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\centerline{Mathematics 352 -- Abstract Algebra II}
\centerline{Midterm Exam 1 Solutions -- February 27, 2008}
\vskip 10pt
\noindent
I.  {\it Terminology.}
\vskip 10pt
\begin{enumerate}
\item[A)] (5) Let $G$ be a group and let $x$ in $G$.
What is the {\it conjugacy class} of $x$ in $G$?
\sol
The conjugacy class of $x$ is
\[
C_x = \{g^{-1}xg \in G : g\in G\}.
\]
(In other words, this is the equivalence class of $x$ for the conjugacy relation
on $G$.)
\item[B)] (5) Let $K$ be an extension field of the field $F$ and 
let $u \in K$.  What does it mean to say that $u$ is {\it transcendental}
over $F$?
\sol
The element $u\in K$ is transcendental over $F$ if there is no 
nonzero polynomial $f(x) \in F[x]$ such that $f(u) = 0$ (or, equivalently,
that $u$ is not algebraic over $F$.) 
\item{C)} (5) Let $K$ be an extension field of the field $F$.  What
does it mean to say that $K$ is {\it normal} over $F$? 
\sol
$K$ is a normal extension of $F$ if $K$ is algebraic over $F$ and
if $p(x)$ is an irreducible polynomial with one root in $K$, then
$p(x)$ splits completely in $K[x]$ (that is, in $K[x]$,
\[
p(x) = c(x - u_1) \ldots (x - u_n),
\]
so all the roots of $p(x)$ are in $K$.)
\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 a direct sum of cyclic subgroups of
prime power order.

\item[B)] (10)  Let $G = \Z_{42} \oplus \Z_{36} \oplus \Z_{84}$.
What are the elementary divisors and invariant factors of $G$?  
\sol
From the factorizations $42 = 2\cdot 3\cdot 7$, $36 = 2^2 \cdot 3^2$, and 
$84 = 2^2 \cdot 3 \cdot 7$, we see that the elementary divisors of $G$
are $2,4,4,3,3,3^2,7,7$, so 
\[
G\simeq \Z_2 \oplus \Z_4 \oplus \Z_4 \oplus \Z_3 \oplus \Z_3 \oplus \Z_9 \oplus \Z_7 \oplus \Z_7.
\]
Regrouping these as usual, we see the invariant factors of $G$ are
$6,84,252$, so that 
\[
G\simeq \Z_6 \oplus \Z_{84} \oplus \Z_{252}.
\]
\item[C)] (5)  What is the Sylow $7$-subgroup of $G$ from part B?  Is this a cyclic group?  
\sol
The Sylow 7-subgroup is isomorphic to $\Z_7 \oplus \Z_7$ from the decompositions
above.  It is not cyclic.  If we take the original direct sum decomposition 
$G = \Z_{42} \oplus \Z_{36} \oplus \Z_{84} = \langle a\rangle \oplus \langle b\rangle \oplus \langle c\rangle$
then the Sylow 7-subgroup is generated by $(a^6,0,0)$ and $(0,0,c^{12})$ (two elements of order $7$).  
\end{enumerate}

\vskip 10pt
\noindent  
III.  
\begin{enumerate}
\item{A)} (20) State and prove the First Sylow Theorem for finite groups.
\sol
The First Sylow Theorem states that if $G$ is a finite group and $p$ is a
prime number such that $p^n | |G|$, then $G$ contains a subgroup 
of order $p^n$.  

The proof is by induction on $|G|$.  When $|G| = 1$, there is nothing to 
prove.  So now for the induction step, assume that the statement of the 
theorem is true for all groups of order $< m$ and consider $G$ of 
order $m$.  By the Class Equation, we have 
\[
|G| = |Z(G)| + \sum_{i=1}^t [G:C(g_i)],
\]
where the $g_i$ are representatives of the distinct conjugacy classes
of size $>1$ and $C(g_i)$ is the centralizer of $g_i$.  

Case 1:  Assume that there is some $i$ such that $[G:C(g_i)]$
is \emph{not} divisible by $p$.  Then $p^n | |C(g_i)|$.  We must 
have $|C(g_i)| < |G|$ since $g_i \notin Z(G)$.  Therefore, by 
the induction hypothesis, $C(g_i)$ has a subgroup of order
$p^n$.  This subgroup is also a subgroup of $G$ and we are done in
this case.  

Case 2:  Now assume that $p|[G:C(g_i)]$ for all $i = 1,\ldots,t$.  Then 
since $p | |G|$, we also have $p||Z(G)|$.  Now $Z(G)$ is a finite abelian
group, so we know by a consequence of the Structure Theorem that
$Z(G)$ contains an element $x$ of order $p$.  Let $C = \langle x\rangle$
be the subgroup generated by this $x$.  Since $C \subseteq Z(G)$, $C$
is normal in $G$ and we can form $G/C$ which has order divisible by 
$p^{n-1}$.  By the induction hypothesis applied to $G/C$, $G/C$ has
a subgroup $H$ of order $p^{n-1}$.  But then the inverse image of $H$
under the quotient map $G \longrightarrow G/C$ is a subgroup of $G$
of order $p^n$.

\item[B)] (10) Use the Sylow Theorems to show that every group of order 33
is cyclic.

\sol
By the Third Sylow Theorem, the number of Sylow 3-subgroups is congruent to 
$1$ mod $3$ and divides $33$.  The only possibility is $1$, so let $H$ be the unique
Sylow $3$-subgroup.  Similarly, there
is exactly one Sylow $11$-subgroup $K$.  This implies that $H$ and $K$ are
normal subgroups.  Moreover $H \cap K = \{e\}$, since if not any $x\ne e$
would simultaneously have order $3$ and order $11$, which is impossible. 
Therefore, $G \simeq H \times K \simeq \Z_3 \times \Z_{11} \simeq Z_{33}$,
where the last isomorphism follows because $(3,11) = 1$.  
\end{enumerate}
\vskip 10pt
\noindent
IV. 
\begin{enumerate} 
\item[A)]  (5) Let $K$ be an extension field of $F$, and let $u\in K$.  Show that
if $u^2$ is algebraic over $F$, then $u$ is also algebraic over $F$.
\sol
(Method 1)  If $u^2$ is algebraic over $F$, then there is some nonzero polynomial
$p(x) \in F[x]$ such that $p(u^2) = 0$.  This implies that the 
polynomial $q(x) = p(x^2)$ has $u$ as a root.  Therefore $u$ is also 
algebraic over $F$.   (Note:  If $p(x) = a_n x^n + \cdots + a_1 + a_0$,
then the polynomial $q(x)$ is just $q(x) = a_n x^{2n} + \cdots + a_1 x^2 + a_0$.)

\sol (Method 2) Since $u^2$ is algebraic over $F$, we know that 
$[F(u^2): F] = n$, the degree of the minimal polynomial of $u^2$ in $F[x]$.
Now consider the polynomial $x^2 - u^2 \in F(u^2)[x]$.  This has $u$ as
a root, so $u$ is algebraic over $F(u^2)$, and $[F(u^2)(u):F(u^2)] = [F(u):F(u^2)] \le 2$.
Therefore $[F(u):F] = [F(u):F(u^2)][F(u^2):F] \le 2n$ is finite, which implies
that $u$ is algebraic over $F$.  

\item[B)]  (10)  Show that $\Q(\sqrt{2},\sqrt{3})$ is a splitting 
field of $f(x) = x^4 + 2x^3 - 8x^2 - 6x - 1$ over $\Q$.  (Hint: Look for
quadratic factors of $f(x)$.)
\sol
The factorization of $f(x)$ in $\Q[x]$ is $f(x) = (x^2 + 4x + 1)(x^2 - 2x - 1)$.
The roots of the first factor are $x = \frac{-4 \pm \sqrt{16 - 12}}{2} = -2 \pm \sqrt{3}$.
The roots of the second factor are $x = \frac{2 \pm \sqrt{4+4}}{2} = 1 \pm \sqrt{2}$.
Hence the splitting field of $f(x)$ over $\Q$ is, by definition,
\[
K = \Q(-2 + \sqrt{3}, -2 - \sqrt{3}, 1 + \sqrt{2}, 1 - \sqrt{2}).
\]
We claim that $K = \Q(\sqrt{2},\sqrt{3})$.  The inclusion $K \subseteq \Q(\sqrt{2},\sqrt{3})$
is clear since each root of $f$ is contained in $\Q(\sqrt{2},\sqrt{3})$.  The other 
inclusion follows since $\sqrt{2} = \frac{1}{2}(1 + \sqrt{2}) + \frac{-1}{2}(1 - \sqrt{2})$
and $\sqrt{3} = \frac{1}{2}(-2 + \sqrt{3}) + \frac{-1}{2}(-2 - \sqrt{3})$ are both in $K$.
So $\Q(\sqrt{2},\sqrt{3}) \subseteq K$.
\end{enumerate}

\noindent
V.  (15) Show that if $F \subseteq E \subseteq K$ are field extensions and 
$[E:F], [K:E]$ are both finite, then $[K:F] = [K:E][E:F]$.  
\sol
Say $\{u_1,\ldots,u_m\}$ is a basis for $E$ over $F$ and $\{v_1,\ldots,v_n\}$
is a basis for $K$ over $F$.  Then each element $a$ in $K$ can be written 
as 
\[
a = c_1 v_1 + \cdots + c_n v_n
\]
for some $c_i \in E$.  Similarly, we have $c_i = a_{i1}u_1 + \cdots + a_{im}u_m$
were $a_{ij} \in F$.  Substituting into the last displayed equation,
\begin{eqnarray*}
a &=& (a_{11}u_1 + \cdots + a_{1m}u_m)v_1 + \cdots +  (a_{n1}u_1 + \cdots + a_{nm}u_m)v_n\\
&=& \sum_{i=1}^n \sum_{j=1}^m a_{ij} v_i u_j.
\end{eqnarray*}
This shows that the $m\cdot n$ elements $v_i u_j$ for $i = 1,\ldots,n$ and $j = 1, \ldots, m$
span $K$ over $F$.  To show that these form a basis of $K$ over $F$, we need to 
show that they are linearly independent.  Suppose there are $a_{ij} \in F$ such that
\begin{eqnarray*}
0 &=& \sum_{i=1}^n \sum_{j=1}^m a_{ij} v_i u_j\\
&=& (a_{11}u_1 + \cdots + a_{1m}u_m)v_1 + \cdots +  (a_{n1}u_1 + \cdots + a_{nm}u_m)v_n.
\end{eqnarray*}
Since the $v_i$ are linearly independent over $E$, this shows that 
\[
a_{i1}u_1 + \cdots + a_{im}u_m = 0
\]
for all $i = 1, \ldots, n$. But then, the $u_j$ are linearly independent over $F$, so 
all $a_{ij} = 0$.  This shows the linear independence so $\{v_j u_i : 1 \le i \le n, 1 \le j \le m\}$
is a basis of $K$ over $F$ and 
\[
[K:F] = m \cdot n = [E:F][K:E].
\]

\noindent
{\it Extra Credit}  (10)  Let $p(x)$ and $q(x)$ be irreducible polynomials
in $F[x]$ such that $\deg p(x)$ and  $\deg q(x)$ are relatively prime integers.
Show that if $u$ is a root of $p(x)$ in
some extension field of $F$, then $q(x)$ is also irreducible in $F(u)[x]$.  
\sol
If $u$ is a root of $p(x)$ in some extension field of $F$, then by Theorem 10.7, 
$[F(u) : F] = \deg p(x) = m$.  Similarly if $v$ is a root of $q(x)$, we have
$[F(v) : F] = \deg q(x) = n$.  Since $m = \deg p(x)$ and $n = \deg q(x)$ are relatively
prime, it follows as in Exercise 11 from Section 10.3 of Hungerford 
that $[F(u,v) : F] = mn$.  From the tower $F \subseteq F(u) \subseteq F(u,v)$,
we get $[F(u,v) : F] = mn = [F(u,v): F(u)][F(u):F]$, so $[F(u,v):F(u)] = n$.
This says that the minimal polynomial of $v$ over $F(u)$ must have degree $n$, 
and it must be a divisor of $q(x)$.  Since $q(x)$ has degree $n$ itself, that
minimal polynomial must be $q(x)$, and hence $q(x)$ is still irreducible 
in $F(u)[x]$.  
\end{document}
\bye
