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\lhead{MATH 521--01}
\chead{Problem Set \#4\answer{ solutions}}
\rhead{}
\lfoot{}
\cfoot{\answer{Page \thepage\ of \pageref{LastPage}}}
\rfoot{due Tuesday, October 21, 2014}

\begin{document}
\begin{enumerate}
\item \question{Let $G$ be a finite group whose order is greater than
    2 but not divisible by 4. Prove that $G$ contains an element of
    odd order (other than the identity).}
  
  \answer{
    \begin{proof}
      There are two possible cases we should address, both of which
      are fairly straightforward.

      \textbf{Case I: $G$ has odd order.} Then, by a simple
      consequence of Lagrange's Theorem, every element of $G$ has
      order which divides the order of $G$, and since an odd number
      has no even divisors, every element of $G$ has odd order.

      \textbf{Case II: $G$ has even order.} Let $|G|=2k$, where $k$ is
      odd (we know $k$ is odd because the order of $G$ is not
      divisible by 4). The orders of elements of $G$ are thus, by
      Lagrange's theorem, divisors of $2k$. If any non-identity
      element has odd order, we're done; if any element $g$ has order
      $2\ell$ for $\ell>2$, then since $\ell|k$, $\ell$ is odd and
      $g^2$ will have order $\ell$. Thus, the only circumstance we
      need to address is the possibility that \emph{every non-identity
        element} in $G$ has order $2$. Let us suppose that to be the
      case, and derive a contradiction.

      Since $|G|>2$, $G$ contains at least the identity element $e$
      and two distinct non-identity elements $a$, and $b$; the latter
      two, by our assumption, have order 2, so $a=a^{-1}$ and
      $b=b^{-1}$. In addition, $ab$ will have order 2, so
      $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$. Thus, $\{e,a,b,ab\}$ will be a
      subgroup of $G$, and so by Lagrange's theorem the order of $G$
      must be divisible by 4, which contradicts our premise.
    \end{proof}
  }
  
\item \question{Let $G$ be an Abelian group of order 15. Prove that
    $G$ is cyclic.}
  
  \answer{
    \begin{proof}
      By Lagrange's Theorem, every element of $G$ has order divisible
      by 15, i.e., the possible orders of elements of $G$ are 1, 3, 5,
      and 15. We know exactly one element, the identity, has order 1,
      and if any elements have order 15, then we are done, since any
      such element would generate the whole group. Let us consider the
      possibilities where $G$ only contains elements of order $3$ and
      $5$.
      
      \textbf{Case I: Every non-identity element of $G$ is of order
        3.} Let $a$ and $b$ be distinct non-identity elements such
      that $a^2\neq b$. Since $G$ is Abelian and each of $a$, $b$, and
      $ab$ has order 3, it is easy to list out all the elements
      generated by $a$ and $b$. So
      $\{e,a,a^2,b,ab,a^2b,b^2,ab^2,a^2b^2\}$ is a subgroup of $G$,
      and since that group has 9 elements, Lagrange's Theorem requires
      that $9\mid 15$, which is a contradiction.

      \textbf{Case II: Every non-identity element of $G$ is of order
        5.} Consider some non-identity element $a$ and the subgroup
      $H=\langle a\rangle=\{e,a,a^2,a^3,a^4\}$ which it
      generates. Consider some $b\notin H$; $b$ also has order 5, so
      let us consider the cosets $H$, $bH$, $b^2H$, $b^3H$, and
      $b^4H$. Since $H$ has index 3, at least two of these four cosets
      must be the same, so that $b^kH=b^\ell H$ for some distinct $k$,
      $\ell$ from 0 to 5. By the equlity criterion for cosets,
      $b^{k-\ell}\in H$, but $b^{k-\ell}\neq e$, so some generator of
      $H$ is a power of $b$, and so $b$ generates $H$, contradicting
      our assumption that $b\notin H$.

      \textbf{Case III: $G$ contains at least one element of order 3
        and at least one element of order 5.} Let $a$ have order 3 and
      $b$ have order 5. A brute force (or number-theoretic) argument
      indicates that since $(ab)^k=a^kb^k$, this power equals the
      identity only when $3\mid k$ and $5\mid k$; i.e., when $15\mid
      k$, so $ab$ would have order 15.
    \end{proof}
  }

\item \question{Suppose $|G|=p^2$ with $p$ prime. For non-identity
    elements $a$ and $b$ of $G$, prove that if $\langle
    a\rangle\neq\langle b\rangle$, then every element of the group can
    be expressed in the form $a^kb^\ell$ for integers $k$ and $\ell$.}

  \answer{By Lagrange's Theorem, every subgroup of $G$ has order 1,
    $p$, or $p^2$; in particular $\langle a\rangle$ and $\langle
    b\rangle$ must have those orders, and since $a$ and $b$ are
    nonidentity elements, they do not generate subgroups of order
    1. If either subgroup (WLOG $\langle a\rangle$) has order $p^2$,
    then $\langle a\rangle=G$ and every element of $G$ can be written
    in the form $a^kb^0$. Thus the only interesting case is when
    $\langle a\rangle$ and $\langle b\rangle$ both have order $p$.

    Note that any non-identity element of a group of prime order
    generates the whole group, so since $\langle a\rangle\neq\langle
    b\rangle$, these two groups do not overlap at all except in the
    identity element. Thus, since $b^k\notin\langle a\rangle$ except
    when $b^k=e$, we may assert that the cosets $\langle a\rangle,
    \langle a\rangle b, \langle a\rangle b^2,\ldots,\langle a\rangle
    b^{p-1}$ are all distinct and disjoint, and thus account for all
    $p^2$ elements of $G$. Since every element of $G$ lies in one of
    these cosets, every element of $G$ may be written as a power of
    $a$ times a power of $b$.}

\item \question{Recall that the \emph{center} $Z(G)$ of a group $G$ is
    the subgroup whose elements are of the form $\{x\in G:xa=ax\text{
      for all }a\in G\}$. Prove that $Z(G_1\oplus G_2)=Z(G_1)\oplus
    Z(G_2)$; that is, that the center of a direct product is the
    direct product of centers.}

  \answer{
    \begin{proof}
      We shall start by showing that every element of $Z(G_1\oplus
      G_2)$ is in $Z(G_1)\oplus Z(G_2)$. Let $(x_1,x_2)$ be an element
      of $Z(G_1\oplus G_2)$; this element is characterized by the fact
      that $(x_1,x_2)(a_1,a_2)=(a_1,a_2)(x_1,x_2)$ for all
      $(a_1,a_2)\in G_1\oplus G_2$. Performing the direct-product
      multiplication on both sides of this criterion, we get that
      $(x_1a_1,x_2a_2)=(a_1x_1,a_2,x_2)$. Since two elements of a
      direct product group are equal only if each term is equal, then
      we require that $x_1a_1=a_1x_1$ and $x_2a_2=a_2x_2$ for any
      $(a_1,a_2)\in G_1\oplus G_2$, which is to say, any $a_1\in G_1$
      and $a_2\in G_2$. Since $x_1a_1=a_1x_1$ for any $a_1\in G_1$,
      $x_1\in Z(G_1)$; likewise $x_2\in Z(G_2)$. Thus this
      $(x_1,x_2)\in Z(G_1)\oplus Z(G_2)$.

      Conversely, let $(x_1,x_2)$ be an element of $Z(G_1)\oplus
      Z(G_2)$. By the construction of a direct product, $x_1\in
      Z(G_1)$ and $x_2\in Z(G_2)$, and thus $x_1a_1=a_1x_1$ for any
      $a_1\in G_1$, and $x_2a_2=a_2x_2$ for any $a_2\in
      G_2$. Assembling these two equalities into a pairwise equality
      of coordinates, our criterion is that, considered as elements of
      $G_1\oplus G_2$, $(x_1a_1,x_2a_2)=(a_1x_1,a_2x_2)$ for any
      $a_1\in G_1$ and $a_2\in G_2$. Rewriting this in terms of the
      product of elements of $G_1\oplus G_2$, our criterion will be
      that $(x_1,x_2)(a_1,a_2)=(a_1,a_2)(x_1,x_2)$ for any
      $(a_1,a_2)\in G_1\oplus G_2$. In other words, $(x_1,x_2)$ must
      commute with any element of $G_1\oplus G_2$, and is thus an
      element of $Z(G_1\oplus G_2)$.
    \end{proof}
  }

\item \question{Prove that $S_3\oplus Z_2$ is isomorphic to $D_6$; you
    may find it helpful to define a specific isomorphism.}

  \answer{Let us define an isomorphism from $D_6$ to $S_3\oplus
    Z_2$. We would want to map $r$ to some element of $S_3\oplus Z_2$
    of order 6; such an element mught be the product of an element of
    $S_3$ of order 3 (i.e. one of the two 3-cycles) with an element of
    $Z_2$ of order 2 (the number 1), so we select
    $\phi(r)=((123),1)$. We then want $f$ to be associated with some
    element of order 2; which one won't matter much, as long as we map
    it to a different element than we map $r^3$ to. We might choose,
    for instance, $\phi(f)=((12),1)$. This induces the rest of the
    isomorphism:
    \begin{align*}
      \phi(e)&=(e,0)\\
      \phi(r)&=((123),1)\\
      \phi(r^2)&=((132),0)\\
      \phi(r^3)&=(e,1)\\
      \phi(r^4)&=((123),0)\\
      \phi(r^5)&=((132),1)\\
      \phi(f)&=((12),1)\\
      \phi(fr)&=((23),0)\\
      \phi(fr^2)&=((13),1)\\
      \phi(fr^3)&=((12),0)\\
      \phi(fr^4)&=((23),1)\\
      \phi(fr^5)&=((13),0)\\
    \end{align*}
  }
\end{enumerate}
\end{document}