,
    12, and 15.}

\item \question{Prove that $S_4$ is not isomorphic to $D_{12}$.}

  \answer{One easy way to demonstrate this is to note that every
    element of $S_4$ has order of 4 or less, whereas $D_{12}$ has
    elements of order 6 and 12. Since they have different orders on
    their elements, these two groups cannot be isomorphic.}

\item \question{Recall that the \emph{centralizer} of an element $a$
    of a group, denoted, $C(a)$, is the subgroup consisting of all $b$
    such that $ab=ba$.

    Prove that for any $a,g\in G$, $C(a)$ is isomorphic to
    $C(gag^{-1})$.}

  \answer{We may attempt to construct an explicit isomorphism. Note
    that, by the definition of a centralizer, $x\in C(gag^{-1})$ if
    and only if $gag^{-1}x=xgag^{-1}$. Multiplying this equation on
    the left by $g^{-1}$ and on the right by $g$, we find that thisd
    criterion is identical to
    \[a(g^{-1}xg)=(g^{-1}xg)a\] which is to say, $x\in C(gag^{-1})$ if
    and only if $g^{-1}xg\in C(a)$. This furnishes a fairly obviouls
    bijection between the two groups: namely, we define the function
    $\varphi:C(gag^{-1})\rightarrow C(a)$ given by
    $\varphi(x)=g^{-1}xg$. We may demonstrate that this is a bijection
    by explicitly constructing its inverse $\varphi^{-1}(x)=gxg^{-1}$.

    Now, all we need to do is show this is actually a homomorphism,
    i.e., that $\varphi(xy)=\varphi(x)\varphi(y)$ for any $x,y\in
    C(gag^{-1})$. This is easily done:
    \begin{align*}
      \varphi(xy)&=g^{-1}(xy)g\\
      &=g^{-1}x(gg^{-1})yg\\
      &=(g^{-1}xg)(g^{-1}yg)\\
      &=\varphi(x)\varphi(y)
    \end{align*}
  }

\item \question{Describe $\Aut(D_5)$ as fully as possible: identify
    its order, its elements, and any multiplicative rules you can find.}

  \answer{We might note that an automorphism of $D_5$ is fully
    described by the images of $r$ and $f$, since $r$ and $f$ generate
    the group. $r$ is an element of order 5 and must be mapped to
    another element of order 5; $f$ is an element of order 2 and must
    be mapped to another element of order 2. There are thus four
    choices of plausible image for $r$ ($r$ itself, $r^2$, $r^3$, and
    $r^4$) while $f$ could be mapped to any of the five elements of
    the form $fr^i$ (with $i$ from 0 to 4). We could thus posit the
    existence of 20 potential isomorphisms, which we might label
    $\varphi_{1,0},\varphi_{1,1},\ldots,\varphi_{4,4}$ with the names indicating
    that $\varphi_{i,j}(r)=r^i$ and $\varphi_{i,j}(f)=fr^j$.

    We may ask, however, if these potential automorphisms actually
    follow the requirements of a homomorphism. Let us experiment with
    how they would act on various rotation-and-flip combinations:
    \[\varphi_{i,j}(r^ar^b)=\varphi_{i,j}(r^{a+b})=(r^i)^{a+b}=(r^i)^a(r^i)^b=\varphi_{i,j}(r^a)\varphi_{i,j}(r^b)\]
    \[\varphi_{i,j}(r^af)=\varphi_{i,j}(fr^{-a})=fr^j(r^i)^{-a}=(r^i)^afr^j=\varphi_{i,j}(r^a)\varphi_{i,j}(f)\]
    \[\varphi_{i,j}(fr^a)=fr^j(r^i)^a=\varphi_{i,j}(f)\varphi_{i,j}(r^a)\]
    so each of these 20 mappings is in fact an automorphism, so
    $\Aut(D_5)=\{\varphi_{10},\ldots,\varphi_{44}\}$, and
    $|\Aut(D_5)|=20$. Note that $\varphi_{10}$ is the identity
    automorphism. We can determine pretty straightforward
    multiplicative rules for these elements by considering the effects
    of compositions on $r$ and $f$:
    \[\varphi_{i,j}\left(\varphi_{i',j'}(r)\right)=\varphi_{i,j}(r^{i'})=(r^i)^{i'}=r^{ii'}\]
    \[\varphi_{i,j}\left(\varphi_{i',j'}(f)\right)=\varphi_{i,j}(fr^{j'})=fr^j(r^i)^{j'}=fr^{j+ij'}\]
    so our multiplication rule is
    $\varphi_{i,j}\varphi_{i',j'}=\varphi_{ii',j+ij'}$, which is more
    than a little messy.}
\end{enumerate}
\end{document}