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\centerline{Mathematics 351 -- Abstract Algebra 1}
\centerline{Solutions for Problem Set 9}
\centerline{November 9, 2007}
\vskip 10pt
\noindent
{\it Section 7.5}
\bigskip
\noindent
5.  (a)  ${\mathbb Z}_{24}$ has order 24, so by Lagrange's
Theorem, the possible orders of subgroups of ${\mathbb Z}_{24}$ are 
$1,2,3,4,6,8,12,24$.  
\vskip 10pt
\noindent
(b)  $S_4$ also has order $4! = 24$, so 
the possible orders of the subgroups of $S_4$ are the same
as in part (a):  $1,2,3,4,6,8,12,24$.  
\vskip 10pt
\noindent
(c)  Since $|D_4| = 8$
and $|{\mathbb Z}_{10}| = 10$, we have $|D_4\times {\mathbb Z}_{10}| = 80$.
Hence the possible orders of subgroups of this group are all
the divisors of 80:  $1,2,4,5,8,10,16,20,40,80$.
\vskip 10pt
\noindent
6.  (a)  Let $G = ({\mathbb Z} \times {\mathbb Z},+)$, and let 
$H = \{(pa,b): a,b \in {\mathbb Z}\}$.  Then $H$ is a subgroup of 
$G$.  There only finitely many distinct cosets, namely
the cosets  
$$(0,0) + H, (1,0) + H, \ldots, (p-1,0) + H$$
(see the solution for problem 12 from Section 6.3 from Problem Set 7).
Hence $[G:H] = p$ is finite.  
\vskip 10pt
\noindent
(b)  Let $G$ be the same group as in part (a), but now let
$H = \{(0,b) : b \in {\mathbb Z}\}$.  The cosets $(a,0) + H$ 
are all distinct.  This follows since if $a \ne a'$, then
$(a,0) - (a',0) = (a - a',0) \notin H$, so $(a,0) + H \ne (a',0) + H$.
Hence $[G:H]$ is infinite in this example.
\vskip 10pt
\noindent
7.  The smallest possible order for such a group $G$ is
the least common multiple of the integers $1,2,\ldots,11,12$.
Thinking of the prime factorizations, $2$, $3$, $4 = 2^2$, 
$5$, $6 = 2\cdot 3$, $7$, $8 = 2^3$, $9 = 3^2$, $10 = 5\cdot 2$,
$11$, $12 = 2^2 \cdot 3$, we see that the least common multiple
will be 
$$N = 2^3\cdot 3^2 \cdot 5 \cdot 7 \cdot 11 = 27720.$$
\noindent
8.  The order $n$ must satisfy $n < 100$ and $25 | n$, 
$10 | n$.  This uniquely determines $n = 50 = {\rm lcm}(25,10)$.  
\vskip 10pt
\noindent
9.  We are given that $H$ and $K$ are subgroups of $G$ with 
$|H| = |K| = p$, $p$ prime.  Moreover $H \ne K$.  We claim first
that $H\cap K$
is a subgroup of $H$ and of $K$.  This follows since $H \cap K$
is not empty ($e \in H\cap K$).  Then 
$a,b \in H\cap K$ implies $a,b \in H$ and $a,b \in K$.  Hence
$ab^{-1} \in H$ and $ab^{-1} \in K$ since $H$ and $K$ are 
subgroups.  Thus, $ab^{-1} \in H\cap K$, so $H\cap K$ is 
a subgroup of $G$.  Now by Lagrange's theorem, $|H\cap K|$
divides $|H| = p$ and $|K| = p$.  But $p$ is assumed
prime, so $|H\cap K|$ is either $1$ or $p$.  If 
$|H\cap K| = p$, then $H\cap K = H = K$.  Since we know
that $H \ne K$, though, this means that $|H\cap K| = 1$.
Hence $H \cap K = \{e\}$. 
\vskip 10pt
\noindent
10.  By the same fact we used in problem 9 above, 
$H\cap K$ is also a subgroup of $G$, hence of $H$ and
$K$ as well.  Then by Lagrange's theorem, we know that
$|H\cap K|$ divides $|H|$ and $|K|$.
\vskip 10pt
\noindent
11.  We will show first that $G$ must be cyclic,
and then deduce that the order of $G$ must be prime.  
Since $G$ has more than one element, let $a\in G$ 
be any element $a \ne e$.  Then $\langle a\rangle$
is a subgroup of $G$ with more than one element.  
But by hypothesis, $G$ has no non-proper subgroups.
This implies that $\langle a\rangle = G$, so that 
$G$ is a cyclic group.  We claim next that $G$ must
be {\it finite}.  This follows because if $G = \langle a\rangle$
was an infinite cyclic group, then $G$ would contain proper
subgroups like $\langle a^2\rangle$.   
This means that $G$ is a finite cyclic group, so
$G \simeq {\mathbb Z}_n$
for some $n > 1$ (Theorem 7.18).  Now we claim that 
$n$ must be prime.  If not, then $n = km$ for some
integers $k,m > 1$.  But then the order of $a^k$ is
$m < n$.  This implies that $|\langle a^k\rangle | = m$.
But then $1 < m < n$ is the order of a subgroup of $G$,
which contradicts the hypothesis that $G$ has no
non-proper subgroups.
\vskip 10pt
\noindent
13.  The order is $30/\gcd(4,30) = 15$.  This can 
be seen directly by computing the powers of $a^4$
and using $a^{30} = e$:
$$e,a^4,a^8,a^{12},a^{16},a^{20},a^{24},a^{28},a^{32} = a^2,
a^6,a^{10},a^{14},a^{18},a^{22},a^{26},a^{30} = e.$$
This shows that the order of $a^4$ is $15$, so $|\langle a^4\rangle | = 15$
as well.  Then the index of $\langle a^4\rangle$ in $\langle a\rangle$
is $|\langle a\rangle|/|\langle a^4\rangle| = 30/15 = 2$.
\vskip 10pt
\noindent
19.  If $|G| = pq$ with $p,q$ primes, then the possible
orders of subgroups of $G$ are the divisors of $pq$,
namely $1,p,q,pq$.  The proper subgroups of $G$ have 
order $p$ or order $q$.  Since both of these are primes,
all the proper subgroups of $G$ must be cyclic
by Theorem 7.28.
\vskip 10pt
\noindent
24.  We will show this by considering two cases.  
If $|G| = 33$ and $G$ has an element $a$ of order
33 (that is, if $G$ is cyclic), then $G$ also has
elements of order $3$, such as $b = a^{11}$.
(Note that 
$$b = a^{11}, b^2 = (a^{11})^2 = a^{22}, b^3 = (a^{11})^3 = a^{33} = e.{\rm )}$$
Now assume that $G$ is not cyclic (there are no elements
of order 33 in $G$).  By Lagrange's theorem, the possible
orders of elements are $1,3,11$.  Suppose that there
are no elements of order $3$ in $G$.  We will show 
that this leads to a contradiction, as follows.  If 
there are no elements of order 3, then every element
other than $e$ has order 11.  Since 
$11$ is prime, by problem 9 above, if $a,b$ are elements
of order $11$, then either $\langle a\rangle = \langle\, b \rangle$
or else $\langle a\rangle \cap \langle b\rangle = \{e\}$.  Take
any one element of order 11.  We have $11$ distinct elements
of $G$ in $\langle a\rangle$.  Since there are $33$ elements
in $G$ altogether, then there must be some $b \in G$ with
$b \notin \langle a\rangle$.  But as above, this implies
$\langle a \rangle \cap \langle b\rangle = \{e\}$.  This
gives 10 additional elements in $G$ in $\langle b\rangle$, 
but not in $\langle a\rangle$.
However, $|G| = 33$, so there are still elements in $G$
not in $\langle a \rangle \cup \langle b\rangle$.  Let
$c$ be any one such element.  Then $\langle c\rangle \cap \langle a\rangle = \{e\}$ and $\langle c\rangle \cap \langle b\rangle = \{e\}$.
Thus $\langle c\rangle$ contains $10$ additional elements of $G$
other than $e$.  But we still have additional elements $G$ since
$|\langle a\rangle  \cup \langle b\rangle \cup \langle c\rangle | =
31$.  However, now we get a contradiction, since if $d$
is one of the elements of $G$ not in the union of these
three subgroups, then $d$ must also have order $11$.  This 
is impossible since there are not 10 additional elements in $G$.
Hence $G$ must have some elements of order 3. 
\vskip 10pt
\noindent
{\it Section 7.6}
\vskip 10pt
\noindent
5. (a) $G$ is clearly nonempty. Let $A = \begin{pmatrix} a & b\\ 0 & d\\ \end{pmatrix}$
and $B = \begin{pmatrix} a' & b'\\ 0 & d'\\ \end{pmatrix}$ be elements in $G$.
Then 
$$AB^{-1} = \begin{pmatrix} a & b\\ 0 & d\\
\end{pmatrix}
\begin{pmatrix} 1/a' & -b'/(a'd')\\
0 & 1/d'\\
\end{pmatrix} = 
\begin{pmatrix} a/a' & (a'b-ab')/(a'd')\\ 0 & d/d'\\
\end{pmatrix}.$$
Since $ad \ne 0$ and $a'd' \ne 0$, this matrix is also in $G$
since $(a/a')(d/d') = (ad)/(a'd') \ne 0$.  This shows that $G$ is
a subgroup of ${\rm GL}(2,{\mathbb R})$, hence a group itself.  
\vskip 10pt
Next, Consider $N$.  $N$ is closed under products:
$$
\begin{pmatrix}
1 & b \\
0 & 1\\
\end{pmatrix} 
\begin{pmatrix}
1 & b' \\
0 & 1\\
\end{pmatrix} 
= 
\begin{pmatrix}
1 & b + b' \\ 0 & 1\\
\end{pmatrix} \in N,$$
and closed under inverses:
$$\begin{pmatrix}
1 & b\\ 0 & 1 \\
\end{pmatrix}^{-1} = 
\begin{pmatrix} 
1 & -b \\ 
0 & 1\\
\end{pmatrix} \in N.$$
Hence $N$ is a subgroup of $G$.  
\vskip 10pt
\noindent
(b) If $A = \begin{pmatrix} a & b\\ 0 & d\\ \end{pmatrix} \in G$ 
and $B = \begin{pmatrix} 1 & b'\\ 0 & 1\\ \end{pmatrix} \in N$, then
\begin{eqnarray*} 
A^{-1}BA &=& 
\begin{pmatrix} 1/a & -b/(ad)\\ 0 & 1/d\\ \end{pmatrix}
\left(\begin{pmatrix} 1 & b'\\ 0 & 1\\ \end{pmatrix}
\begin{pmatrix} a & b\\ 0 & d\\ \end{pmatrix}\right) \\
&=& \begin{pmatrix} 1/a & -b/(ad)\\ 0 & 1/d\\ \end{pmatrix}
\begin{pmatrix} a & b + b'd \\ 0 & d\\ \end{pmatrix}\\
&=& \begin{pmatrix} 1 & (b'd)/a\\ 0 & 1\\ \end{pmatrix}.
\end{eqnarray*}
By definition, this is an element of $N$.  By equivalent statement
(2) in Theorem 7.34, this implies that $N$ is a normal subgroup 
of $G$. 
\vskip 10pt
\noindent
8.  The group table for the quaternion group $Q$ from problem
14 in Section 7.1 is:
\[
\begin{array}{rrrrrrrrrr}
           & \vline&1 & -1 & i & -i & j & -j & k & -k\\ \hline
          1& \vline &1 & -1 & i & -i & j & -j & k & -k\\
         -1& \vline &-1&  1 & -i& i & -j & j &-k & k  \\
          i& \vline &i & -i & -1& 1 & k & -k & -j & j \\
          -i &\vline & -i & i & 1 & -1 & -k & k & j & -j\\ 
          j &\vline & j & -j & -k & k & -1 & 1 & i & -i\\
          -j &\vline & -j & j & k & -k & 1 & -1 & -i & i\\
          k &\vline & k & -k & j & -j & -i & i & -1 & 1\\
          -k &\vline & -k & k & -j & j & i & -i & 1 & -1\\
\end{array}
\]

          \noindent
          (a)  The cyclic subgroups of $Q$ are:
          \begin{eqnarray*} 
      \langle 1\rangle &=& \{1\}\\
                      \langle -1\rangle &=& \{1,-1\}\\
          \langle i\rangle &=& \{1,i,-1,-i\} = \langle -i\rangle\\
          \langle j\rangle &=& \{1,j,-1,-j\} = \langle -j\rangle\\
          \langle i\rangle &=& \{1,k,-1,-k\} = \langle -k\rangle\\
\end{eqnarray*}

\noindent
(b) The cyclic subgroups $\langle i\rangle$, $\langle j\rangle$, 
$\langle k\rangle$
are all normal by a result we proved in class, since they have order $4 = 8/2$.  The other two cyclic subgroups are normal since $1,-1$ commute
with all elements in $Q$, so $aH = Ha$ if $H = \{1\}$ or 
$H = \{1,-1\}$.

\vskip 10pt
\noindent
16.  First, $K\cap N$ is a subgroup of $G$ by the argument given
in the solution for problem 9 from Section 7.5 above.  Now we
will show that $K\cap N$ is a normal subgroup of $G$ by using
equivalent statement (2) from Theorem 7.34.  Let $g \in K\cap N$
and $a\in G$.  Then $g \in K$ and $g\in N$.  Since $K$ is normal in $G$,
we have $a^{-1}ga \in K$.  Similarly, since $N$ is normal in $G$,
$a^{-1} g a \in N$.  Therefore $a^{-1} g a \in K\cap N$.
It follows that $a^{-1} (K \cap N) a \subset K\cap N$, so
$K\cap N$ is a normal subgroup of $G$.
\vskip 10pt
\noindent
17.  We will prove that $N\cap K$ is normal in $K$
using the equivalent statement (2) of Theorem 7.34.  So
let $g \in N\cap K$ and let $a \in K$.  Then
$a^{-1} g a \in K$ because $a,g \in K$ and $K$ is
a subgroup.  On the other hand, $a^{-1} g a\in N$
since $g\in N$, $a\in K \subseteq G$, and $N$ is a normal subgroup of $G$.
Hence $a^{-1} g a \in N\cap K$ whenever $g\in N\cap K$ and $a \in K$.
Therefore $N\cap K$ is a normal subgroup of $K$.
\vskip 10pt
\noindent
22.  This follows from the fact proved in Linear Algebra
that $f(AB) = \det(AB) = \det(A)\det(B) = f(A)f(B)$.
\vskip 10pt
\noindent
23.  Let $A \in {\rm SL}(2,{\mathbb R})$, so $\det(A) = 1$.
If $B \in {\rm GL}(2,{\mathbb R})$ is an arbitrary invertible
matrix, then by problem 22,
$$\det(B^{-1}AB) = \frac{1}{\det(B)} \det(A) \det(B) = \det(A) = 1.$$
Hence for all $B$,
$$B^{-1} {\rm SL}(2,{\mathbb R}) B \subseteq {\rm SL}(2,{\mathbb R}).$$
Hence ${\rm SL}(2,{\mathbb R})$ is a normal subgroup of
${\rm GL}(2,{\mathbb R})$ by Theorem 7.34 (equivalent statement (2)).
\end{document}
