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\centerline{Mathematics 352 -- Abstract Algebra II}
\centerline{Solutions for Problem Set 3}
\centerline{February 4, 2008}
\vskip 10pt
\noindent
{\it Section 8.4}
\vskip 10pt
\noindent
6.  Let $H \subseteq K$ be subgroups of $G$ such that $H$ is normal as
a subgroup of $K$.  This means that for all $x \in K$, $x^{-1}Hx = H$.
By the definition, this means $x \in N(H)$, so $K \subseteq N(H)$.
\vskip 10pt
\noindent
7.  (a)  Let $A \subseteq G$ be a subgroup.  Let $x \in A$.  Then 
$x^{-1}Ax \subseteq A$ since $A$ is closed under inverses and products.
On the other hand if $y \in A$ is an arbitrary element, then 
$xyx^{-1} \in A$ since $A$ is a subgroup and hence
$x^{-1} (xyx^{-1}) x = y \in x^{-1}Ax$.  Therefore $A \subseteq x^{-1}Ax$,
so combining the two inclusions, $A = x^{-1}Ax$.  This shows $A \subset N(A)$.

(b)  We have 
\begin{eqnarray*}
g \in N(A) &\Longleftrightarrow& g^{-1}Ag = A \quad \text{by definition}\\
           &\Longleftrightarrow& Ag = gA \quad \text{ (multiply by $g$ on left)}.
\end{eqnarray*}
\vskip 10pt
\noindent
12.  Let $K$ be a Sylow $p$-subgroup of $G$, and $N$ be a normal subgroup 
of $G$.  Assume that $K$ is a normal subgroup of $N$.  We claim first that
this means that $K$ is also a Sylow $p$-subgroup of $N$.  
Assume $|G| = p^t m$ with $(p,m) = 1$, so $|K| = p^t$.  But
by Lagrange's theorem, $|N|$ divides $|G|$, and so the largest power
of $p$ that can divide $N$ is $p^t$.  Since $K \subseteq N$, we must 
have $|N| = p^t m'$ where $(p,m') = 1$, and hence $K$ is also a Sylow $p$-subgroup 
of $G$.  Now let $x$ be any element of $G$.  Since $N$ is normal in $G$, $x^{-1}Nx = N$.
Since $K \subseteq N$, this implies $x^{-1}Kx \subseteq N$.  But
since $|x^{-1}Kx| = |K|$, it follows that
$x^{-1}Kx = K$ is also a Sylow $p$-subgroup of $N$.  
Since $K$ is normal in $N$, Sylow 2 implies that there is 
just one such Sylow $p$-subgroup of $N$, namely $K$.  This means that $x^{-1}Kx = K$.
Therefore, $K$ is also normal in $G$.  
\vskip 10pt
\noindent
{\it Section 8.5}
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\noindent
1.  
We have $|G| = p^2 q$ with $p,q$ prime, $p < q$ and $q \not\equiv 1 \bmod p$.
The only hypothesis we are missing to apply Theorem 8.30 is the one that 
says $p^2 \not\equiv 1\bmod q$.
Suppose that $q | (p^2 - 1)$.  Then since $q$ is prime, $q | (p + 1)$ or $q | (p - 1)$.
The second is impossible since $p < q$ so $p - 1 < q$ as well.  The first is also impossible.  
Suppose that $q|(p+1)$.  There are two 
cases to consider.  First, if $p = 2$, then $q = 3$ and this contradicts $q \not\equiv 1 \bmod p$.
Second, if $p$ is an odd prime, then $p+1$ is even, $p + 1 = 2\ell$ for some $\ell < p$.
It must be the case that $q|\ell$ since $q > p$, so $q$ is also an odd prime.  
But this is also impossible since $p < q$.  Hence the hypothesis
for Theorem 8.30 is satisfied, and $G$ must be abelian. 
\vskip 10pt
\noindent
6.  The $n$ for which these theorems are sufficient are 
\begin{eqnarray*}
\text{Theorem 7.28 (prime order)} & \text{applies to} & n = 2,3,5,7,11,13,17,19,23,29,31,37,41,\\
                                 & \text{applies to} & \quad  43,47,53,59,61,67,71,73,79,83,89,97\\
\text{Corollary 8.29 (order $p^2$)} & \text{applies to} & n = 4,9,25,49\\
\text{Theorem 8.33 (order $2p$)} & \text{applies to} & n = 6,10,14,22,26,34,38,46,58,62,74,82,86,94\\
\text{Corollary 8.18 (order $pq$)} & \text{applies to} & n = 15,33,35,51,65,69,77,85,87,91,95\\
\text{Theorem 8.30 (order $p^2q$)} & \text{applies to} & n = 45,99\\
\text{Theorem 8.34 (order $8$)} & \text{applies to} & n = 8\\
\text{Theorem 8.35 (order $12$)} & \text{applies to} & n = 12\\
\text{none of these} & \text{apply to} & n = 16,18,20,21,24,27,28,30,32,36,39,\\
                                      && \quad 40,42,44,48,50,52,54,55,56,57,60,63\\
                                      && \quad 64,66,68,70,72,75,76,78,80,81,84,88\\
                                      && \quad 90,92,93,96,98,100.
\end{eqnarray*}
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\noindent
8.  (a)  We have
\[
\begin{pmatrix} \pm 1 & a \\
                0 & 1\\
\end{pmatrix}
\begin{pmatrix} \pm 1 & b \\
                0 & 1\\
\end{pmatrix}
= 
\begin{pmatrix} \pm 1 & a \pm b \\
                0 & 1\\
\end{pmatrix}.
\]
Moreover,
\[
\begin{pmatrix} \pm 1 & a \\
                 0 & 1\\
\end{pmatrix}^{-1}
=
\begin{pmatrix} \pm 1 & \mp a \\
                0 & 1\\
\end{pmatrix}
\]
Hence this set of matrices is closed under products and inverses and 
forms a subgroup of ${\rm GL}(2,\Z_n)$.

(b)  Let $R = \begin{pmatrix} 1 & 1 \\
                0 & 1\\
\end{pmatrix}.$
Then $R^n = I$ (the identity matrix).  Similarly let
$D = \begin{pmatrix} - 1 & 0 \\
                        0 & 1\\
\end{pmatrix} $
so $D^2 = I$.  We have 
\[
DR = \begin{pmatrix} - 1 & 0 \\
                        0 & 1\\
\end{pmatrix}
\begin{pmatrix}  1 & 1 \\
                 0 & 1\\
\end{pmatrix}
=
\begin{pmatrix}  -1 & -1 \\
                   0 & 1\\
\end{pmatrix}
=
\begin{pmatrix}  1 & -1 \\
                 0 & 1\\
\end{pmatrix}
\begin{pmatrix}  -1 & 0 \\
                 0 & 1\\
\end{pmatrix}
=
R^{-1}D.
\]
Therefore the mapping given by $R^i D^j \mapsto r^i d^j$ 
is an isomorphism from this group of matrices to $D_n$. 
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\noindent
10.  We have $D_n = \langle r,d : d^n = e, r^2 = e, dr = r^{-1}d\rangle$, and 
every element can be written uniquely as $r^id^j$ with $0 \le i \le n - 1$
and $0 \le j \le 1$.
(a)  If $n = 2k$, then 
$r^i r^k = r^{i+k} = r^k r^i$, so $r^k$ commutes with all the elements of the 
subgroup generated by $r$.  Moreover, applying the relation $dr = r^{-1}d$ repeatedly,
\[
(r^i d) r^k = r^i(dr^k) = r^i (r^{-1})^k d = r^i (r^k)^{-1} d = (r^i r^k) d = (r^k r^i) d = r^k (r^i d).
\]
Therefore $r^k \in Z(D_n)$.   

(b) We show that every other element of $D_n$ besides $e,r^k$
does not commute with something, 
which will imply that $Z(D_n) = \{e,r^k\}$.  Consider the elements $r^i$ for 
$i \ne 0, k$ first.  We have $d r^i = (r^{-1})^i d = (r^i)^{-1} d = r^{n - i}d$.
Since $i \ne k$, $i \not\equiv n - i \bmod n$, and this shows that $r^i \notin Z(D_n)$.  
Similarly, the element $r^i d$ does not commute with $r$:
\[
(r^i d) r = r^i (dr) = r^i r^{-1} d = r^{i-1}d \quad \text{but} 
\quad
r (r^id) = r^{i+1} d.
\]  
Therefore $r^i d \notin Z(D_n)$ for all $i$.  

(c) The same reasoning as in part (b) shows that when $n$ is odd, $Z(D_n) = \{e\}$
since there is no element such that $(r^i)^{-1} = r^i$ except $r^0 = e$.
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\noindent
11.  By consulting the group table, $Z(Q) = \{1,-1\}$.
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