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\title{Group Theory: Math30038, Sheet 6}
\date{Solutions GCS}
\author{}
\begin{document}
\maketitle

\begin{enumerate}

\item {\em Consider the group $D$ of rigid symmetries of a regular
$n$-gon (which may be turned over). Prove that this group has
order $2n$, is non-abelian, and can be generated by two elements 
each of order 2. Show that $D$ has a cyclic subgroup of
index 2.} \newline
\noindent{\bf Solution }\ Each axis of symmetry of a regular
$n$-gon either passes through a vertex and the midpoint of the
opposite side (if $n$ is odd) or a pair of opposite vertices or
mid-points of sides (if $n$ is even), Either way, these give rise
to $n$ elements of order $2$ in $D$ which are
relections in these axes. Label the vertices with the numbers
from 1 to $n$ in clockwise order. Suppose that $n = 2r+1$ is odd. In this
event let $x = (2,n)(3,n-1) \cdots (r+1,r+2)$ and 
$y = (1,2)(n,3)\cdots (r+1,r+3)$. Now 
$z= xy = (1,2,3, \ldots, n)$ and $yx = z^{-1} \not = z$ since $n \geq 3$
for non-triviality. Thus $D$ is non-abelian. In the degenerate case $n=2$
the group is in fact abelian.
Now $\langle z \rangle$ is the cyclic group
of order $n$ consisting of rotations, and $x\langle z \rangle$ is the
set of $n$ reflections mentioned earlier. This exhausts $D$ and 
$D = \langle x, y\rangle.$ Next suppose that 
$n = 2r$ is even. This time let \[x = (1,2)(n,3)\cdots(r+2,r+3)\]
and \[y = (1,3)(n,4) \cdots (r+1,r+3)\] so $xy 
= (1,2,3, \ldots, n)$ and the same story unfolds.

\item {\em Consider the group $D$ of rigid symmetries of the integers:
so $D$ is the group of all bijections $\theta$ from $\mathbb Z$ to $\mathbb Z$ 
which preserve distance. Thus $\theta$ must have the property that
if $x, y \in \mathbb Z$, then $|x-y| = |(x)\theta - (y)\theta|.$
Prove that this group has infinite order, 
is non-abelian, and can be generated by two elements
each of order 2. Show that $D$ has a cyclic subgroup of
index 2.} \newline
\noindent{\bf Solution }\ Let \[x = (1,-1)(2,-2)(3,-3) \cdots\] and
\[y = (0,1)(-1,2)(-2,3)\cdots.\] Now $xy$ is the map which sends
$z$ to $z+1$ for every integer $z$. Now $\langle xy \rangle$
is a cyclic subgroup of $D$. Any element $t$ of $D$ which is not
in $\langle xy \rangle$ must send $0$ to $c$ and $1$ to $c-1$.
Now $t(xy)^{-c+1}$ sends $0$ to $1$ and $1$ to $0$. The definition
of $D$ ensures that $t(xy)^{-c+1} = y$ and
so $t = y (xy)^{c-1} \in y \langle xy \rangle.$ Thus the index of 
$\langle xy \rangle$ in $D$ is at most 2. However, $y \not \in 
\langle xy \rangle$ since $y$ reverses the direction of the integers.
Thus $|D: \langle xy \rangle | = 2.$ 

\item {\em Let $D = \langle x, y \rangle$ where $o(x) = o(y) = 2$
and $x \not = y$. Let $z = xy$ and put $H = \langle z \rangle$.}
\begin{enumerate}
\item {\em Prove that $x^{-1}zx = y^{-1}zy = z^{-1}.$} \newline
\noindent{\bf Solution }\ $x^{-1}zx = x^{-1}xyx = yx = y^{-1}x^{-1}
= z^{-1}.$ Also $y^{-1}zy = y^{-1}xy^2 =  y^{-1}x^{-1} =  z^{-1}.$
\item {\em Prove that $x, y  \not \in H$.} \newline
\noindent{\bf Solution }\ If $x \in H,$ then $x^{-1}zx = z^{-1}$
but also $x^{-1}zx = z$ since $H$ is cyclic and therefore abelian.
Thus $z^2 = 1$. A cyclic group can contain at most one subgroup of order $2$,
and so 
$x = z = xy$ so $y = 1$ which contradicts $o(y) = 2$.
Therefore $x \not \in H$. If $y \in H$ then $x = zy \in H$ which
we know is not the case, so $y  \not \in H$.
\item {\em Prove that $|G:H| = 2$.}  \newline
\noindent {\bf Solution }\ Since $x^{-1} = x$ and $y^{-1} = y$,
and both $x$ and $y$ square to $1$, any element of $D$ must be 
either $1$, or $xyx\cdots xyx$ or $yxy\cdots yxy$ or a power of $z$.
Now $xyx\cdots xyx \in x \langle z \rangle$ and $yxy\cdots yxy
 = xxyxy\cdots yxy \in x \langle z \rangle$. Thus the index of $H$
is at most 2. Since $x \not \in H$ it follows that $|G:H| = 2$.
\item {\em Let $n = o(z) \in \mathbb N \cup \{ \infty\}.$ For each
possible value of $n$ let $G$ be called $D_n$. Show that the
multiplication in $D_n$ is completely determined (i.e. the number
$n$ nails down the group).}  \newline
\noindent {\bf Solution }\ We have shown that in each group
$D$ there is a cyclic subgroup $H = \langle z \rangle$ of index $2$,
and that $xH \not = H$. Therefore every element of $D_n$ is uniquely expressible
at $x^\varepsilon h$ where $\varepsilon \in \{0,1\}$ and $h \in H$. Note that
$z^x = z^{-1}$ so $h^x = h^{-1}$ for all $h \in H$. 
Multiplication is as follows: suppose that $h, k \in H = \langle z \rangle.$
We have \[
\begin{array}{rcl}
h \cdot k & = & hk\\
h \cdot xk k & = & x h^x k = xh^{-1}k\\
xh \cdot k & = & xhk \\  
xh \cdot xk & = & h^xk = h^{-1}k
\end{array}\]
Thus the only issue is how multiplication happens in $H$,
but $H = \langle s \rangle$ is cyclic and its multiplication
is determined by the single number $n = o(s)$.

\item {\em For each $n \in \mathbb N \cup \{ \infty\}$, determine the
centre of $D_n$.}  \newline
\noindent {\bf Solution }\ There is no group $D_1$. The group $D_2$
is of order 4 and so is abelian. Let us suppose that $n \geq 3$.
Now $\langle xy \rangle$ contains at most one element of order 2.
If $n$ is odd or infinite, there is no such element, but if 
$n = 2r$ is even, there is such an element $w=z^{r}$ and
since $w^x = w^y = w$, $w$ is central. If $t = xz^m$ is in 
$D_n$ but not in $H = \langle xy \rangle$ for some integer $m$, then 
$t^y = yxyyz^my = xxyxyyz^my = xz^2z^{-m}.$ Now $t^y = 1$
if and only if $z^{2m} = z^2$ so $o(z)$ divides $2m-2$. 
However $t^x = t$ if and only if $xz^m = xz^{-m}$ and so 
$o(z)$ divides $2m$. Thus if $t$ is central then $o(z)$ must
divide both $2m-2$ and $2m$ and so must divide $2$. Thus
for $n > 2$ the centre of $D_n$ is the trivial subgroup unless
$n$ is finite and even, in which case the centre of $D_n$
is $\{ 1, z^{n/2}\}$, a cyclic subgroup of order $2$.

\item {\em Determine the conjugacy classes of $D_8$.} \newline
\noindent {\bf Solution }\ I meant to ask about
$D_4$, so we will do both cases. Let $G$ be the group under discussion.
It is easy to verify that the centralizer of $x$ has order 4, and
is $\{1, z^2, x, xz^2\}.$ Therefore the conjugacy class of 
$x$ in $D_4$ has size $2$. Note that $x^y = yxy = xxyxy = xz^2$.
Thus the conjugacy class of $x$ is $\{x, xz^2\}$. Now $y = xz$
and the centralizer of $xz$ at least contains $1, xz, z^2$ and $xz^3$
so the conjugacy class of $y$ has size at most 2. However
$y^x = xyx = zx = xz^{-1} = xz^3.$ Thus the conjugacy class 
of $y$ ($=xz$) is $\{ xz, xz^3 \}$. The centre of $G$ is $\{1,z^2\}$
and so each of the elements of this set is in a conjugacy class of size 1.
Finally, the elements $z$ and $z^3$ are non-central and therefore
are in conjugacy classes of size greater than 1. The remaining
conjugacy class is therefore $\{z, z^3\}$. In summary, the
conjugacy classes of $D_4$ are 
\[ \{1\},\ \{z^2\},\ \{z,z^3\},\ \{x, xz^2\}\] and  \[\{xz, xz^3\}.\]
Let us work with $D_8$. The centralizer of $x$ contains $1, x, z^4$ and $xz^4$
so the conjugacy class of $x$ has size at most $16/4 = 4$. 
Now $x^y = yxy = xz^2$, and $x^{yx} = xz^6$ and $x^{yxy} =
xz^2z^2 = xz^4$. Thus the conjugacy class of $x$ is
$\{ x, xz^2, xz^4, xz^6\}.$ Similar calculations reveal that
the conjugacy class of $y = xz$ is $\{xz, xz^3, xz^5, xz^7\}.$
The elements of $H$ are all centralized by $H$ and so are either
central, or are in conjugacy classes of size 2. Each element of 
$H$ is conjugate to its inverse. Therefore the conjugacy classes 
of $D_8$ are
\[ \{x,xz^2,xz^4,xz^6\},\ \{xz, xz^3, xz^5, xz^7\},\ \{1\},\ \{z^4\},\
\{z,z^7\},\ \{z^2,z^6\}\] and  \[\{z^3,z^5\}.\]
\item {\em Do you recognize $D_6$?} \newline
\noindent {\bf Solution }\ I meant to ask about $D_3$, a non-abelian
group of order 6, which is a copy of $S_3$. Notice that $S_3$ is
generated by two distinct elements of order 2 (as one should expect).
\end{enumerate}

\item Suppose that $G$ is a non-abelian finite group of
order $2p$ where $p$ is a prime number. Prove that
$G$ is generated by two elements 
order 2. \newline
\noindent {\bf Solution }\ The fact that $G$ is non-abelian forces
the prime $p$ to be odd. The group $G$ contains an element $x$ of 
order $2$ and an element $t$ of order $p$ by Cauchy's theorem. Now
$G = \langle x, t \rangle$ is non-abelian so $x$ and $t$ do not
commute. Now $y = x^t \not = 1$ and $y^2 = t^{-1}xxt = 1$ so
$y$ has order $2$.  Let $T = \langle x,y \rangle$ so $T$ has even order
more than 2, and  dividing $2p$. Therefore $T = G$ and by previous theory,
$G$ is a copy of $D_p$.

\item We define a subgroup $Q$ of $S_8$ by letting 
$i = (1,2,3,4)(5,6,7,8)$, $j = (1,5,3,7)(2,8,4,6)$ 
({\bf and NOT $(2,6,4,8)$ as earlier stated})
and put $Q = \langle i, j \rangle$. Let $k = ij$
and $z = i^2$. {\em This group was the basis of William Rowan Hamilton's
generalization of the complex numbers called the
{\rm Quaternions.}}
\begin{enumerate}
\item Show that $i^2 = j^2 = k^2  = z$ and $z^2 = 1$.\newline
\noindent {\bf Solution }\ This is routine. 
\item Show that $ij = k$, $jk = i$ and $ki = j$.\newline
\noindent {\bf Solution }\ This is routine.
\item Show that $ji = zk$, $kj = zi$ and $ik = zj$.\newline
\noindent {\bf Solution }\ This is routine.
\item Show that $z$ is in the centre of $Q$.\newline
\noindent {\bf Solution }\ $z = i^2 = j^2$ so $z$ commutes with both 
$i$ and $j$, so $z$ is central. 

\item Show that $Q = \langle i \rangle \cup z \langle i \rangle$. \newline
\noindent {\bf Solution }\ This is nonsense. $z \in \langle i \rangle$
and therefore  $\langle i \rangle \cup z \langle i \rangle = \langle i \rangle.$
However, for any element $u \not \in \langle i \rangle$ it will be
the case that $Q = \langle i \rangle \cup u \langle i \rangle$. The choices
for $u$ are $j, zj, k$ and $zk$. Let us choose $u$ to be $k$, and demonstrate
this decomposition in this case. Any element of $Q$ is a word in $i$ and $j$.
Now $j = ik^{-1} = ik^3$ so every element of $Q$ is a word in $i$ and $k$
(with positive exponents). Now any occurrence of $ik$ can be replaced by 
$kiz$ Using this repeatedly the $k$s can
migrate to the left. When this is done use $z = i^2$  to eliminate $z$.
If the power of $k$ 
involved is even, then use $k^{2n } = i^{2n}$ to eliminate $k$.
If the power of $k$
involved is odd, all but one $k$ can be eliminated. Thus
$Q = \langle i \rangle \cup k \langle i \rangle.$ 
\item Show that $|Q| = 8$.\newline
\noindent {\bf Solution }\ 
This follows immediately from the previous part.
\item Show that $Q$ and $D_8$ (in Question 1) are both non-abelian
groups of order 8, but they contain different numbers of elements of order
4. .\newline
\noindent {\bf Solution }\ I meant to write `Show that $Q$ and $D_4$ 
(in Question 1) are both non-abelian
groups of order 8,' I fear that $D_8$ has order 16. 
In $D_4$ there are 5 involutions; there are 4 reflections
and one rotation through $\pi$. In $Q$ however, only $z$ has order
2. One may verify tha t the remaining six non-identity elements of
$Q$ all have order 4. Now $Q$ is non-abelian since $ij \not = ji$.
$D_4$ is non-abelian because $xy \not = yx$.
\item {\em Determine the conjugacy classes of $Q$.} \newline
\noindent {\bf Solution }\ They are
\[ \{ 1 \},\ \{ z\},\ \{i,zi\},\ \{j,zj\} \mbox{ and }\{k,zk\}.\]
There are any number of ways to demonstrate this.

\item {\em On which bridge are the quaternions inscribed?} \newline
\noindent {\bf Solution} The Sir William Rowan Hamilton Bridge, Dublin. 
\end{enumerate}

\item Let $G$ denote the set of invertible $n$ by $n$ matrices
with complex entries. This is a group under multiplication of
matrices. Give a transversal for the conjugacy
classes of $G$. {\em Hint: the course MA20012 
does this (and not much else).}
\newline \noindent{\bf Solution }\ Jordan normal forms with ``$\lambda$''
non-zero.
\item Show that if $N$ is a normal subgroup of $G$, then $N$ must be
a union of conjugacy classes of $G$ including the conjugacy
class of the identity element. Deduce that the only normal subgroups
of $A_5$ are $1$ and $A_5$, but that $A_4$ has a normal subgroup $M$ which
is neither 1 nor $A_4$.\newline
\noindent{\bf Solution }\ Since $g^{-1}Ng = N$ for all $g \in G$,
it follows that if $n \in B$, then the conjugacy class of $n$ is 
a subset of $N$.  
The 
conjugacy classes of $A_5$ are the conjugates of $id$, $(1,2,3)$, 
$(1,2)(3,4)$, $(1,2,3,4,5)$ and $(1,2,3,5,4)$. These conjugacy classes
have sizes $1,20,15,12$ and $12$ respectively. The only sums these sizes
which include 1 and 
which divide 60 are 1 and 60, so $A_5$ is a simple group.
An analogous calculation for $A_4$ yields that the conjugacy classes
of $A_4$ have sizes $1, 3, 4$ and $4$. In this case $1+3$ is a divisor of 12,
so it is possible that there is a normal subgroup of size 4.
It so happens that \[\{ id,\ (1,2)(3,4),\ (1,3)(2,4),\ (1,4)(2,3) \}\]
is a (normal) subgroup of $A_4$ of this order.
\end{enumerate}
\end{document}