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\centerline{Mathematics 352 -- Abstract Algebra II}
\centerline{Solutions for Problem Set 4}
\centerline{February 11, 2008}
\vskip 10pt
\noindent
{\it Section 8.5}
\vskip 10pt
\noindent
17.  Let $G$ be a group of order 21.
We have $21 = 3\cdot 7$.  Since 
$7 \equiv 1 \bmod 3$, Corollary 8.18 does not apply.
However, by Sylow 3, there is one Sylow 7-subgroup $H$, which 
is normal by Sylow 2.  Moreover, there are either
1 or 7 Sylow 3-subgroups. 

In the case that the Sylow 3-subgroup $K$
is unique the reasoning of the proof
of Corollary 8.18 applies.  We have $H\cap \{e\}$ and $HK = G$
since $HK$ contains 21 distinct elements of $G$.  Therefore 
by Theorem 8.3,
\[
G \simeq H \times K \simeq \Z_7 \times \Z_3 \simeq \Z_{21},
\]
and $G$ is cyclic. 

We now consider the case where there are 7 Sylow $3$-subgroups
$K_1,\ldots,K_7$.  We claim that there is \emph{only one} possible
structure for $G$, up to isomorphism: We will show
that $G$ is generated by two elements $a,b$ with $a^7 = e$, 
$b^3 = e$ and $b^{-1}ab = a^2$.  Let $H = \langle a\rangle$
so $|a| = 7$.  Similarly, let $K_1 = \langle b\rangle$, so 
$|b| = 3$.  Note that since $H \cap K = \{e\}$, 
we have $G = \{b^j a^i : 0 \le k \le 6, 0\le j \le 2\}$.
Since $H$ is normal, the conjugate $b^{-1} a b = a^k$
for some $k \in \{1,2,3,4,5,6\}$.  
Moreover, 
\begin{eqnarray*}
b^{-2} a b^2 &=& b^{-1} (b^{-1}ab) b = b^{-1} a^k b = (b^{-1}ab)^k = (a^k)^k = a^{k^2}\\
a = b^{-3} a b^3 &=& a^{k^3},
\end{eqnarray*}
so 
\begin{equation}\label{key}
k^3 \equiv 1 \bmod 7.
\end{equation}
As in the dihedral group case,
knowing $k$ in (\ref{key}), or the equivalent form $ab = ba^k$
lets us determine the products 
\[
(b^ja^i)(b^{j'}a^{i'}) = b^{j+j'} a^{i k^{j'} + i'}.
\]
This gives closure under the product operation.  Associativity follows from 
this also since 
\begin{eqnarray*}
\left((b^ja^i)(b^{j'}a^{i'})\right)(b^{j''}a^{i''}) &=&
(b^{j+j'} a^{i k^{j'} + i'})(b^{j''} a^{i''}) = b^{(j+j') + j''} a^{(i k^{j'} + i')k^{j''} + i''}\\
(b^ja^i)\left((b^{j'}a^{i'})(b^{j''}a^{i''})\right) &=&
(b^ja^i)(b^{j'+j''} a^{i' k^{j''} + i''}) = b^{j+(j' + j'')} a^{i k^{j'+j''} + i' k^{j''} + i''},
\end{eqnarray*}
which are the same.  The identity element is $e = b^0a^0$.  The inverse of 
$b^j a^i$ is $b^{j'} a^{i'}$ where $j' + j \equiv 0 \bmod 3$ and $i k^{j'} + i' \equiv 0 \bmod 7$.
Note that $i k^{j'} + i' \equiv 0 \bmod 7$ implies that 
$i' k^j + i \equiv 0 \bmod 7$ also when $j' + j \equiv 0 \bmod 3$ because of (\ref{key}), so 
this is a well-defined two-sided inverse. 
So we get a group of order $21$ whenever
the condition (\ref{key}) holds.  

\begin{itemize}
\item If $k = 1$, then $G$ is abelian and we are back in the first case.
\item We cannot have $k = 3,5,6$ since then $3^3 = 27 \equiv 6 \bmod 7$,
$5^3 = 125 \equiv 6 \bmod 7$, and $6^3 = 216 \equiv 6 \bmod 7$.  
\item This leaves $k = 2,4$.  We claim these two groups are actually 
isomorphic.  This follows since if $b^{-1}ab = a^2$, then $b^{-2}ab^2 = a^4$.
The element $b^2$ also generates the Sylow $3$-subgroup $K_1$, so 
in fact this is the same group up to isomorphism.
\end{itemize}   
\vskip 10pt
\noindent
{\it Section 10.2}
\vskip 10pt
\noindent
17.  (a)  We have $x = \sqrt{1 + \sqrt{5}}$ so $x^2 = 1 + \sqrt{5}$ and 
$(x^2 - 1)^2 = 5$, so $x^4 - 2x^2 - 4 = 0$.  It is not hard to check that
this polynomial is irreducible over $\Q$, so this is the minimal polynomial.

(b)  For $x = i\sqrt{3} + \sqrt{2}$, $x^2 = -1 + 2i\sqrt{6}$,  
$x^3 = -3i\sqrt{3} - 7\sqrt{2}$ and $x^4 = -23 - 4i\sqrt{6}$.
Hence $x^4 + 2x + 25 = 0$.    It is not hard to check that
this polynomial is irreducible over $\Q$, so this is the minimal polynomial.

\vskip 10pt
\noindent
19.  By Theorem 10.4, we have $[F(u):F] = [F(u):E][E:F]$.
However, by Theorem 10.7, $[F(u):F] = p$, which is prime.
Therefore either $[F(u):E] = p$ and $[E:F] = 1$, in which case $E = F$, or
$[F(u):E] = 1$ and $[E:F] = p$, in which case $E = F(u)$.

\vskip 10pt
\noindent
20.  We have $F \subseteq F(u^2) \subseteq F(u)$  since $u^2 \in F(u)$.
Therefore, $[F(u):F] = [F(u):F(u^2)][F(u^2):F]$ by Theorem 10.4.  
By Theorem 10.7, $[F(u) : F(u^2)]$ is either $1$ or $2$
since the polynomial $f(x) = x^2 - u^2$ 
with coefficients in $F(u^2)$ has $x = u$ as a root.  But the minimal polynomial
of $u$ over $F$ has odd degree so $[F(u):F]$ is odd by Theorem 10.7.  This 
implies that $[F(u):F(u^2)] = 1$.  Hence $F(u) = F(u^2)$.

\vskip 10pt
\noindent
23.  If $u \in K$ but $u \notin \Q$, then have $\Q \subseteq \Q(u) \subseteq K$
By Theorem 10.4 and the hypothesis $[K:\Q] = 2$, this implies $[\Q(u):\Q] = 2$
and $[K:\Q(u)] = 1$, so $K = \Q(u)$.  By Theorem 10.7, the minimal polynomial
of $u$ is of the form $x^2 + bx + c$ for some $b,c \in \Q$.  Applying the 
quadratic formula, 
\[
u = \frac{-b \pm \sqrt{b^2 - 4c}}{2} = \frac{-b}{2} \pm \frac{\sqrt{b^2-4c}}{2}.
\]   
First, we claim that $\Q(u) = \Q(\sqrt{b^2 - 4c})$.  The $\subseteq$ inclusion
here is clear since $u = \frac{-b}{2} \pm \frac{1}{2} \sqrt{b^2 - 4c} \in \Q(\sqrt{b^2 - 4c})$.  Hence
$\Q(u) \subseteq \Q(\sqrt{b^2 - 4c})$.  On the other hand 
$\sqrt{b^2 - 4c} = \mp(2u + b)$ which implies $\sqrt{b^2 - 4c} \subset \Q(u)$
and hence $\Q(\sqrt{b^2 - 4c}) \subseteq \Q(u)$.  

Next, write $b^2 - 4c = \frac{m^2}{n^2}\cdot\frac{e}{f}$ where $m,n,e,f \in \Z$ are
\emph{integers} and $e,f$ are not divisible by the square of any prime, and  
assume $b^2 - 4c = \frac{m^2e}{n^2f}$ is in lowest terms.  Then 
$\sqrt{b^2 - 4c} = \frac{m}{n}\sqrt{\frac{e}{f}}$ and reasoning similarly to the 
above, we have 
\[
\Q(u) = \Q(\sqrt{b^2-4c}) = \Q\left(\sqrt{\frac{e}{f}}\right).
\]

Finally, note that $\sqrt{\frac{e}{f}} = \frac{\sqrt{ef}}{f}$, where $f \in \Z$, and
$d = ef$ is not divisible by the square of any prime in $\Z$.  Then by similar
reasoning to the above, we have 
\[
\Q(u) = \Q\left(\sqrt{\frac{e}{f}}\right) = \Q(\sqrt{d}),
\]
which is what we wanted to show.
\vskip 10pt
\noindent
25.  Aiming for a contradiction, suppose that $u$ is transcendental
over $F$, but some element of $F(u)$ 
not in $F$ is algebraic over $F$.  Then we have some 
\[
v = \frac{a_0 + a_1 u + \cdots + a_n u^n}{b_0 + b_1 u + \cdots + b_m u^m} \in F(u)
\]
(but not in $F$, so $v \ne 0$)
which is a root of some nonzero polynomial $f(x) = c_0 + c_1 x + \cdots + c_k x^k \in F[x]$. 
In the expressions for $v$ and for $f(x)$, we may assume that $a_n \ne 0$,
$b_m \ne 0$, and $c_0, c_k \ne 0$ in $F$.  (The reason we can assume $c_0 \ne 0$
is that $v \ne 0$.  So if $v$ is a root of any polynomial in $F[x]$, then $v$
is a root of some polynomial in $F[x]$ with nonzero constant term.)   We may also assume that 
$f(x) = 0$ \emph{has no roots in $F$}.  If there are any such roots $a \in F$, then we 
know $f(x)$ will factor in $F[x]$ as $f(x) = (x - a)q(x)$ for some $q(x) \in F[x]$.
Then $f(v) = 0$ implies $q(v) = 0$ since $v \ne a \in F$.  So we can remove any 
factors producing roots in $F$ and we will be left with $f(x)$ having no roots in $F$.  

Substitute $x = v \ne 0$ in $f(x)$ and put the terms over a common denominator and 
collect like powers of $u$ on the top.  The result is an equation 
\[
0 = c_0 + c_1 v + \cdots + c_k v^k = c_0 + \cdots + c_k \left(\frac{a_0 + a_1 u + \cdots + a_n u^n}{b_0 + b_1 u + \cdots + b_m u^m}\right)^k = 
\frac{A_0 + A_1 u + \cdots + A_N u^N}{(b_0 + b_1 u + \cdots + b_m u^m)^k},
\]
where $A_i \in F$ for all $i$.  This implies 
\[
0 = A_0 + A_1 u + \cdots + A_N u^N.  
\]
If $n > m$ then $N = nk$ and $A_N = c_k a_n^k \ne 0$ so the polynomial $g(x) = A_0 + A_1 x + \cdots 
+ A_N x^N$ is not zero in $F[x]$.  Similarly, if $m > n$, then $N = mk$ and $A_N = c_0 b_m^k \ne 0$,
so $g(x)$ is not zero in $F[x]$.  Finally, if $m = n$, then $N = mk = nk$ and the coefficient
$A_N$ is 
\[
A_N = c_0 b_n^k + c_1 b_n^{k-1} a_n + \cdots + c_{k-1}b_n a_n^{k-1} + c_k a_n^k.
\]
If $A_N = 0$, then dividing through by $b_n^k$ we get
\[
0 = c_0 + c_1 \left(\frac{a_n}{b_n}\right) + \cdots + c_k \left(\frac{a_n}{b_n}\right)^k.
\]
But this implies that $f(x) = 0$ has $x = \frac{a_n}{b_n} \in F$ as a root.  This
contradicts what we said above about $f(x)$.  Hence $A_N \ne 0$ in this case too.
It follows that $u$ is a root of a nonzero polynomial in $F[x]$, which is a contradiction
to the hypothesis that $u$ is transcendental over $F$.  
\end{document}
