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\lhead{MATH 521--01}
\chead{Problem Set \#2\answer{ solutions}}
\rhead{}
\lfoot{}
\cfoot{\answer{Page \thepage\ of \pageref{LastPage}}}
\rfoot{due Thursday, September 18, 2014}
\begin{document}
\begin{enumerate}
\item \question{Identify each of the following algebraic structures as
either a group or not a group, and justify your answer. For each
structure which is a group explain if it is Abelian.}
\begin{itemize}
\item \question{The real numbers $\mathbb R$ under the operation
$a\odot b=a+b+1$.}
\answer{This structure is an Abelian group: for any two real
numbers $a$ and $b$, $a\odot b$ is real, so we have closure;
$a\odot b\odot c=a+b+c+2$ regardless of the grouping, so we have
associativity; $a\odot(-1)=(-1)\odot a=a$, so we have an
identity element (specifically $-1$); finally,
$a\odot(-2-a)=(-2-a)\odot a=-1$, so every element $a$ has an
inverse. Finally, note that $a\odot b=b\odot a$, so this group
is commutative.}
\item \question{The integers $\mathbb Z$ under the operation
$a\text{ max }b$, which returns the larger of $a$ and $b$
(e.g. $7\text{ max }-3=7$.}
\answer{This is not a group, because it has no identity --- there
is no number $e$ whose maximum with any number $a$ is equal to
$a$.
A variation on this algebraic structure, the maximum operation
on $\mathbb Z\cup\{-\infty\}$, \emph{does} have an identity
element, since the artificial element $-\infty$ is defined to be
less than every integer so $-\infty\text{ max }a=a$ for all
$a$. However, even this algebra lacks an inverse.
The above operation is in fact closed, associative, and
commutative.}
\item \question{The set of subsets of $\{1,2,3,4,5\}$ under the
union operation $S\cup T$.}
\answer{This operation lacks an inverse. The identity element is
the set $\emptyset$, since $S\cup\emptyset=S$ for all $S$, but
for any nonempty set, e.g. $S=\{1,2\}$, $S\cup T$ will be at
least as large as $S$ and thus $S\cup T\neq\emptyset$.
This operation actually has all properties of an Abelian group
except for the inverse.}
\item \question{The set of strings of the symbols $a$, $b$, and
$b^{-1}$ under the operation of concatenation with the rule that
two adjacent $a$s or $b$s cancel,
e.g. $(ababa)(ab)=abab\cancel{aa}b=aba\cancel{bb}=aba$.}
\answer{This is a group: concatenating and canceling two strings
results in a string, so this structure is closed; concatenating
three strings does not depend on the order in which they are
``glued together'', so the operation is associative; the empty
string is an identity; and the reversal of a string is its
inverse.
Note that this operation is not commutative, so the group is
non-Abelian: $(ab)b=a$, while $b(ab)=bab$.}
\end{itemize}
\item \question{Let $D_6$ be the group of symmetries of a hexagon. Identify a subgroup of $D_6$ with each of the following orders: $1$, $2$, $3$, $6$.}
\answer{Let $r$ be a 60-degree rotation and $f$ an (arbitrary)
flip. We may denote the elements of $D_6$ canonically as such:
$\{e,r,r^2,r^3,r^4,r^5,f,fr,fr^2,fr^2,fr^3,fr^4,fr^5\}$, and
then identify the given subgroups.
There is only one subgroup of order 1; unsurprisingly, it's the
trivial group $\{e\}$.
There are seven different subgroups of order
2. Generally, a group of order 2 is always $\{e,x\}$ with $x^2=e$,
and the possible stand-in values for $x$ are $r^3$ or any of the
six flips.
A group of order three basically has to be $\{e,x,x^2\}$ where
$x^3=e$; you might think $\{e,x,y\}$ in general should work, but
then $x^2$, $xy$, and $y^2$ all need to be in the group, and that
gets messy. The only possible subgroup matching this template is
$\{e,r^2,r^4\}$.
Groups of order 6 are much more versatile in their form, but
finding a few in $D_6$ isn't hard; there are at least three
possibilities: $\{e,r,r^2,r^3,r^4,r^5\}$,
$\{e,r^2,r^4,f,fr^2,fr^4\}$, and $\{e,r^2,r^4,fr,fr^3,fr^5\}$.}
\item \question{Prove that if $G$ is a finite group with order 7,
then $G$ is cyclic.}
\answer{We wish to show that $G$ contains an element of order 7;
this element would then generate the whole group. We shall show
specifically that $G$ \emph{cannot} contain elements of orders
from 2 to 6 ($G$ does, of course, contain an element of order 1,
namely, the identity). We shall address each case individually:
\begin{prop}
If $|G|=7$, then $G$ contains no element of order 2.
\end{prop}
\begin{proof}
Suppose counterfactually that $G$ contains an element $a$ of
order 2; we may call the elements of $G$ by the names $e$, $a$,
$b$, $c$, $d$, $f$, and $g$. Since $a\neq e$, we can be certain
that $ba\neq b$, $ca\neq c$, etc. Without loss of generality we
can assert that $ba=c$, and then $ca=ba^2=b$. Likewise, if we
assert $da=f$, then $fa=d$ as well. However, now there is no
possible value for $ga$: $ga$ cannot equal $xa$ for $x\neq g$,
and we have already assigned the values of $a$, $b$, $c$, $d$,
$e$, and $f$ to $ea$, $ca$, $ba$, $fa$, $aa$, and $da$
respectively, so that $ga$ must equal $g$, but this cannot be
the case because $a$ is not the identity.
\end{proof}
\begin{prop}
If $|G|=7$, then $G$ contains no element of order 3.
\end{prop}
\begin{proof}
Suppose counterfactually that $G$ contains an element $a$ of
order 3; we may call the elements of $G$ by the names $e$, $a$,
$a^2$, $b$, $c$, $d$, and $f$. Since $a\neq a^2\neq e$, we can
be certain that $b$, $ba$, and $ba^2$ are distinct. Without loss
of generality we can assert that $ba=c$ and $ba^2=d$. Then
$ca=d$ and $da=b$. However, at this point since we know: $ea=a$,
$aa=a^2$, $(a^2)a=e$, $ba=c$, $ca=d$, and $da=b$. It must be the
case that $fa$ is distinct from all of these, so $fa=f$, which
is impossible since $a\neq e$.
\end{proof}
\begin{prop}
If $|G|=7$, then $G$ contains no element of order 4.
\end{prop}
\begin{proof}
If $a$ was an element of $G$ of order 4, then $a^2$ would have
order 2; we saw above that that is impossible.
\end{proof}
\begin{prop}
If $|G|=7$, then $G$ contains no element of order 5.
\end{prop}
\begin{proof}
If $a$ was an element of $G$ of order 5, we could build a near-complete Cayley table for $G$, positing the existence of sixth and seventh elements of $G$ called $b$ and $c$:
\begin{center}
\begin{tabular}{|c||c|c|c|c|c|c|c|}
\hline
&$e$&$a$&$a^2$&$a^3$&$a^4$&$b$&$c$\\\hline\hline
$e$&$e$&$a$&$a^2$&$a^3$&$a^4$&$b$&$c$\\\hline
$a$&$a$&$a^2$&$a^3$&$a^4$&$e$&?&?\\\hline
$a^2$&$a^2$&$a^3$&$a^4$&$a$&$e$&?&?\\\hline
$a^3$&$a^3$&$a^4$&$a$&$a^2$&$e$&?&?\\\hline
$a^4$&$a^4$&$a$&$a^2$&$a^3$&$e$&?&?\\\hline
$b$&$b$&?&?&?&?&?&?\\\hline
$c$&$c$&?&?&?&?&?&?\\\hline
\end{tabular}
\end{center}
However, to complete this table following the group laws, $ab$
and $ac$ must be $b$ and $c$ in some order; to not conflict with
the identity, $ab=c$ and $ac=b$. But then $a^2b$ and $a^2c$ must
\emph{also} be $b$ and $c$ in some order, and since $eb=b$ and
$ab=c$, $a^2b$ can be neither $b$ nor $c$, leading to a
contradiction.
\end{proof}
\begin{prop}
If $|G|=7$, then $G$ contains no element of order 6.
\end{prop}
\begin{proof}
If $a$ was an element of $G$ of order 6, then $a^3$ would have
order 2; we saw above that that is impossible.
\end{proof}
}
\item \question{Let $G$ be a group such that for any $a,b,c,d,x\in G$,
if $axb=cxd$, then $ab=cd$. Prove that $G$ is Abelian.}
\answer{For any $a$ and $b$, consider $x=a^{-1}b^{-1}$, $c=ba$, and
$d=e$. Note that $axb=e=cxd$, so $ab=cd$. Thus $ab=ba$.
Other choices also work, like letting $x=a^{-1}$, $c=b$, and
$d=a$.}
\item \question{Which elements of $Z_{200}$ have order 5? Explain how
you know.}
\answer{ By the Fundamental Theorem of Cyclic Groups, the unique
subgroup of $Z_{200}$ of order 5 is
$\langle\frac{200}5\rangle=\{0,40,80,120,160\}$, Each of the
elements of this group except $0$ generates this subgroup, and
since the order of an element of $G$ is equal to the order of the
subgroup it generates, these four numbers (40, 80, 120, and
160) are the only elements of $Z_{200}$ with order 5.}
\end{enumerate}
\end{document}