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\title{Group Theory: Math30038, Sheet 1}
\author{GCS: Solutions}
\date{}
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\maketitle

{\it The course web site is available via
{\tt http://www.bath.ac.uk/$\sim$masgcs/}}

\begin{enumerate}
\item Let $G$ be a group. Suppose that $x \in G$.
Let \[ C_G(x) = \{ g \in G \mid  gx = xg \} \subseteq G.\]
Prove that $C_G(x) \leq G$.\newline
\noindent {\bf Solution: } $1 \in C_G(x) \not = \emptyset.$
Moreover if $g \in G$ and $gx = xg$, then $xg^{-1} = g^{-1}x$
so if $a, b \in C_G(x)$, then $ab^{-1}x = axb^{-1} = xab^{-1}$
and therefore $ab^{-1} \in C_G(x)$.
\item {\em Let $G$ be a group. Suppose that $S \subseteq G$.
Let \[C_G(S)  = \{ g \in G \mid  gs = sg \forall s \in S\}.\]
Prove that $C_G(S) \leq G$.} \newline
\noindent {\bf Solution: } We have
\[ C_G(S) = \cap_{s \in S} C_G(s)\] and since the intersection
of subgroups is a subgroup, we are done.
\item Let $G$ be a group. Suppose that $S \subseteq G$.
Let \[N_G(S) = \{ g \in G \mid gS = Sg \}\] where
$gS = \{ gs \mid s \in S\}$ and $Sg = \{ sg \mid s \in S \}$.
Prove that $C_G(S) \leq N_G(S)  \leq G$. \newline
\noindent {\bf Solution: } The fact that $N_G(S)$ is a group
follows the outline of the proof that $C_G(x)$ is a group.
The condition to be in $C_G(S)$ is stronger than that to be
in $N_G(S)$ so $C_G(S) \leq N_G(S)  \leq G$.
\item {\em Let $G$ be a group. Suppose that $S \subseteq G$
and that $x \in N_G(S)$. Prove that
$C_G(S)x = xC_G(S)$.}  \newline
\noindent {\bf Solution: } We will show that
$C_G(S) = x^{-1}C_G(S)x$ which will suffice. 
Suppose that $s \in S$, then $xs = s'x$ for some $s' \in S$.
Now shoose any $c \in C_G(S)$, then
\[ x^{-1}cxs = x^{-1}cs'x = x^{-1}s'cx = x^{-1}(xsx^{-1})cx
= sx^{-1}cx\] and therefore $sx^{-1}c \in C_G(S)$
and so
$x^{-1}C_G(S)x \leq C_G(S).$ Replacing $x$ by $x^{-1}$ in 
this argument yields that $x C_G(S)x^{-1} \leq C_G(S).$
Premultiplying by $x^{-1}$ and postmultiplying by $x$
gives
$C_G(S) \leq x^{-1}C_G(S)x$. Now we have two mutually
reverse inclusions
so $C_G(S) = x^{-1}C_G(S)x$ for all $x \in N_G(S).$
\item {\em Let $G$ be a finite group. Suppose that
$\emptyset \not = H \subseteq G$ has the property that
if $a,b \in H$, then $ab \in H$. Does it follow that $
H  \leq G$? What happens if we relax the condition that
$G$ is finite?}  \newline
\noindent {\bf Solution: } If $h \in H$, then 
$\langle h \rangle \leq H$ and so $o(h)$ is a natural
number $n$. If $h = 1$ then $h^{-1} = h \in H$. Otherwise
$n > 1$ and then $h^{-1} = h^{n-1} \in H$ so $H$ is a 
subgroup of $G$. In the event that $G$ is infinite,
things fall apart. For example perhaps 
$G = \mathbb Z$ under addition, the $\mathbb N$ is a non-empty
additively closed subset which is not a subgroup since
$-1 \not \in \mathbb N$.
\item Suppose that $G$ is a group and that $x \in G$.
Prove that $(x^{-1})^{-1} = x$.  \newline
\noindent {\bf Solution: } $x^{-1} x = 1 = x^{-1} (x^{-1})^{-1}.$
Premultiplying by $x$ gives the result.
\item {\em Does there exist a group $G$ containing 
elements $a,b$ such that $a^2 = b^2 = (ab)^3 = 1$?} \newline
\noindent {\bf Solution: } Yes, the trivial group will do it.
More interestingly, let $G$ be the group $S_3$, and put
$a = (1,2)$ and $b = (2,3)$. Now the orders of $a, b$ and $ab$%
are exactly ``as advertized''.
\item {\em Suppose that $G$ is a group with the property that
$x^2 = 1$ whenever $x$ is an element of $G$. Show that
$G$ must be abelian.} \newline
\noindent {\bf Solution: } Suppose that $a, b \in G$,
then $abab = 1 = baab$. Postmultiplying by $ba$ yields
$ab = ba$.
\item {\em (Challenge) Suppose that $G$ is a group with the property that
$x^3 = 1$ whenever $x$ is an element of $G$. Show that
$G$ need not be abelian.} \newline
\noindent {\bf Solution: } Consider the set of 3 by 3 upper triangular
matrices with entries in $F = \mathbb Z_3$, the field of integers modulo
3 which have 1s on the leading diagonal. It is easy to verify that
these 27 matrices form a group. Each matrix is of the form $I + \Delta$
wheer $I$ is the 3 by 3 identity matrix and $\Delta$ is strictly
upper triangular. Note that $\Delta^3$ is the zero matrix, so
the inverse of $I + \Delta$ is $I - \Delta + \Delta^2$ and 
moreover $(I + \Delta)^3 = I + 3\Delta + 3\Delta^2 = I$
so ev ery element of the group has order dividing 3. 
The matrices $I + E_{12}$ and $I + E_{23}$ do not commute
as may be verified by direct calculation; here
$E_{ab}$ is the 3 by 3 matrix with entries in $F$ where
the entrie in the $i$-th row and $j$-th column
is $\delta_{ia}\delta_{jb}$ where $\delta$ is
the Kronecker delta.
\end{enumerate}
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