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\centerline{Mathematics 351 -- Abstract Algebra 1}
\centerline{Solutions for Problem Set 8}
\centerline{November 1, 2007}
\vskip 10pt
\noindent
{\it Section 7.1}
\vskip 10pt
\noindent
9. 
\begin{enumerate}
\item[(a)] The order of $U_n$ is the number of $k$, $0 \le k \le n - 1$
that are relatively prime to $n$.  So $U_{10} = \{1,3,7,9\}$ has
order $4$, $U_{12} = \{1,5,7,11\}$ has order $4$ and $U_{24} = \{1,5,7,11,13,17,19,23\}$ 
has order $8$.  {\it Comment:} The numbers of elements here are values of a
famous function called ``Euler's $\phi$-function.''  It can be shown that
the number of $k$ as above relatively prime to $n$ is given by the following
formula if the prime factorization of $n = p_1^{e_1} \cdots p_s^{e_s}$:
$$\phi(n) = p_1^{e_1 - 1} (p_1 - 1) \cdots p_s^{e_s - 1}(p_s - 1).$$
For instance $24 = 2^3 \cdot 3$, so $\phi(24) = 2^2(2 - 1) \cdot 3^0 (3 - 1) = 8$.
\item[(b)]  $U_{20} = \{1,3,7,9,11,13,17,19\}$ (order 8).  The orders of the 
elements are
\begin{eqnarray*}
|1| &=& 1 \text{ (it's the identity element)}\\
|9| = |11| = |19| &=& 2 \text{ since for instance } 9^2 = 81 \equiv 1 \bmod 20\\
|3| = |7| = |13| = |17| &=& 4 \text{ since } 3^2 \equiv 7^2 \equiv 13^2 \equiv 17^2 \equiv 9 \bmod 20
\end{eqnarray*} 
\end{enumerate}
\vskip 10pt
\noindent
10.
\begin{enumerate}  
\item[a.]  In $\mathbb{Z}_4$, $|0| = 1, |1| = 4, |2| = 2, |3| = 4$.
\item[b.]  In $\mathbb{Z}_4 \times \mathbb{Z}_2$, $|(0,0)| = 1$, 
$|(1,0)| = |(1,1)| = |(3,0)| = |(3,1)| = 4$, and $|(0,1)| = |(2,0)| = |(2,1)| = 2$.
\item[c.]  In $S_3$, $|()| = 1$, $|(123)| = |(132)| = 3$, and $|(12)| = |(13)| = |(23)| = 2$.
\item[d.]  (Using Hungerford's notation), $|r_0| = 1$, $|r_1| = |r_3| = 4$, and
$|r_2| = |d|,|t|,|h|,|v| = 2$.
\item[e.] In $\mathbb{Z}$, $0$ has order 1, but every other element has
infinite order.
\end{enumerate}
\vskip 10pt
\noindent
16.  We have $x = a_1 a_2 \cdots a_n$, so $x^2 = (a_1 a_2 \cdots a_n)(a_1 a_2\cdots a_n)$.
$G$ is a group, so each of the $a_i$ has an inverse in $G$.  This inverse must
be one of the factors appearing in the second product.  Since $G$ is abelian, we
can rearrange the factors to write 
$x^2 = (a_1a_1^{-1})(a_2a_2^{-1})\cdots (a_na_n^{-1}) = e$, which is what
we wanted to show.
\vskip 10pt
\noindent
22.  If $(ab)^2 = a^2b^2$, then we have $abab = aabb$ in $G$.  Since $G$
is a group each element has an inverse in $G$, so we can multiply the given
equation on both sides by $a^{-1}$ on the left and $b^{-1}$ on the right:
$a^{-1}(abab)b^{-1} = a^{-1}(aabb)b^{-1}$.  After simplifying, we get
$ba = ab$.  Since this is true for all $a,b \in G$, $G$ is abelian.
\vskip 10pt
\noindent
25.  We assume that every $a \ne e$ in $G$ has order 2, so $a^2 = a$, and 
hence $a = a^{-1}$.  It follows that {\it every element of} $G$
satisfies $a = a^{-1}$, since this is true of $e$ as well.
If $a,b$ are any two elements of $G$, then by the ``reverse order''
rule for the inverse of a product:
$ab = a^{-1}b^{-1} = (ba)^{-1} = ba$.  Hence $G$ is abelian.
\vskip 10pt
\noindent
30.  Consider the mapping $\iota : G \to G$ where $\iota(g) = g^{-1}$.  
Then $\iota(e) = e$, and if $g \in G$, then $\iota(\iota(g)) = (g^{-1})^{-1} = g$.
If $g$ is an element of order $> 2$ then $g$ and $\iota(g)$ are distinct, and 
are interchanged by $\iota$. 
Hence, ``pairing up'' elements with their inverses via $\iota$, since $|G|$
is even, there must be at least one other element $g \in G$ other than $e$ 
with $\iota(g) = g$.  Since $g^{-1} = g$, $g^2 = e$ but $g\ne e$. 
This shows that $G$ must contain an element of order $2$.  
\vskip 10pt
\noindent
{\it Section 7.3}
\vskip 10pt
\noindent
6.  The definition of $H$ is equivalent to saying that $H = \{a \in G : a^k = e\}$
for the fixed integer $k$.  We use Theorem 7.10 to show
that $H$ is a subgroup of $G$ when $G$ is abelian.
Notice that $H$ is nonempty since $e^k = e$ implies
$e \in H$.  Next, if $a,b \in H$, then $a^k = b^k = e$. Consider
$ab \in G$.  Since $G$ is assumed abelian we can reorder the factors 
to rewrite $(ab)^k = (ab)(ab)\cdots (ab) = (aa \cdots a)(bb \cdots b) = a^kb^k$.
Since $a,b \in H$, this shows $(ab)^k = e$, so $ab \in H$.  Finally, let
$a\in H$, so $a^k = e$.  This implies $e = (a^k)^{-1} = (a^{-1})^k$.  Hence
$a^{-1} \in H$ as well.   It follows that $H$ is a subgroup of $G$.
\vskip 10pt
\noindent
7.  We reason as in problem 6 above, but now note that if $a \in T$, then
the power $k > 0$ such that $a^k = e$ will vary with which element $a$ we
are considering.  Notice that $T$ is nonempty since $e^k = e$ implies
$e \in H$.  Next, if $a,b \in T$, then $a^k = b^\ell = e$ for some 
integers $k, \ell$.  Consider
$ab \in G$, and let $m = k\ell$. Since $G$ is assumed abelian we can reorder the factors 
to rewrite $(ab)^m = (ab)(ab)\cdots (ab) = (aa \cdots a)(bb \cdots b) = a^mb^m$.
Then $a^m = (a^k)^\ell = e$ and $b^m = (b^\ell)^k = e$.
This shows $(ab)^m = e$, so $ab \in T$.  Finally, let
$a\in T$, so $a^k = e$ for some $k > 0$.  This implies $e = (a^k)^{-1} = (a^{-1})^k$.  Hence
$a^{-1} \in T$ as well.   It follows that $T$ is a subgroup of $G$.
\vskip 10pt
\noindent 
15.  This is {\it false}.  The group $S_3$ gives a counterexample since
every proper subgroup of $S_3$ is cyclic of order 2 or 3, but $S_3$
is not cyclic itself. There are no elements of order 6 in $S_3$.
\vskip 10pt
\noindent 
16.  We must show that $\mathbb{Q}^{**}$ is not equal to $\langle r\rangle$
for any $r \in \mathbb{Q}^{**}$.  Certainly $r = 1$ does not work since
$\{1^k : k \in \mathbb{Z}\} = \{1\}$ is the trivial subgroup.  So consider
$r \ne 1$ and the subgroup $\langle r \rangle$.
Without loss of generality, we may assume $r > 1$, since if $r < 1$, then 
$r^{-1} = \frac{1}{r} > 1$, and $\langle r\rangle = \langle r^{-1} \rangle$.
If $r > 1$, then the elements of $\langle r\rangle$ are ordered as follows
in $\mathbb{R}$:
\[
0 < \cdots r^{-2} < r^{-1} < 1 < r < r^2 < r^3 < \cdots
\]
Recall from Principles of Analysis (MATH 242) that the set of rational numbers is 
{\it dense} in $\mathbb{R}$.  In other words, in every interval of positive
length in $\mathbb{R}$, there are rational numbers.  In particular, there
must be rational numbers $s$ in the open interval $(1,r)$.  Any such rational
number $s$ is in $\mathbb{Q}^{**}$, since $s > 0$.  But $s \ne r^k$ for
any $k$ because of the way the powers of $r$ are ordered as above.  This
shows that $\langle r\rangle$ is not all of $\mathbb{Q}^{**}$.  Since this
is true for all $r \ne 1$, $\mathbb{Q}^{**}$ is not a cyclic group.
\vskip 10pt
\noindent
20.  First note that if $(a,b) \in \mathbb{Z} \times \mathbb{Z}$, then 
$(a,b) = a(1,0) + b(0,1)$.  So the set $\{(1,0), (0,1)\}$ generates
$\mathbb{Z}\times\mathbb{Z}$.  On the other hand,
\[
(1,0) = (3,1) + (-2,-1) \text{ and } (0,1) = 2(-2,-1) + (4,3).
\]
Hence for any $(a,b) \in \mathbb{Z}\times \mathbb{Z}$, we have
\begin{eqnarray*}
(a,b) & = & a (1,0) + b(0,1)\\
& = & a((3,1) + (-2,-1)) + b( 2(-2,-1) + (4,3))\\
& = & a(3,1) + (a + 2b) (-2,-1) + b(4,3).
\end{eqnarray*}
This shows $\mathbb{Z}\times \mathbb{Z}$ is contained in 
the subgroup generated by $\{(3,1),(-2,-1),(4,3)\}$.
The opposite inclusion is true by definition.  Hence
$\{(3,1),(-2,-1),(4,3)\}$ generates $\mathbb{Z}\times \mathbb{Z}$.
\vskip 10pt
\noindent
21.  (a)  Check that the element $(1,1) \in \mathbb{Z}_2 \times \mathbb{Z}_3$ has
order $6$:  
\[(0,0), (1,1), 2\cdot (1,1) = (0,2), 3\cdot (1,1) = (1,0), 4\cdot (1,1) = (0,1), 
5\cdot (1,1) = (1,2).
\]
Since we have all 6 elements of $\mathbb{Z}_2 \times \mathbb{Z}_3$, this 
group is the same as the cyclic subgroup $\langle (1,1)\rangle$, hence
cyclic.

\noindent
(b)  The cyclic subgroups are
\begin{eqnarray*}
\langle (0,0)\rangle &=& \{(0,0)\}\\
\langle (1,0)\rangle &=& \{(0,0),(1,0)\}\\
\langle (0,1)\rangle &=& \{(0,0),(0,1),(0,2),(0,3)\} = \langle (0,3)\rangle\\
\langle (1,1)\rangle &=& \{(0,0),(1,1),(0,2),(1,3)\} = \langle (1,3)\rangle\\
\langle (0,2)\rangle &=& \{(0,0),(0,2)\}\\
\langle (1,2)\rangle &=& \{(0,0),(1,2)\}.
\end{eqnarray*}
Since none of these is of order $8$, $\mathbb{Z}_2 \times \mathbb{Z}_4$ is not cyclic.
However, $\mathbb{Z}_2 \times \mathbb{Z}_4$ is generated by $(1,0)$ and $(0,1)$.  
\vskip 10pt 
\noindent
30. (a)  We use Theorem 7.10 to show that $HK$ is a subgroup when $G$ is abelian.
First, $HK \ne \emptyset$ since $e\in H$ and $e\in K$, so $e\cdot e = e \in HK$.
If $a,b \in HK$, then $a = h_1k_1$ and $b = h_2k_2$, where $h_1,h_2 \in H$ 
and $k_1,k_2 \in K$.  Since $G$ is abelian,
\[
ab = (h_1k_1)(h_2k_2) = h_1(k_1h_2)k_2 = h_1(h_2k_1)k_2 = (h_1h_2) (k_1k_2).
\]
Since $H$ and $K$ are subgroups, $h_1h_2 \in H$ and $k_1k_2 \in K$.  Therefore
$ab \in HK$ and $HK$ is closed under products.  
Finally if $a = hk \in HK$, then since $G$ is abelian again, 
\[
a^{-1} = (hk)^{-1} = k^{-1} h^{-1} = h^{-1}k^{-1}.
\]
Then $h^{-1} \in H$ and $k^{-1} \in K$ since $H,K$ are subgroups of $G$.  
Therefore $HK$ is also closed under inverses.  
\vskip 10pt
\noindent
(b)  Let $G = S_3$, and let $H = \{(),(12)\}$, $K = \{(),(13)\}$.  
Then 
\[
HK = \{()(), ()(13), (12)(), (12)(13)\} = \{(), (13), (12), (132)\}.
\]
This is not a subgroup of $S_3$ since it is not closed under products
or inverses.  ({\it Comment:}  Note, it even violates the necessary 
condition from the statement of Lagrange's Theorem!)
\vskip 10pt
\noindent
{\it Section 7.4}
\vskip 10pt
\noindent
3.  (a)  The group of invertible $2\times 2$ matrices with entries in
$\mathbb{Z}_2$ is $GL(2,\mathbb{Z}_2) =$
\[
 \left\{
I = \begin{pmatrix}
1&0\\
0&1\\
\end{pmatrix}, 
R = \begin{pmatrix}
0&1\\
1&1\\
\end{pmatrix},
R^2 = \begin{pmatrix}
1&1\\
1&0\\
\end{pmatrix},
S = \begin{pmatrix}
1&1\\
0&1\\
\end{pmatrix},
T = \begin{pmatrix}
1&0\\
1&1\\
\end{pmatrix},
U = \begin{pmatrix}
0&1\\
1&0\\
\end{pmatrix}\right\}.
\] 
(These are all the $2\times 2$ matrices of determinant $1$ with $\mathbb{Z}_2$-entries, 
a group under matrix multiplication).

(b)  Under the mapping $I \mapsto ()$, $R \mapsto (123)$, $R^2 \mapsto (132)$,
$S \mapsto (13)$, $T\mapsto (12)$, $U \mapsto (23)$, we see that the group tables
will ``match up'':
\[
\begin{array}{cccccccc}
\cdot & \vline & I & R & R^2 & S & T & U \\ \hline
I    & \vline & I & R & R^2 & S & T & U \\
R    & \vline & R & R^2 & I & U & S & T \\
R^2  & \vline & R^2 & I & R & T & U & S \\
S    & \vline & S & T & U & I & R & R^2\\
T    & \vline & T & U & S & R^2 & I & R \\
U    & \vline & U & S & T & R & R^2 & I\\
\end{array}
\]
and
\[
\begin{array}{cccccccc}
\circ & \vline & () & (123) & (132) & (13) & (12) & (23) \\ \hline
()    & \vline & () & (123) & (132) & (13) & (12) & (23) \\
(123) & \vline & (123) & (132) & () & (23) & (13) & (12) \\
(132) & \vline & (132) & () & (123) & (12) & (23) & (13) \\
(13)  & \vline & (13) & (12) & (23) & ()   & (123)& (132)\\
(12)  & \vline & (12) & (23) & (13) & (132) & ()   & (123) \\
(23)  & \vline & (23) & (13) & (12) & (123) & (132) & ()\\
\end{array}
\]
\vskip 10pt
\noindent
5.  $U_5$ is the group $\{1,2,3,4\}$ under multiplication mod 5, while
$U_{10}$ is the group $\{1,3,7,9\}$ under multiplication mod 10.
Note that $U_5$ is cyclic with generator $2$, since $2^2 = 4$, and 
$2^3 = 8 \equiv 3 \bmod 5$.  Similarly $U_{10}$ is cyclic since
$3^2 = 9$ and $3^3 = 27 \equiv 7 \bmod 10$.  The mapping 
$\varphi : U_5 \to U_{10}$ defined by 
\begin{eqnarray*}
\varphi(1) &=& 1\\
\varphi(2) &=& 3\\
\varphi(4) &=& 9\\
\varphi(3) &=& 7
\end{eqnarray*}
is an isomorphism since it is injective and surjective, and 
$\varphi(2^k) = 3^k$ for all $k$,
so 
\[
\varphi(2^k\cdot 2^\ell) = \varphi(2^{k+\ell}) = 3^{k+\ell} = 3^k \cdot 3^\ell = 
\varphi(2^k)\cdot \varphi(2^\ell).
\]
\vskip 10pt
\noindent
10.  Let $f : G \to H$ be a surjective homomorphism and assume $G$ is
abelian.  Then for all $a,b \in H$, $a = f(x)$ and $b = f(y)$ for 
some $x,y \in G$ and therefore
\[
ab = f(x)f(y) = f(xy) = f(yx) = f(y)f(x) = ba.
\]
This establishes the desired fact that $H$ is abelian, since
the equation holds for all $a,b \in H$. 
\vskip 10pt
\noindent
11.  We will prove this for $n \ge 1$ by induction on $n$ first.
For the base case, we have $f(a^1) = f(a) = (f(a))^1$, so there
is nothing to prove.  Now assume that the statement is true
for $n = k$.  Since $f$ is a group homomorphism, using the induction
hypothesis at the second to last step:
\[
f(a^{k+1}) = f(a^k \cdot a) = f(a^k) \cdot f(a) = (f(a))^k f(a) = (f(a))^{k+1}.
\]
This establishes the statement for $n \ge 1$.  The formula is clearly true
for $n = 0$ since $f(a^0) = f(e) = e = (f(a))^0$.  Finally, we show that
the formula is true for $n = -m$ where $m\ge 1$ by another induction proof.  
For the base case here we need to see
\[
f(a^{-1}) = (f(a))^{-1}.
\]
This is true by part 2 of Theorem 7.19.  Now assume the formula has been
established for $m = k$.  Then
\[
f(a^{-(k+1)}) = f(a^{-k} \cdot a^{-1}) = f(a^{-k}) \cdot f(a^{-1}) = 
(f(a))^{-k} (f(a))^{-1} = (f(a))^{-(k+1)}.
\]
This establishes the statement for $n \le -1$ as well.
\vskip 10pt
\noindent
13. (a)  We will ignore the hint and prove the parts separately.  
First we show that $a^{-1}Ha$ is a subgroup of $G$.  First, $a^{-1}Ha$
is not empty since $a^{-1}ea = e \in a^{-1}Ha$.  Next, let $x = a^{-1}ha$ 
and $y = a^{-1}ka$ be elements of $a^{-1}Ha$.  Then 
\[
xy = (a^{-1}ha)(a^{-1}ka) = a^{-1}hka.
\]
Since $H$ is a subgroup, $hk \in H$.  So $xy \in a^{-1} H a$.
This shows $a^{-1}Ha$ is closed under products.  Last, if $x = a^{-1}ha \in a^{-1}Ha$,
then $x^{-1} = (a^{-1}ha)^{-1} = a^{-1}h^{-1}(a^{-1})^{-1} = a^{-1}h^{-1}a \in a^{-1}Ha$,
since $h^{-1} \in H$ (a subgroup).  
Hence $a^{-1}Ha$ is also a subgroup of $G$.  

Now, we show that $a^{-1}Ha$ is isomorphic to $H$ as a group.  Consider
the mapping $\varphi : G \to G$ defined by $\varphi(x) = a^{-1}xa$.  (This 
is the ``inner automorphism'' mentioned in the problem.)  We check that 
\[
\varphi(xy) = a^{-1}xya = a^{-1}x(aa^{-1})ya = (a^{-1}xa)(a^{-1}ya) = \varphi(x) \varphi(y).
\]
This shows that $\varphi$ is a group homomorphism.  It is injective and surjective 
because $\varphi$ has an inverse function $\psi(x) = axa^{-1}$:  For all $x\in G$,
\[
\varphi(\psi(x)) = a^{-1}(axa^{-1})a = x \text{ and }
\psi(\varphi(x)) = a(a^{-1}xa)a^{-1} = x.
\]
Then $\varphi$ restricts to a mapping with domain $H$ and range $a^{-1}Ha$, 
and similarly $\psi$ restricts to a mapping with domain $a^{-1}Ha$ and range
$H$.  The restricted mappings are still inverses.   
This shows $H$ and $a^{-1}Ha$ are isomorphic via $\varphi$.  
\vskip 10pt
\noindent
(b)  If $H$ is finite then $a^{-1}Ha = \varphi(H)$ where $\varphi$ is as
in part a.  Since $\varphi$ is injective and surjective, $|H| = |a^{-1}Ha|$.
\vskip 10pt
\noindent
18.   Let $G = \langle a\rangle$ and let $f : G \to H$ be a surjective
group homomorphism.  Then every element $b \in H$ is $b = f(c)$ for
some $c \in H$.  But by hypothesis, $c = a^n$ for some integer $n$.
Hence using problem 11 above,
$b = f(c) = f(a^n) = (f(a))^n$.  This shows that every element of $H$
is a power of $f(a)$, so $H \subset \langle f(a)\rangle$.
The other inclusion is true by definition, hence $H = \langle f(a) \rangle$ is cyclic, generated
by $f(a)$.   
\end{document} 
