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\centerline{Mathematics 351 -- Abstract Algebra 1}
\centerline{Solutions for Problem Set 10}
\centerline{December 5, 2007}
\vskip 10pt
\noindent
{\it Section 7.7}
\vskip 10pt
\noindent
4.  First notice that $G = \mathbb{Z}_4\times\mathbb{Z}_4$ 
is abelian, so $N$ and every other subgroup is 
normal.  This implies that $G/N$ has the structure of a group.
We can determine this structure most directly by constructing the group 
table for the quotient group.  First, the elements in $N$ are 
$N = \{(0,0),(3,2),(2,0),(1,2)\}$.  The distinct cosets are
\begin{eqnarray*}
&&(0,0) + N, (0,1) + N = \{(0,1),(3,3),(2,1),(1,3), \\
&&(0,2) + N = \{(1,0),(0,2),(3,0),(2,2)\}\},
(0,3) + N = \{(1,1),(0,3),(3,1),(2,3)\}.
\end{eqnarray*}
Then the group table for the quotient group $G/N$ is constructed using the 
coset addition operation (omitting $+N$'s for simplicity): 
\[
\begin{array}{cccccc}
+ & \vline & (0,0) & (0,1) & (0,2) & (0,3) \\ \hline
(0,0)& \vline & (0,0) & (0,1) & (0,2) & (0,3) \\
(0,1)& \vline & (0,1) & (0,2) & (0,3) & (0,0) \\
(0,2)& \vline & (0,2) & (0,3) & (0,0) & (0,1) \\
(0,3)& \vline & (0,3) & (0,0) & (0,1) & (0,2) \\
\end{array}
\]
For example, $(0,2) + N + (0,3) + N = (0,1) + N$ from the above.
By inspection, $G/N$ is isomorphic to the cyclic group of order 4,
$\mathbb{Z}_4$, since the coset $(0,1) + N$ is a generator for the whole group.
\vskip 10pt
\noindent 
5.  We proceed as in 4 above.  $N = \{(0,0),(1,1),(2,0),(3,1),(4,0),(5,1)\}$
Hence $(1,1)$ is an element of order $6$, and by Lagrange's Theorem $|G/N| = 2$.  There are two cosets 
of $N$ in $G$, $N = (0,0) + N$ and $(0,1) + N = \{(0,1),(1,0),(2,1),(3,0),(4,1),(5,0)\}$.
The quotient group is isomorphic to $\mathbb{Z}_2$ (every group of order 2 is 
isomorphic to $\mathbb{Z}_2$ since $2$ is prime).  
\vskip 10pt
\noindent
6.  This could also be done by constructing the group tables of $U_{32}/N$ and $U_{16}$
and comparing their structures.  We will illustrate a different method in this 
solution.  We have that $U_{32}=\{1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31\}$
is an abelian group of order $16$
under {\it multiplication} mod 32.  The subgroup $N = \{1,17\}$ 
has $8$ distinct multiplicative cosets, one for each $k = 1,3,5,7,9,11,13,15$.
This true because if $k = 2\ell + 1$, then $17k = (16 + 1)(2\ell + 1) = 32\ell + 16 + (2\ell + 1)
\equiv 16 + k \bmod 32$.  For instance, this says $N\cdot 5 = \{5,21\}$, etc.  
We construct
\begin{eqnarray*} 
\varphi : U_{32}/N & \to & U_{16}\\
           N\cdot k &\mapsto& [k]_{16}\\
\end{eqnarray*}
This is well defined since if $N\cdot k = N\cdot \ell$, then by the above $k \equiv \ell\bmod 16$, 
so they define the same class in $U_{16}$.  The mapping $\varphi$ is one-to-one and onto
by construction.  It is a group homomorphism because
$$\varphi((N\cdot k)(N\cdot \ell)) = \varphi(N\cdot(k\ell)) = 
[k\ell]_{16} = [k]_{16} [\ell]_{16} = \varphi(N\cdot k)\varphi(N\cdot\ell).$$
Therefore $U_{32}/N$ and $U_{16}$ are isomorphic groups.
  
\vskip 10pt
\noindent
8.  Let $Nx\in G/N$.  Then by the definition of the coset product
operation and the fact that $x^2 \in N$, $(Nx)^2 = (Nx)(Nx) = Nx^2 = N = Ne$.
Since $Ne$ is the identity element in $G/N$.  This shows that if 
$Nx \ne Ne$, then $Nx$ has order 2 in $G/N$.

\vskip 10pt
\noindent
9.  $N$ is automatically normal in $G$, so $G/N$ is a group.  Let $Na$ and $Nb$
be any two cosets.  Then by the definition of the coset
product and the fact that $G$ is abelian, 
$$(Na)(Nb) = N(ab) = N(ba) = (Nb)(Na).$$
This shows that $G/N$ is an abelian group.
\vskip 10pt
15.  (a) Let $p/q + \mathbb{Z} \in \mathbb{Q}/\mathbb{Z}$.  The operation is addition, so
$$q\cdot(p/q + \mathbb{Z}) = (p/q + p/q + \cdots + p/q) + \mathbb{Z} = p + \mathbb{Z} = \mathbb{Z}
= 0 + \mathbb{Z}.$$
This shows that $p/q + \mathbb{Z}$ has finite order (equal to some divisor of the integer
$q$).  

(b)  Let $p/q$ be a rational number in lowest terms (that is, $(p,q) = 1$).
Then when we form $q(p/q + \mathbb{Z}) = 0 + \mathbb{Z}$, 
no smaller multiple of $p/q + \mathbb{Z}$ equals $0 + \mathbb{Z}$ 
since the smallest positive multiple of $p$ that is also a multiple of $q$ is $pq$.  
This shows that $\mathbb{Q}/\mathbb{Z}$ contains elements of every order $q \ge 1$.  
\vskip 10pt
\noindent
16.  The result in part (a) of Exercise 15 also shows
that every element of the subgroup $\mathbb{Q}/\mathbb{Z} \subset \mathbb{R}/\mathbb{Z}$
has finite order in $\mathbb{R}/\mathbb{Z}$.  Now let $r \in \mathbb{R}/\mathbb{Z}$
and assume that there is some positive integer $q$ such that 
$0 + \mathbb{Z} = q(r + \mathbb{Z}) = qr + \mathbb{Z}$.
This is only true if $qr = p\in \mathbb{Z}$.  But then $r = p/q \in \mathbb{Q}$.
Hence the set of elements of finite order in $\mathbb{R}/\mathbb{Z}$ is 
contained in $\mathbb{Q}/\mathbb{Z}$.  Since both containments hold, the two 
subgroups are equal.  
\vskip 10pt
\noindent
23.  (a)  Let $a,b \in G$, then following the hint, for all $g \in G$, 
$$g^{-1}aba^{-1}b^{-1}g = (g^{-1}ag)(g^{-1}bg)(g^{-1}a^{-1}g)(g^{-1}b^{-1}g) = 
(g^{-1}ag)(g^{-1}bg)(g^{-1}ag)^{-1}(g^{-1}bg)^{-1} \in G'.$$
If we have any element $c$ of $G'$, then $c = c_1c_2\cdots c_n$ is a product of elements 
$c_i = a_ib_ia_i^{-1}b_i^{-1}$ of
the form above for $i = 1,\ldots,n$. Hence 
$g^{-1}cg = (g^{-1}c_1g)(g^{-1}c_2g)\cdots (g^{-1}c_ng)$ is also 
a product of commutators, hence an element of $G'$.
This shows $g^{-1}G'g \subseteq G'$ for all $g\in G$.    Hence $G'$ is normal by part (2) of Theorem 7.34.

(b)  In the quotient group $G/G'$, consider any two cosets $G'a$ and $G'b$.
By the definition of the coset product, we have 
$$(G'a)(G'b)(G'a)^{-1}(G'b)^{-1} = G'(aba^{-1}b^{-1}) = G' = G'e,$$
since $aba^{-1}b^{-1} \in G'$.  This implies $(G'a)(G'b) = (G'b)(G'a)$
for all $a,b$.  Hence
$G/G'$ is an abelian group. 
\vfil\eject
\noindent
{\it Section 7.8}
\vskip 10pt
\noindent
6.  (a) Every subgroup of $\mathbb{Z}_{12}/H$ is $K/H$ for a subgroup 
$K \subset \mathbb{Z}_{12}$ containing $H$ (Theorem 7.44, part (3)).  
These subgroups are $K_0 = H$ itself, $K_1 = \{0,3,6,9\}$, $K_2 = \{0,2,4,6,8,10\}$, 
and $K_3 = \mathbb{Z}_{12}$.  The corresponding subgroups of $\mathbb{Z}_{12}/H$
are $K_0/H = \{0 + H\}$, $K_1/H = \{0+H,3+H\}$, $K_2/H = \{0+H,2+H,4+H\}$, and 
$K_3/H = \mathbb{Z}_{12}/H = \{0+H,1+H,2+H,3+H,4+H,5+H\}$.  

(b)  Similar to part (a).  This time there are only three such subgroups, 
$L_i/K$ where $L_0 = K$, $L_1 = \{0,2,4,6,8,10,12,14,16,18\}$, and $L_2 = \mathbb{Z}_{20}$.
\vskip 10pt
\noindent
10.  (a)  The set $K$ is the set of all elements of order 2 or less.  Since $e\in G$ 
has order 1, $K$ is nonempty.  Moreover, if $a,b \in K$, then using the abelian
hypothesis on $G$,
$(ab)^2 = (ab)(ab) = a^2b^2 = e$.  Hence $ab$ has order at most 2 and is an element in $K$.  Finally,
if $a\in G$, then the order of $a^{-1}$ is the same as the order of $a$, so
$a^{-1}\in K$ as well.  Thus $K$ is a subgroup of $G$.

(b)  The set $H$ is the set of squares of elements in $G$.  This is nonempty
since $G$ is nonempty.  If $a = x^2$ and $b = y^2$ are in $H$, then 
$ab = x^2 y^2 = (xy)(xy) = (xy)^2$ is also in $H$ since $x$ and $y$ commute.  
Finally, if $a = x^2$ is in $H$, then $a^{-1} = (x^2)^{-1} = (x^{-1})^2 \in H$.
Hence $H$ is a subgroup of $G$.

(c)  Let $\varphi : G \to H$ be the mapping $\varphi(x) = x^2$.  This is 
a group homomorphism under the hypothesis that $G$ is abelian:
$\varphi(xy) = (xy)^2 = (xy)(xy) = x^2 y^2 = \varphi(x)\varphi(y)$.  
By definition $\varphi$ is onto $H$.  The kernel of $\varphi$ is precisely
the subgroup $K$ from part (a).  Hence by the First Isomorphism Theorem,
$H \simeq G/K$.  
\vskip 10pt
\noindent
15.  The group $\mathbb{R}^*$ is the multiplicative group of nonzero reals.
The mapping 
\begin{eqnarray*}
\det : {\rm GL}(2,\mathbb{R}) &\to& \mathbb{R}^*\\
                      A &\mapsto& \det(A)
\end{eqnarray*}
is a group homomorphism by Exercise 22 of section 7.6.  The kernel of 
$\det$ is the subgroup ${\rm SL}(2,\mathbb{R})$ (this was one way 
to show that ${\rm SL}(2,\mathbb{R})$ is a normal subgroup of
${\rm GL}(2,\mathbb{R})$ -- see Exercise 23 of section 7.6).  Hence by the First 
Isomorphism Theorem, ${\rm GL}(2,\mathbb{R})/{\rm SL}(2,\mathbb{R}) \simeq \mathbb{R}^*$.
\vskip 10pt
\noindent
16. (a)  By the distributive law in $R$, $f(x+y) = r(x+y) = rx + ry = f(x) + f(y)$.
Hence $f : R \to R$ is a group homomorphism for the additive group structure on $(R,+)$.  

(b) Let $R = \mathbb{Z}$ and $r = 4\in \mathbb{Z}$.  Then $f(1\cdot 2) = 4\cdot(1\cdot 2) = 8$
but $f(1)\cdot f(2) = (4\cdot 1)\cdot (4\cdot 2) = 32$.  Hence $f(1\cdot 2) \ne f(1)\cdot f(2)$.
So $f$ is not a ring homomorphism from $\mathbb{Z}$ to itself.  

(c) Suppose that $R$ is a commutative ring and $r$ is an {\it idempotent} element
(that is $r^2 = r$).  Then 
$f(xy) = r(xy) = r^2(xy) = (rx)(ry) = f(x)f(y)$.
So in this case $f$ is a ring homomorphism.  ({\it Comment:}  Another correct answer would be
to say that $r = 0,1$ always work!  Any ring $R$ contains idempotents $r = 0,1$.  Some
rings have others too.  For instance, $R = \mathbb{Z}\times \mathbb{Z}$ under
the component-wise sum and product operations also 
has idempotents $r = (1,0)$ and $r = (0,1)$. ) 
\vskip 10pt
\noindent
17.  (a)  $G$ is clearly nonempty since $a,b,c\in \mathbb{Q}$ are arbitrary.  Let 
$$A = \begin{pmatrix} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{pmatrix}, 
 B = \begin{pmatrix} 1 & a' & b'\\ 0 & 1 & c'\\ 0 & 0 & 1 \end{pmatrix}.$$
Then 
\begin{equation}
\label{prod}
AB = \begin{pmatrix} 1 & a+a' & b + ac' + b'\\ 0 & 1 & c+c'\\ 0 & 0 & 1 \end{pmatrix}.
\end{equation}
Since this matrix is also in $G$, $G$ is closed under matrix multiplication.
The computation above also shows that $AB = I$, the $3\times 3$ identity matrix,
if $a' = -a, c' = -c$, and $b' = -b - ac' = -b + ac$.  The matrix
$$ B = A^{-1}= \begin{pmatrix} 1 & -a & -b + ac \\ 0 & 1 & -c\\ 0 & 0 & 1 \end{pmatrix}$$
is also in $G$.  Hence $G$ is a subgroup of ${\rm GL}(3,\mathbb{Q})$, hence a group.  

(b)  From the equation for $AB$ above and the reverse product
$$BA = \begin{pmatrix} 1 & a'+a  & b' + a'c + b\\ 0 & 1 & c'+c\\ 0 & 0 & 1 \end{pmatrix},$$
we see that $AB = BA$ if and only if $a'c = ac'$.  Hence $A$ is in the center of $G$
if and only if $a'c = ac'$ {\it for all choices of $a',c' \in \mathbb{Q}$}.  For 
$a' = 1, c'=0$ we get $c = 0$ and then since $c' \ne 0$ is also possible, $a = 0$
as well.  The center of $G$ is 
$$Z(G) = \left\{A \in G : A = \begin{pmatrix} 1 & 0 & b\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}, b \in 
\mathbb{Q}\right\}.$$
To show the last statement in this part, consider the mapping
$$\varphi : Z(G) \to \mathbb{Q}$$
defined by 
$$\varphi\left(\begin{pmatrix} 1 & 0 & b\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}\right) = b.$$
This is clearly one-to-one and onto, and moreover by the product formula (\ref{prod}) above in
the case $a = c = a' = c' = 0$, we see if $A$ and $B$ are as in part A, 
$\varphi(AB) = b + b' = \varphi(A) + \varphi(B)$.    This shows that $Z(G)$ and $(\mathbb{Q},+)$
are isomorphic groups.

(c)  Now define $\psi : G \to \mathbb{Q}\times\mathbb{Q}$ (a group under component-wise sums) by 
$$\psi \left(\begin{pmatrix} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{pmatrix}\right) = (a,c).$$
By the product formula (\ref{prod}), $\psi$ is a group homomorphism:
$\psi(AB) = (a + a',c + c') = (a,c) + (a',c') = \psi(A) + \psi(B)$.  Then
we note that $\ker(\psi)$ is the set of matrices $A$ in $G$ with $a = c =0$.  
This is precisely $Z(G)$ by part (b).  Hence by the First Isomorphism
Theorem, $\mathbb{Q}\times\mathbb{Q} \simeq G/Z(G)$.  
\vskip 10pt
\noindent
{\it Section 7.9}
\vskip 10pt
\noindent
3. 
(a)  This is $(12)(45)(679)$.  (b)  $(13)(254)(789)$.  (c)  $(13)(254)(69)(78)$.
(d) $(1573)(24)$.  (e)  $(123)(456)(78)$.  
\vskip 10pt
\noindent
4. (a) One way (there are many other correct ones too -- see for instance
the different factorization from Corollary 7.48!): 
$(12)(45)(79)(69)$.  (b) $(13)(54)(24)(89)(69)$.
(c) $(13)(54)(24)(69)(78)$.  (d) $(73)(53)(13)(24)$.  (e)  $(23)(13)(56)(46)(78)$.  
\vskip 10pt
\noindent
5.  (a) $(2468) = (68)(48)(28)$ is odd.  (b) even.  (c)  even.   
\vskip 10pt
\noindent
23.  Let $\tau = (ij)$.  Then we claim that for any $\sigma\in S_n$,
$\rho = \sigma \tau\sigma^{-1} = (\sigma(i)\sigma(j))$.  To see this note that
$\rho(\sigma(i)) = \sigma(\tau(i)) = \sigma(j)$ and 
$\rho(\sigma(j)) = \sigma(\tau(j)) = \sigma(i)$.  Moreover if $k$ is any
element of $\{1,2,\ldots,n\}$ besides $\sigma(i), \sigma(j)$ then we 
can write $k =\sigma(\ell)$ for some $\ell \ne i,j$.  Then
$\rho(k) = \sigma(\tau(\ell)) = \sigma(\ell) = k$.  Hence $\rho$ fixes
every other element of $\{1,2,\ldots, n\}$ besides $i,j$.  Therefore
$\rho = \sigma \tau \sigma^{-1} = (\sigma(i) \sigma(j))$ is a transposition.

\vskip 10pt
\noindent
24.  We combine Exercise 23 and the factorization of the cycle $\tau$ as
a product of transpositions as in Corollary 7.48.  We have
$(a_1 a_2 \cdots a_k) = (a_1 a_k) (a_1 a_{k-1}) \cdots (a_1 a_2)$.
Hence 
\begin{eqnarray*}
\sigma (a_1 a_2 \cdots a_k) \sigma^{-1} &=& \sigma (a_1 a_k) (a_1 a_{k-1}) \cdots (a_1 a_2) \sigma^{-1}\\
&=&  (\sigma (a_1 a_k)\sigma^{-1})(\sigma (a_1 a_{k-1})\sigma^{-1})\cdots (\sigma (a_1 a_2) \sigma^{-1})
\\
&=& (\sigma(a_1) \sigma(a_k))(\sigma(a_1)\sigma(a_{k-1})) \cdots )(\sigma(a_1)\sigma(a_2))\\
&=& (\sigma(a_1)\sigma(a_2)\cdots \sigma(a_k))\\
\end{eqnarray*}
\vskip 10pt
\noindent
25.  Consider $H = G\cap A_n$, which is a subgroup of $G$.  $H$ consists of all the even
permutations that are in $H$.  But now since there is some $\sigma \in G$ that is an odd permutation,
we can consider the right coset $H\sigma$.  Every element of the coset $H\sigma$ is an odd permutation
that is in $G$.
Moreover every odd $\rho$ permutation in $G$ has the form $\rho = (\rho\sigma^{-1})\sigma$.  Since $\rho$
and $\sigma$ are odd, $\rho\sigma^{-1}$ is even, so it is an element of $H$.  Therefore
every odd permutation in $G$ belongs to the coset $H\sigma$.  
Hence $H \cap H\sigma = G$, since every permutation in $G$ is either even or odd.  This 
shows that $H$ has index 2 in $G$.  
\vskip 10pt
\noindent
{\it Section 8.1}
\vskip 10pt
\noindent
5.  The Klein 4-group 
$\mathbb{Z}_2\times\mathbb{Z}_2$ has order $4$, but no elements of order 4 (only 
orders 1 and 2).  Therefore $\mathbb{Z}_2 \times \mathbb{Z}_2$ is not cyclic even 
though $\mathbb{Z}_2 = \langle 1\rangle$ is a cyclic group.
\vskip 10pt
\noindent
6.  (a) Every element of $\mathbb{Z}_{12}$ can be written
uniquely as a sum of an element of $H = \{0,3,6,9\}$ and
$K = \{0,4,8\}$. (the cosets of $H$ for the elements in $K$ fill out all
of $\mathbb{Z}_{12}$). Hence by Theorem 8.1,  $\mathbb{Z}_{12} \simeq H \times K$.

(b)  Every element of $\mathbb{Z}_{15}$ can be written
uniquely as a sum of an element of $H = \{0,5,10\}$ and
$K = \{0,3,6,9,12\}$ (the cosets of $H$ for the elements in $K$ fill out all
of $\mathbb{Z}_{15}$).  Hence by Theorem 8.1,  $\mathbb{Z}_{15} \simeq H \times K$.
7.

(c)  Finally, every element of $\mathbb{Z}_{30}$ can be written
uniquely as a sum of an element of $H = \{0,5,10,15,20,25\}$ and
$K = \{0,6,12,18,24\}$.  Hence by Theorem 8.1,  $\mathbb{Z}_{30} \simeq H \times K$.
Similarly, $\mathbb{Z}_{30} \simeq \{0,10,20\} \times \{0,3,6,9,12,15,18,21,24,27\}$.
and 
$$\mathbb{Z}_{30} \simeq \{0,15\} \times \{0,2,4,6,8,10,12,14,16,18,20,22,24,26,28\}.$$
Note that we also have 
$$\mathbb{Z}_{30} = \{0,15\}\times \{0,10,20\}\times \{0,6,12,18,24\}.$$
\vskip 10pt 
\noindent
7.  $\Rightarrow$:  If $G_1\times G_2 \times \cdots \times G_n$ is an abelian group,
then for each $i$, $1 \le i\le n$, and all $a,b \in G_i$, 
$$(e_1,\ldots,e_{i-1},a,e_{i+1},\ldots,e_n)(e_1,\ldots,e_{i-1},b,e_{i+1},\ldots,e_n) = 
(e_1,\ldots,e_{i-1},ab,e_{i+1},\ldots,e_n)$$
and 
$$(e_1,\ldots,e_{i-1},b,e_{i+1},\ldots,e_n)(e_1,\ldots,e_{i-1},a,e_{i+1},\ldots,e_n) = 
(e_1,\ldots,e_{i-1},ba,e_{i+1},\ldots,e_n).$$
 The left sides of these equations are equal, hence the right sides are equal
too, so $ab = ba$.  This shows $G_i$ is abelian.

$\Leftarrow$:  Suppose each $G_i$ is abelian.  Then in the direct product
for all $(a_1,\ldots,a_n)$ and $(b_1,\ldots,b_n)$,
$$(a_1,\ldots,a_n)(b_1,\ldots,b_n) = (a_1b_1,\ldots,a_nb_n) = 
(b_1a_1,\ldots,b_na_n) = (b_1,\ldots,b_n)(a_1,\ldots,a_n)$$
This shows that the direct product is also an abelian group.
\vskip 10pt
\noindent
9.  The answer is {\it No}.  One reason we can see this is that 
$\mathbb{Z}_4\times\mathbb{Z}_2$ contains only elements of order $1,2,4$
(none of order 8):
\begin{eqnarray*}
1 &=& |(0,0)|\\
2 &=& |(2,0)| = |(2,1)| = |(0,1)|\\
4 &=& |(1,0)| = |(1,1)| = |(3,0)| = |(3,1)|.
\end{eqnarray*}
This shows that $\mathbb{Z}_4\times\mathbb{Z}_2$ cannot be isomorphic to the 
cyclic group $\mathbb{Z}_8$.  

\end{document}
