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\centerline{Mathematics 352 -- Abstract Algebra II}
\centerline{Solutions for Final Exam -- May 3, 2008}
\vskip 10pt
\noindent
I.  {\it Terminology.}  Answer \emph{any five of the following}.  You can earn 
\emph{up to 10 points Extra Credit} for answering more than five.
\vskip 10pt
\begin{enumerate}
\item[(A)] (5) What is a \emph{Sylow $p$-subgroup} of a finite group $G$?
\sol
If $p$ is prime and $|G| = p^n m$ where $p \nd m$, 
then a Sylow $p$-subgroup of $G$ is a subgroup $H \subseteq G$ with $|H| = p^n$.
\item[(B)] (5) What is the \emph{class equation} of a finite group?
\sol
The class equation of a group is the equation that says $|G|$ is the 
sum of the cardinalities of the distinct conjugacy classes.  Specifically,
since each $c \in Z(G)$ (the center) is the only element in its conjugacy
class, we can write the class equation as
\[
|G| = |Z(G)| + \sum_{i=1}^t |C_{g_i}| = |Z(G)| + \sum_{i=1}^t [G:C(g_i)],
\]
where $g_1,\ldots,g_t$ are any elements, one from each of the conjugacy
classes of size $\ge 2$.
\item[(C)] (5) What does it mean to say that a point $Q$ in the plane 
is \emph{constructible}, starting from two given points $O = (0,0)$, 
and $P = (1,0)$?
\sol
It means that $Q$ is obtained after a finite number of steps
consisting of constructing new points as the intersection of two lines, or
a line and a circle, or two circles, where all lines are formed by joining
two previously constructed points (using a straightedge), and all 
circles have radii with center and opposite endpoint at two previously constructed points. 
\item[(D)] (5) What is the \emph{degree} of an extension field $K$
of a field $F$ (written $[K:F]$)?
\sol
The degree of $K$ over $F$ is the dimension of $K$ as a vector space
over the field of scalars $F$.
\item[(E)] (5) What is the \emph{characteristic} of a field?
\sol
The characteristic of a field is the smallest strictly positive integer $n$ such 
that 
\[
n\cdot 1 = 1 + 1 + \cdots + 1 = 0
\]
in $F$, or $0$ if there are no such integers $n > 0$.
\item[(F)] (5) What is the \emph{Galois group}, ${\rm Gal}_F K$, of an
extension $K$ of a field $F$?
\sol
The Galois group is the set of $F$-automorphisms of $K$, under the 
operation of function composition.
\item[(G)] (5) What does it mean to say that an extension field
$K$ of a field $F$ is \emph{Galois} over $F$?
\sol
$K$ is Galois over $F$ if $[K:F] < \infty$, and $K$ is normal and separable over $F$.
\item[(H)] (5) What does it mean to say that a finite group $G$ is 
\emph{solvable}?
\sol
The group $G$ is solvable if there exists a decreasing sequence of subgroups
\[
G = G_0 \supset G_1 \supset G_2 \supset \cdots \supset G_{n-1} \supset G_n = \{e\}
\]
such that for each $i = 0,\ldots, n-1$, $G_{i+1}$ is a normal subgroup of
$G_i$, and $G_i/G_{i+1}$ is abelian (equivalently there exists a sequence such
that the quotients are cyclic, equivalently, there exists a sequence such that
the quotients are cyclic of prime order).   
\end{enumerate}
\vskip 10pt
\noindent
II. 
\begin{enumerate}
\item[(A)] (10) State the Structure Theorem for finite abelian groups. 

\sol Every finite abelian group is the direct sum of cyclic subgroups, each
of order a power of a prime.

\item[(B)] (20) Classify all finite abelian groups of order $n = 300$.  How many different
isomorphism classes are there?  Give both the elementary divisors and the invariant factors 
for each isomorphism class.

\sol
Since $300 = 2^2\cdot 3 \cdot 5^2$, there are four different isomorphism classes.  In the 
diagram below, the left column shows the classification by the elementary divisors and 
the right shows the classification by the invariant factors:
\begin{eqnarray*}
\Z_4 \oplus \Z_3 \oplus \Z_{25} &\simeq & \Z_{300}\\
(\Z_2 \oplus \Z_2) \oplus \Z_3 \oplus \Z_{25} &\simeq & \Z_2 \oplus \Z_{150}\\
\Z_4 \oplus \Z_3 \oplus (\Z_5\oplus\Z_5) &\simeq & \Z_5 \oplus \Z_{60}\\
(\Z_2\oplus \Z_2) \oplus \Z_3 \oplus (\Z_5\oplus \Z_5) &\simeq & \Z_{10} \oplus \Z_{30}.
\end{eqnarray*}
\end{enumerate}
\vskip 10pt
\noindent
III.
\begin{enumerate}
\item[(A)] (15) State the three Sylow Theorems for finite groups.
\sol
Let $G$ be a finite group
\begin{itemize}
\item Sylow I:  If  and $p^n | |G|$, then $G$ has a subgroup 
of order $p^n$.
\item Sylow II:  If $|G| = p^n m$, where $p \nd m$, then all subgroups of $G$
of order $p^n$ (the Sylow $p$-subgroups as in I (A)) are conjugate in $G$.
\item Sylow III:  The number of Sylow $p$-subgroups is $\equiv 1 \bmod p$ and 
divides $|G|$.
\end{itemize}

\item[(B)] (10) Show using the Sylow Theorems
that every group $G$ of order $n = 99$ has a non-trivial
normal subgroup $N$ (that is, a normal subgroup $N \ne \{e\}, G$). 

\sol
Since $99 = 3^2\cdot 11$, consider the Sylow $3$-subgroups in $G$.  By Sylow III,
the number of such subgroups must be congruent to 1 mod $3$ and must divide $99$. 
Since the only divisors of $99$ are $1,3,9,11,33,99$, the only possibility is that
there is a unique subgroup of order $9$.  This is normal because of Sylow II.
\item[(C)] (10 Extra Credit)  Show that every group of order $n = 99$
is isomorphic either to $\Z_9 \times \Z_{11}$ or to $\Z_3 \times \Z_3 \times \Z_{11}$.  

\sol  The same type of argument done in part (B) works with the Sylow $11$-subgroups 
too and shows there is a unique, normal Sylow $11$-subgroup.  Let $H$ be the Sylow
$3$-subgroup and let $K$ be the Sylow 11-subgroup.  We see that $H \cap K = \{e\}$
since any element of the intersection must have order simultaneously a power of 3 and 
a power of 11.  It follows that $HK = G$ since the 99 elements of $HK$ are distinct.
Hence $G \simeq H \times K$ by Theorem 8.3 in Hungerford.  
Now $H$ is a group of order $9 = 3^2$, so Corollary 8.29 in Hungerford
shows that $H \simeq \Z_9$ or $H \simeq \Z_3 \times \Z_3$.  Since $K$ has
prime order, $K$ is cyclic, isomorphic to $\Z_{11}$.  This shows what we wanted. 
\end{enumerate}
\vskip 10pt
\noindent
IV. (10)  Show that if $K$ is an extension field of $F$ and $[K:F] < \infty$, 
then every element of $K$ is algebraic over $F$. 

\sol
Say $[K : F] = n$.  Let $u \in K$ and consider the set $\{1,u,u^2,\cdots,u^n\} \subset K$.
Since this set has cardinality $n+1$ and it lies in an $n$-dimensional vector space over
the field $F$,
it must be linearly \emph{dependent}.  That is, there are scalars $c_0,c_1,\ldots,c_n \in F$,
not all zero, such that $c_0 + c_1 u + c_2 u^2 + \cdots + c_n u^n = 0$.
But this says that $u$ is a root of the polynomial $f(x) = c_0 + c_1 x + c_2 x^2 + \cdots + c_n x^n \in F[x]$,
and $f(x)$ is not the zero polynomial.  Therefore $u$ is algebraic over $F$.
\vskip 10pt
\noindent
V.  (15)  It can be shown, using some slightly tedious but elementary
manipulations with trig identities, 
that the number $x = 2\cos\left(\frac{2\pi}{7}\right)$ is a root of
the equation $x^3 + x^2 - 2x - 1 = 0$.  What does this imply about the constructibility
of the regular $7$-gon?  Explain.  

\sol It says that the regular $7$-gon is \emph{not constructible} by straightedge
and compass.  The reason is that this cubic polynomial is \emph{irreducible} in $\Q[x]$.
Since it is a polynomial of degree 3, to show irreducibility, it is enough
to show the polynomial has no rational roots.  But the only possible rational roots
(by the Rational Roots Test) would be $x = \pm 1$, and neither of them gives zero.  
Hence $\left[\Q\left(\cos\left(\frac{2\pi}{7}\right)\right) : \Q\right] = 3$ (by Theorem 10.7).
This is not a power of 2, hence $\cos\left(\frac{2\pi}{7}\right)$ is not constructible,
and the regular $7$-gon is not constructible.
\vskip 10pt
\noindent
VI.  Let $K$ be the splitting field of $x^{64} - x \in \Z_2[x]$, an extension
field of $F = \Z_2$.
\begin{enumerate}
\item[(A)] (10) Show that the mapping $\varphi : K \to K$ given by 
$\varphi(x) = x^2$ is an $F$-automorphism of $K$.

\sol
$K$ is a finite field of order 64 by our main theorem about finite fields.
Since $64 = 2^6$, $K$ has characteristic $2$.  
The mapping $\varphi$ satisfies 
\begin{eqnarray*}
\varphi(a + b) &=& (a + b)^2 = a^2 + b^2 = \varphi(a) + \varphi(b) \text{ (the ``freshman dream'')}\\
\varphi(ab) &=& (ab)^2 = a^2b^2 = \varphi(a)\varphi(b) \text{ (since multiplication is commutative in $K$)}.
\end{eqnarray*}
In addition, each $a \in K$ satisfies $a^{64} = a$, so $(a^{32})^2 = \varphi(a^{32}) = a$.
This shows that $\varphi$ is onto.  Since $K$ is a finite set, this implies that $\varphi$
is also one-to-one.  Finally, if $c = 0,1$ in $\Z_2$, then $\varphi(c) = c^2 = c$.
Therefore, $\varphi$ is an $F$-automorphism of $K$.
\item[(B)] (10) Show that the mapping $\varphi$ from part (A) is an element
of order 6 in the Galois group ${\rm Gal}_F K$.  
(In fact, ${\rm Gal}_F K = \langle \varphi \rangle$ is cyclic of order 6.)

\sol
If $a \in K$ is an arbitrary element, then consider $\varphi^6(a) = (\varphi\circ\varphi\circ \cdots \circ\varphi)(a)$:
\[
\varphi^6(a) = \varphi(\varphi( \cdots \varphi(a) \cdots ) ) = \left((\cdots a^2 \cdots )^2 \right)^2 = a^{64}.
\]
But every element of $K$ satisfies $a^{64} = a$, so $\varphi^6$ is the identity mapping.
\item[(C)] (5) Find the orders of all the subfields of $K$.  

\sol
The subfields of $K$ have orders $2^d$ where $d|6$, so we have subfields of order
$2^1 = 2$ (i.e. $F$), $2^2 = 4$, $2^3 = 8$, and $2^6 = 64$ (i.e $K$).
(\emph{Comment:} Note that these are in one-to-one correspondence with the subgroups
of the cyclic group $\langle \varphi\rangle$, so the statements of the 
Fundamental Theorem of Galois Theory are holding in this example.  In fact
$K$ is a Galois extension of $F$.) 
\end{enumerate}
VII.  
\begin{enumerate}
\item[(A)] (15) State the Fundamental Theorem of Galois Theory. 

\sol
Let $K$ be a Galois extension of $F$.
\begin{itemize}
\item[(1)]  The subgroups $H$ of $G = {\rm Gal}_F K$ are in one-to-one, 
inclusion-reversing correspondence with the intermediate fields $F \subseteq E \subseteq K$
via the mappings $H \mapsto E_H = \{a \in K : \sigma(a) = a \text{ for all } \sigma \in H\}$
and $E \mapsto {\rm Gal}_E K$.  In this correspondence, $[E_H:F] = [G:H]$ and 
$[K:E_H] = |H|$.  
\item[(2)] In the correspondence from part (1), the intermediate field $E$ is 
a normal extension of $F$ if and only if $H = {\rm Gal}_E K$ is a normal
subgroup of ${\rm Gal}_F K$ and if so,
\[
{\rm Gal}_F K/{\rm Gal}_E K \simeq {\rm Gal}_F E.
\]
\end{itemize}

\item[(B)] (20) Suppose we know that $K$ is a Galois extension of $F$ with
${\rm Gal}_F K$ an \emph{abelian} group of order $18$ whose Sylow $2$-
and $3$-subgroups are both cyclic.   How many intermediate fields $E$, 
$F \subseteq E \subseteq K$, are there and 
what are each of their degrees over $F$?

\sol
By the Structure Theorem for finite abelian groups, we must have 
${\rm Gal}_F K \simeq \Z_2 \oplus \Z_9 \simeq \Z_{18}$, 
hence this Galois group is cyclic, say
${\rm Gal}_F K = \langle\sigma\rangle$.
Every subgroup of a cyclic group of order $n$ is cyclic, with order some $d | n$.
Moreover, there is exactly one such subgroup for each $d$.  Here $n = 18$, 
so there are subgroups of orders $1,2,3,6,9,18$:
\begin{itemize}
\item order 1:  $\langle {\rm id}\rangle$
\item order 2:  $\langle \sigma^9\rangle$
\item order 3:  $\langle \sigma^6\rangle$
\item order 6:  $\langle \sigma^3\rangle$
\item order 9:  $\langle \sigma^2\rangle$
\item order 18: $\langle \sigma\rangle = {\rm Gal}_F K$.
\end{itemize}  
The following diagram shows the inclusions:
\[
\SelectTips {cm}{}
\xymatrix{
&&\{{\rm id}\} \ar[dl] \ar[dr] & \\
&\langle \sigma^6\rangle \ar[dl]\ar[dr]&& \langle \sigma^9\rangle\ar[dl]\\
\langle \sigma^2\rangle \ar[dr]&& \langle \sigma^3\rangle\ar[dl]& \\
&\langle \sigma \rangle = {\rm Gal}_F K&&
}
\]
The corresponding diagram of intermediate fields is
\[
\SelectTips {cm}{}
\xymatrix{
&&K& \\
&E_1\ar[ur]&& E_2\ar[ul]\\
E_3\ar[ur] && E_4\ar[ul]\ar[ur]& \\
&F\ar[ul]\ar[ur]&&
}
\]
where $E_1 = E_{\langle\sigma^6\rangle}$, $E_2 = E_{\langle\sigma^9\rangle}$,
$E_3 = E_{\langle\sigma^2\rangle}$, and $E_4 = E_{\langle\sigma^3\rangle}$.
By part (1) of the FTGT, we have $[E:F] = [{\rm Gal}_F K: {\rm Gal}_E K]$ for
each intermediate field $E$.  Here 
\begin{itemize}
\item $[E_1 : F] = [\langle\sigma\rangle : \langle\sigma^6\rangle] = 18/3 = 6$,
\item $[E_2 : F] = [\langle\sigma\rangle : \langle\sigma^9\rangle] = 18/2 = 9$,
\item $[E_3 : F] = [\langle\sigma\rangle : \langle\sigma^2\rangle] = 18/9 = 2$,
\item $[E_4 : F] = [\langle\sigma\rangle : \langle\sigma^3\rangle] = 18/6 = 3$.
\end{itemize}
\end{enumerate}
\vskip 10pt
\noindent
VIII.  Let $p$ be a prime number and let 
\[
\zeta_p = \cos\left(\frac{2\pi}{p}\right) + i \sin\left(\frac{2\pi}{p}\right),
\]
be the primitive $p$th root of 1 in $\C$.  
\begin{enumerate}
\item[(A)] (10) Show that if $F$ is a field containing $\zeta_p$ and 
$K$ is the splitting field of the polynomial $x^p - a \in F[x]$ for some
$a\in F$ that is not the $p$th power of any element of $F$, then 
${\rm Gal}_F K$ is isomorphic to the additive group $(\Z_p,+)$ (cyclic of order $p$).  

\sol
Let $u$ be any one root of $x^p - a$ in $K$.  Then the roots are
$\{u,\zeta_p u, \zeta_p^2 u, \ldots, \zeta_p^{p-1} u\}$.  Since $\zeta_p \in F$,
this implies that 
\[
K = F(u,\zeta_p u, \zeta_p^2 u, \ldots, \zeta_p^{p-1} u) = F(u),
\]
and the simple extension $K = F(u)$ is actually the splitting field, hence normal.
If $\sigma \in {\rm Gal}_F K$, then $\sigma(u) = \zeta_p^i u$ for some $i = 0,1,\ldots,p-1$.
Assume $i > 0$, and consider the composition $\sigma^k$.  We have 
\[
\sigma^k(u) = (\zeta_p^i)^k u = \zeta_p^{ki} u.
\]
Since the smallest positive power of $\zeta_p$ that equals 1 is $\zeta_p^p = 1$, and $p$ is prime,
this implies the first time we get $\sigma^k(u) = u$ is for $k = p$.
In other words, $\sigma$ is an element of order $p$ in ${\rm Gal}_F K$.  But since 
$x^p - a$ has degree $p$, the dimension $[K:F] \le p$.  It follows that 
$[K:F] = |{\rm Gal}_F K| = p$.   Hence ${\rm Gal}_F K \simeq \Z_p$.

\item[(B)] (10) Show that ${\rm Gal}_\Q \Q(\zeta_p)$ is isomorphic to 
the multiplicative group $(\Z_p^*,\cdot)$ (cyclic of order $p-1$).  

\sol
The primitive $p$th root of unity $\zeta_p$ is a root of 
\[
f(x) = \frac{x^p - 1}{x - 1} = x^{p-1} + \cdots + x + 1 \in \Q[x].
\]
This is an irreducible polynomial, so $[\Q(\zeta_p) : \Q] = p-1$.  The roots of 
$f(x)$ are $\zeta_p,\zeta_p^2,\ldots,\zeta_p^{p-1}$, and there are elements 
of the Galois group mapping $\zeta_p$ to each of these powers.  Say $\sigma_i(\zeta_p) = \zeta_p^i$.
Then $\sigma_i(\sigma_j(\zeta_p)) = (\zeta_p^j)^i = \zeta_p^{(ij)}$.  Hence $\sigma_i\circ\sigma_j
=\sigma_{ij \bmod p}$.  This shows that the mapping 
\begin{eqnarray*} 
\varphi : \Z_p^* &\to& {\rm Gal}_\Q \Q(\zeta_p)\\
            i\bmod p &\mapsto & \sigma_i
\end{eqnarray*}
is an isomorphism of groups.  

\end{enumerate}
\vskip 10pt
\noindent  
IX.  Let $\zeta_p$ be as in question VIII, and let $K$ be the splitting field
of $x^p - 2 \in \Q[x]$.  
\begin{enumerate}
\item[(A)] (10) Show carefully that $[K:\Q] = p(p-1)$.  

\sol
First, note that by the arguments in VIII (A), the field $K = \Q(\alpha,\zeta_p)$
where $\alpha = {\root p\of 2}$ (the real, positive root).  
Note that $\gcd(p,p-1) = 1$ automatically.  Hence the desired statement about the degree 
follows directly from the result we proved in Exercise 11 of Section 10.3.
Here is the way this result follows ``from first principles.''
We consider the towers of extensions 
\begin{eqnarray*}
&&\Q \subset \Q(\alpha) \subset \Q(\alpha,\zeta_p)\\
&&\Q \subset \Q(\zeta_p) \subset \Q(\alpha,\zeta_p)
\end{eqnarray*}
The polynomial $x^p - 2$ is irreducible in $\Q[x]$ (by Eisenstein for the prime 2).  Therefore
we know $[\Q(\alpha):\Q] = p$.  The degree $[\Q(\zeta_p) : \Q] = p - 1$ by VIII (B).
Hence $[K:\Q]$ must be divisible by both $p$ and $p-1$.  However, since $\gcd(p,p-1) = 1$,
this implies that the degree must be divisible by $p(p-1)$.  But this implies that
$[K:\Q] = p(p-1)$ for the following reason.
The degree $[K:\Q(\zeta_p)] \le p$ because the minimal
polynomial of $\alpha$ over $\Q(\zeta_p)$ must be some factor
of $x^p - a$.  Hence $[K:\Q] = [K:\Q(\zeta_p)][\Q(\zeta_p):\Q] \le p(p-1)$.
Since $[K:\Q] \le p(p-1)$, and is divisible by $p(p-1)$, it must equal $p(p-1)$.

\item[(B)] (5)  Is ${\rm Gal}_\Q K$ a solvable group?  Why or why not?
\end{enumerate}

\sol 
Yes, ${\rm Gal}_\Q K$ \emph{is solvable}.  A ``fast'' proof is to use the 
Galois Criterion:  The Galois group of the splitting field of a 
polynomial is solvable if and only if the polynomial is solvable
by radicals.  This is clearly true for the polynomial $f(x) = x^p - a$.
We have that the distinct roots are $x = \zeta_p^i {\root p \of a}$ for 
$i = 0,1,\ldots,p-1$ (!)

Here is an alternate proof ``from first principles.''
Consider the second tower in the solution 
for part A.  The intermediate field $\Q(\zeta_p)$ is normal over $\Q$, since
it is the splitting field of the polynomial $x^p - 1$ in $\Q[x]$.  Hence
by the second part of the FTGT, we have that the subgroup $G_1$ corresponding
to $\Q(\zeta_p)$ in $G = {\rm Gal}_\Q K$ is normal and by VIII (B)
\[
G/G_1 \simeq {\rm Gal}_\Q \Q(\zeta_p) \simeq \Z_p^*,
\]
which is an abelian group.  Moreover by the first part of the FTGT, 
$G_1 = {\rm Gal}_{\Q(\zeta_p)} K$, and this group is isomorphic to
$\Z_p$ by VIII (A).  This is also abelian.  Hence the chain of subgroups
\[
G = G_0 \supset G_1 \supset \{id\}
\]
shows that $G$ is a solvable group.

\end{document}
