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\centerline{Mathematics 351 -- Abstract Algebra 1}
\centerline{Solutions for Final Examination}
\centerline{December 12, 2007}
\vskip 10pt
\noindent
I.  {\it Terminology}
\medskip
\begin{itemize}
\item[A)] (5)  What does it mean to say that $R$ is
a {\it ring with identity}?
\sol
$R$ is a ring with identity if there is a {\it multiplicative identity}
$1\in R$ -- an element $1$ such that $a\cdot 1 = a = 1\cdot a$ for all 
$a\in R$.
\medskip
\item[B)] (5) What does it mean to say that $a \in F$
is a {\it root} of $f(x) \in F[x]$?
\sol
The element $a \in F$ is a root of $f(x) \in F[x]$ if 
$f(a) = 0_F$.
\medskip
\item[C)] (5) What does it mean to say that $f(x) \in F[x]$
is {\it irreducible}?
\sol
The polynomial $f(x)$ is irreducible in $F[x]$ if in 
every factorization $f(x) = g(x)h(x)$ in $F[x]$, one
of $g(x), h(x)$ is a unit in $F[x]$ (a nonzero constant),
and the other is an associate of $f(x)$.
\medskip 
\item[D)] (5) What does it mean to say that a subset $I$ of 
a ring $R$ is an {\it ideal} in $R$?
\sol 
A nonempty subset $I$ of a ring $R$ is an ideal if it is
a subring of $R$ which has the property that $a\cdot r, r\cdot a \in I$
whenever $a \in I$ and $r \in R$ ($I$ ``absorbs all products with elements
of $I$).
\medskip
\item[E)] (5) What does it mean to say that a permutation
$\sigma \in S_n$ is {\it even}?
\sol
The permutation $\sigma$ is even if it is the product of 
an even number of transpositions (or equivalently if the
determinant of the associated permutation matrix is $+1$.)
\end{itemize}

\vskip 10pt
\noindent
II.  
\begin{itemize}
\item[A)] (15) State and prove the Eisenstein Irreducibility Criterion 
in ${\mathbb Q}[x]$.
\sol
The statement is:  Let $f(x) = a_n x^n + \cdots + a_1 x + a_0$ be a nonconstant
polynomial in ${\mathbb Z}[x]$.  Assume there is some prime $p \in {\mathbb Z}$
such that $p\nd a_n$, $p|a_{n-1}, \ldots, p|a_0$, but
$p^2\nd a_0$.  Then $f(x)$ is irreducible in ${\mathbb Q}[x]$.
\medskip
\item{} Proof: Arguing by contradiction, suppose $f(x) = g(x)h(x)$
where $g(x)$ and $h(x)$ both have positive degree.  By Theorem 4.22, 
we can assume that $g(x),h(x) \in {\mathbb Z}[x]$ as well so write:
\begin{eqnarray*}
f(x) &=& a_n x^n + \cdots + a_1 x + a_0 \\
&=& (b_k x^k + \cdots + b_1 x + b_0)(c_\ell x^\ell + \cdots + c_1 x + c_0)
\end{eqnarray*}
for some $k,\ell\ge 1$.  By hypothesis $p|a_0$ so $p|(b_0c_0)$.  However
$p^2\nd a_0$, so by renaming the polynomials $g(x),h(x)$ if necessary
we may assume $p|b_0$ but $p\nd c_0$.  Now look at the coefficient of $x$:
$a_1 = b_0c_1 + b_1c_0$.  We know $p|a_1$ and $p|b_0$, so $p|(b_1c_0)$ as
well.  Since $p\nd c_0$, this implies that $p | b_1$ as well.  We continue
in the same way to show that $p|b_0, b_1, \ldots, b_k$.  (Note that the 
process we started here will continue until the coefficient 
$a_k = b_0c_k + b_1c_{k-1} + \cdots + b_k c_0$ since $k < n$.)  But now
notice that $p|b_k$ implies $p|(b_kc_\ell)$, so $p|a_n$.   This is a contradiction
to the hypothesis that $p$ does not divide the leading coefficient of $f(x)$, $a_n$.
Hence $f(x)$ must be irreducible in ${\mathbb Q}[x]$.  
\medskip
\item[B)] (10) Use the Eisenstein criterion to show that
there are irreducible polynomials of every degree $\ge 1$
in ${\mathbb Q}[x]$. 
\sol
We just need to ``cook up'' polynomials of each degree satisfying
the hypotheses in the Eisenstein Criterion.  One such way, using the
prime $p = 2$:    For all degrees $n \ge 1$, take
$$f(x) = x^n + 2.$$
These are all irreducible since all the conditions for Eisenstein are met. 
\end{itemize}

\vskip 10pt
\noindent
III.  
\begin{itemize}
\item[A)] (10) Show that $a\in F$ is a root of 
$f(x) \in F[x]$ if and only if $(x - a)|f(x)$ in
$F[x]$.  
\sol
If $(x-a)|f(x)$, then $f(x) = q(x)(x - a)$ for some $q(x) \in F[x]$.
Hence $f(a) = q(a)(a - a) = 0$.  This shows $a$ is a root of $f(x)$.  
Conversely, assume that $x = a$ is a root of $f(x)$ and use the 
division algorithm to divide $x - a$ into $f(x)$ and 
write $f(x) = q(x)(x - a) + r$.  Since $deg(x - a) = 1$, $r\in F$ 
is a constant.  However, if $a$ is a root of $f(x)$, then $0 = f(a) = q(a)(a - a) + r = 0 + r$.
Hence $r = 0$, so $x - a$ divides $f(x)$. 

\item[B)]  (10) Express $f(x) = x^4 - 16$ as a 
product of irreducible polynomials in ${\mathbb Q}[x]$, in
${\mathbb R}[x]$, and finally in ${\mathbb C}[x]$.  
\sol
\begin{itemize}
\item[$\bullet$] In ${\mathbb Q}[x]$:  $x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$.
\item[$\bullet$] In ${\mathbb R}[x]$:  $x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$ (same)
\item[$\bullet$] In ${\mathbb C}[x]$:  $x^4 - 16 = (x - 2)(x + 2)(x - 2i)(x + 2i)$.
\end{itemize}
We know that $x^2 + 4$ is irreducible over ${\mathbb Q}[x]$ and ${\mathbb R}[x]$ since
that polynomial has degree 2 but has no roots in ${\mathbb Q}$ or ${\mathbb R}$.

\item[C)]  (5) In what sense are your factorizations
from part B {\it unique}?
\sol
They are unique in the sense that any other irreducible factorization would
consist of the same number of factors and each polynomial would be a nonzero
constant times one of the polynomials in the given factorization.  (The factors
would match up after rearrangement, and up to multiplication by units.)
\end{itemize}

\vskip 10pt
\noindent
IV.  Let $p(x) = x^2 + x + 1\in {\mathbb Z}_2[x]$.  
\begin{itemize}
\item[A)] (15)  Determine the addition and multiplication
tables for the quotient ring $R = {\mathbb Z}_2[x]/(p(x))$.
\sol
The addition table is (writing, for instance, $1$ for the coset $1 + (p(x))$):
\[
\begin{array}{cccccc}
+ & \vline & 0 & 1 & x & x+1 \\ \hline
0 & \vline & 0 & 1 & x & x+1 \\
1& \vline & 1 & 0 & x+1 & x \\
x& \vline & x & x+1 & 0 & 1 \\
x+1& \vline & x+1 & x & 1 & 0 \\
\end{array}
\]
The multiplication table is
\[
\begin{array}{cccccc}
\cdot & \vline & 0 & 1 & x & x+1 \\ \hline
0 & \vline & 0 & 0 & 0 & 0 \\
1& \vline & 0 & 1 & x & x+1 \\
x& \vline & 0 & x & x+1 & 1 \\
x+1& \vline & 0 & x+1 & 1 & x \\
\end{array}
\]

\item[B)]  (5) Is $R$ a field?  Why or why not?
\sol
The answer is {\it yes}.  This can be seen directly by examining
the multiplication table.  Note that every nonzero element has a multiplicative
inverse.  It also follows from the result of Theorem 5.10 since the polynomial
$p(x) = x^2 + x + 1$ is irreducible in ${\mathbb Z}_2[x]$.  

\item[C)]  (5) Show that $x + 1\in{\mathbb Z}_2[x]/(p(x))$ 
is a root of the polynomial $y^2 + y + 1 \in {\mathbb Z}_2[y]$. 
\sol 
We compute directly with the addition and multiplication tables: If $y = x+1$,
then $y^2 = x$, so $y^2 + y + 1 = x + (x + 1) + 1 = 0$ since the coefficient
arithmetic is done in ${\mathbb Z}_2$.
\end{itemize}

\vskip 10pt
\noindent
V.  
\begin{itemize}
\item[A)] (10) State the First Isomorphism Theorem for
rings.
\sol The First Isomorphism Theorem states that if $\varphi : R \to S$
is a ring homomorphism, then ${\rm im}(\varphi)$ is isomorphic to 
$R/\ker(\varphi)$ as a ring.  
\item[B)]  (10)  Let $\varphi$ be the mapping
\begin{eqnarray*}
\varphi : {\mathbb Z}[x] &\to& {\mathbb Z}_3\\
                  f(x) &\mapsto& f(0) \bmod 3.
\end{eqnarray*}
(That is, $\varphi$ takes each polynomial
$f(x) = a_nx^n + \cdots + a_1x + a_0$ to the 
class of $a_0$ mod 3).  Show that $\varphi$ is a ring homomorphism
and show that its kernel is the ideal $(x,3)\subset {\mathbb Z}[x]$.
\sol
$\varphi$ is a ring homomorphism since 
\begin{eqnarray*}
\varphi(f(x) + g(x)) &=& [f(0) + g(0)] (\text{class in } {\mathbb Z}_{3}) \\
&=& [f(0)] + [g(0)],
\end{eqnarray*}
and 
\begin{eqnarray*}
\varphi(f(x) \cdot g(x)) &=& [f(0)\cdot g(0)]\\
&=& [f(0)] + [g(0)] \bmod 3.
\end{eqnarray*}
The kernel of $\varphi$ is the set of all $f(x)$ such that $\varphi(f(x)) = 0$
in ${\mathbb Z}_3$.  Any polynomial in $(x,3)$ is contained in the kernel 
since if $f(x) = xg(x) + 3h(x)$ for some $g(x),h(x)$, then 
$[f(0)] = [0\cdot g(0) + 3h(0)] = [3h(0)] = [0]$.  Conversely, if 
$[f(0)] = 0$, then $f(x) \in {\mathbb Z}[x]$ is a polynomial whose
constant term is a multiple of $3$.  This means that we can write
$f(x) = xg(x) + 3h(x)$ for polynomials $g(x),h(x)$.  For instance,
if $f(x) = a_n x^n + \cdots + a_1 x + a_0$, then $[f(0)] = [a_0] = 0$
so $a_0 = 3m$ for some $m \in {\mathbb Z}$.  Hence
$f(x) = x(a_n x^{n-1} + \cdots + a_1) + 3m \in (x,3)$.

\item[C)] (5) Is the ideal $(x,3)$ in ${\mathbb Z}[x]$ maximal?
Why or why not?
\sol
The answer is {\it yes}.  The homomorphism from part B
is onto.  Hence 
$${\mathbb Z}_3 \simeq {\mathbb Z}[x]/\ker(\varphi) = {\mathbb Z}[x]/(x,3).$$
Since $3$ is a prime number ${\mathbb Z}_3$ is a field, hence
$(x,3)$ is maximal.
\end{itemize}

\vskip 10pt
\noindent
VI.  Let $R$ be a ring with identity.  
\begin{itemize}
\item[A)] (7.5) Show that the set of units in $R$ is a group
under the multiplication operation from $R$.
\sol
We let $U$ be the set of units in $R$, that is 
$$U = \{r \in R : rs = 1 = sr \text{ for some } s \in R\}.$$
We must show that the gorup axioms hold in $U$.  First $U$ 
is closed under products since $(rs)^{-1} = s^{-1}r^{-1} \in U$ 
if $r \in U$ and $s \in U$.  Associativity of multiplication 
follows from the ring axioms.  $U$ has an identity element
since the ring multiplicative identity $1$ serves that purpose.
Finally, every element of $U$ has a multiplicative inverse
by definition.  Hence $U$ is a group under multiplication.

\item[B)] (10) Determine the multiplication table for the
 group of units in the ring
${\mathbb Z}_8 \times {\mathbb Z}_4 = \{(r,s) : r\in {\mathbb Z}_8
{\rm\ and\ } s \in {\mathbb Z}_4\}$ (componentwise
sum and product).
\sol    
The units in ${\mathbb Z}_8$ are $\{1,3,5,7\}$ and the 
units in ${\mathbb Z}_4$ are $\{1,3\}$.  In order for 
$(a,b) \in {\mathbb Z}_8 \times {\mathbb Z}_4$ 
to be a unit, there must be some $(c,d)$ such that
$(a,b)(c,d) = (ac,bd) = (1,1)$.  Hence $a$ and $b$
must both be units themselves in their respective
rings.  The operation table is 
\[
\begin{array}{cccccccccc}
\cdot & \vline & (1,1) & (3,1) & (5,1) & (7,1) &  (1,3) & (3,3) & (5,3) & (7,3) \\ \hline
(1,1) & \vline & (1,1) & (3,1) & (5,1) & (7,1) &  (1,3) & (3,3) & (5,3) & (7,3) \\
(3,1) & \vline & (3,1) & (1,1) & (7,1) & (5,1) &  (3,3) & (1,3) & (7,3) & (5,3) \\
(5,1) & \vline & (5,1) & (7,1) & (1,1) & (3,1) &  (5,3) & (7,3) & (1,3) & (3,3) \\
(7,1) & \vline & (7,1) & (5,1) & (3,1) & (1,1) &  (7,3) & (5,3) & (3,3) & (2,3) \\
(1,3) & \vline & (1,3) & (3,3) & (5,3) & (7,3) &  (1,1) & (3,1) & (5,1) & (7,1) \\
(3,3) & \vline & (3,3) & (1,3) & (7,3) & (5,3) &  (3,1) & (1,1) & (7,1) & (5,1) \\
(5,3) & \vline & (5,3) & (7,3) & (1,3) & (3,3) &  (5,1) & (7,1) & (1,1) & (3,1) \\
(7,3) & \vline & (7,3) & (5,3) & (3,3) & (1,3) &  (7,1) & (5,1) & (3,1) & (1,1) \\
\end{array}
\]

\item[C)] (7.5) Is your group from part B {\it cyclic}? Why
or why not?
\sol
By inspection, we see that every element of this group has order $2$.  Hence
it cannot be cyclic(!)
\end{itemize}

\vskip 10pt
\noindent
VII.  {\it Note: This problem had a small but
unfortunate error in its statement. It should have said:}
\vskip 10pt
\noindent
Let $G$ be a group, and let $H$ be a subgroup of $G$.
Define 
$$N_H = \{x\in G: xHx^{-1} = H\},$$
called the {\it normalizer of $H$} in $G$.
\vskip 10pt
The condition $xHx^{-1} \subseteq H$ is enough to imply $xHx^{-1} = H$ when
$H$ is a finite group and in some other cases, but not in complete generality.
The condition $xHx^{-1} = H$ is needed for one part below.

\begin{itemize}
\item[A)] (10) Let $G = {\rm GL}(2,{\mathbb R})$ be the group 
of $2\times 2$ invertible matrices under matrix multiplication
and let
$$H =  \left\{\begin{pmatrix} 1 & r\\ 0 & 1\\ \end{pmatrix} : r \in {\mathbb R}\right\}.$$
Show that $H$ is a subgroup of $G$ and determine $N_H$ in this case.
\sol
First we show $H$ is a subgroup.  $H$ is clearly nonempty since
$H$ contains one matrix for each real $r$.  $H$ is closed under matrix
products since
$$\begin{pmatrix} 1 & r\\ 0 & 1\\ \end{pmatrix}\begin{pmatrix} 1 & r'\\ 0 & 1\\ \end{pmatrix}
=\begin{pmatrix} 1 & r+r\\ 0 & 1\\ \end{pmatrix} \in H.$$
Also, $H$ is closed under inverses since 
$$\begin{pmatrix} 1 & r\\ 0 & 1\\ \end{pmatrix}^{-1} = \begin{pmatrix} 1 & -r\\ 0 & 1\\ \end{pmatrix}
\in H.$$
This shows that $H$ is a subgroup of ${\rm GL}(2,{\mathbb R})$.
\vskip 10pt
\item[] Next, we show that $N_H$ is the set of upper-triangular invertible matrices,
$$B =  \left\{\begin{pmatrix} a & b\\ 0 & d\\ \end{pmatrix} : a,b,d \in {\mathbb R}, ad \ne 0\right\}.$$
First, we see that
$$\begin{pmatrix} a & b\\ 0 & d\\ \end{pmatrix}^{-1} = \begin{pmatrix} \frac{1}{a} & \frac{-b}{ad}\\ 0 & \frac{1}{d}\\ \end{pmatrix}.$$
So
$$
\begin{pmatrix} a & b\\ 0 & d\\ \end{pmatrix}
\begin{pmatrix} 1 & r\\ 0 & 1\\ \end{pmatrix}
\begin{pmatrix} \frac{1}{a} & \frac{-b}{ad}\\ 0 & \frac{1}{d}\\ \end{pmatrix}
=\begin{pmatrix} 1 & \frac{ar}{d}\\ 0 & 1\\ \end{pmatrix}$$
which is in $H$.  Moreover every matrix in $H$ can be obtained.  
Hence $B\subseteq N_H$.  Conversely, if
$\begin{pmatrix} a & b\\ c & d\\ \end{pmatrix}$ is in $N_H$, then 
$$\begin{pmatrix} a & b\\ c & d\\ \end{pmatrix}
\begin{pmatrix} 1 & r\\ 0 & 1\\ \end{pmatrix} \cdot \frac{1}{ad-bc}
\begin{pmatrix}  d & -b\\ -c & a\\ \end{pmatrix}
= \begin{pmatrix} * & *\\ \frac{-c^2 r}{ad-bc} & *\end{pmatrix} $$
is in $H$ for all $r$. Looking at the lower left entry, we see that $c$ must be $0$.  
Hence the matrix $\begin{pmatrix} a & b\\ c & d\\ \end{pmatrix}$ is in $B$,
so $N_H\subseteq B$.   

\item[B)] (10) Show that for any subgroup of a general 
group, $H \subseteq G$, 
$N_H$ is a subgroup of $G$ containing $H$, and that $H$ is a normal
subgroup of $N_H$.
\sol
First, $N_H$ contains
$H$ (and hence is nonempty).  Indeed, if $y \in H$, since $H$ is a subgroup
(closed under products and inverses), $yxy^{-1} \in H$ for all $x \in H$.  
Hence $H \subseteq N_H$.  Next, suppose that $y,y'$ are both in $N_H$.
Then $H = yHy^{-1} = H$, so $H = y'(yHy^{-1})(y')^{-1} = (y'y)H(y'y)^{-1}$.
Hence $y'y \in N_H$.  Finally, if $y \in N_H$, then 
$yHy^{-1} = H$.  It follows that $y^{-1}Hy = H$ 
as well.  This shows that $y^{-1} \in N_H$, so $N_H$ is subgroup of $G$.  

\item[] $H$ is a normal subgroup of $N_H$ since for all $y \in N_H$, $yHy^{-1} = H$.
Hence one of the equivalent conditions of Theorem 7.34 holds.

\item[C)] (5) Show that the quotient group $N_H/H$
for $G, H$ as in part A is isomorphic to the group 
${\mathbb R}^*\times {\mathbb R}^*$, where
${\mathbb R}^*$ denotes the multiplicative group 
of nonzero real numbers.  
\sol
Define 
\begin{eqnarray*}
\varphi: N_H &\to& {\mathbb R}^*\times {\mathbb R}^*\\
         \begin{pmatrix} a & b \\ 0 & d \\ \end{pmatrix} &\mapsto & (a,d).
\end{eqnarray*}
The mapping $\varphi$ is clearly onto.  It is also a group homomorphism since
\begin{eqnarray*}
\varphi\left(
 \begin{pmatrix} a & b \\ 0 & d \\ \end{pmatrix} 
 \begin{pmatrix} a' & b' \\ 0 & d' \\ \end{pmatrix}\right)
&=& \varphi \begin{pmatrix} aa' & ab'+bd' \\ 0 & dd' \\ \end{pmatrix}\\
&=& (aa',dd')\\
&=& (a,d)(a',d')\\
&=& \varphi\left(
 \begin{pmatrix} a & b \\ 0 & d \\ \end{pmatrix} \right) 
\varphi\left(
 \begin{pmatrix} a' & b' \\ 0 & d' \\ \end{pmatrix} \right)
\end{eqnarray*}
The kernel of $\varphi$ is the subset of $N_H$ consisting of matrices
with $a = d = 1$, which is just $H$.
Then the statement we want follows from the First Isomorphism Theorem for groups.

\end{itemize}
\vskip 10pt
\noindent
VIII.  
\begin{itemize}
\item[A)] (5) By Lagrange's theorem, what are the possible
orders of subgroups of $S_4$?
\sol
These are the divisors of $|S_4| = 24$:
$$|H| \in \{1,2,3,4,6,8,12,24\}.$$
\item[B)] (10) Show that the subset $G \subset S_4$ defined by 
$$G = \{(),(1234),(13)(24),(1432),(24),(1234)(24),(13)(24)(24),(1432)(24)\}$$
is a subgroup of $S_4$.  Note that the second four elements are
equal to the compositions of the first four with the two-cycle $(24)$.
\sol
To do this in a reasonable way (i.e. without having to write out the 
whole $8\times 8$ group table which is routine but tedious), 
note first that the first four elements 
are the elements of the cyclic subgroup $H = \langle (1234)\rangle \subset S_4$.
The second four elements are these composed with $(24)$.  
To show that $G$ is closed under products, consider the four cases
\begin{itemize}
\item[$\bullet$] $\sigma \tau$, 
\item[$\bullet$] $\sigma (\tau (24))$, 
\item[$\bullet$] $(\sigma (24)) \tau$, and 
\item[$\bullet$] $(\sigma (24))(\tau (24))$
\end{itemize}
for $\sigma,\tau \in H$.  
In the first case we have an element of $H \subset G$ since $H$ is closed under
products.  In the second case, by associativity, $\sigma(\tau (24)) = (\sigma \tau)(24)$
and $\sigma\tau \in H$ so we have an element of $G$ again. 
In the third case, note that $(\sigma (24))\tau = \sigma ((24)\tau (24))(24)$ since
$(24)$ is an element of order 2 (equals its own inverse).  The element $(24) \tau (24)$
the permutation obtained by interchanging 2,4 in $\tau$.  If $\tau = ()$
this changes nothing.  If $\tau = (1234)$ then $(24)(1234)(24) = (1423) \in H$.
If $\tau = (13)(24)$, then $(24)(13)(24)(24) = (13)(24) \in H$ again.  If $\tau = (1432)$,
then $(24)(1432)(24) = (1234)$.  Hence all these elements give elements of $G$ too
(elements of the form $\sigma (24)$ for $\sigma \in H$).  
Finally in the fourth case, the reasoning used in case three shows that we obtain
an element in $H$ from $(24)\tau(24)$ so $\sigma (24)\tau (24) \in H \subset G$.
Hence $G$ is closed under products.  

\item[] To show that $G$ is closed under inverses, note that $H$ is a subgroup
of $S_4$ so the inverse of each element in $H$ is also in $H$.  The inverse
of $\sigma (24)$ is $(24)\sigma^{-1} = (24)\sigma^{-1}(24)(24) = ((24)\sigma^{-1}(24))(24)$.
This equals an element of $H$ times (24) as in cases three and four of the proof for
products.  
 
\item{C)} (10) Determine the left and right cosets of $G$
in $S_4$. Is $G$ a normal subgroup of $S_4$?  Why or why not?
\sol
The left cosets are (with elements written as products of disjoint cycles for easy comparison)
\begin{eqnarray*}
()G = G &=& \{(), (1234), (13)(24), (1432), (24), (12)(34), (13), (14)(23)\}\\
(12)G &=& \{(12), (234), (1324), (143), (124), (34), (132), (1423)\}\\
(14)G &=& \{(14), (123), (1342), (243), (142), (1243), (134), (23)\}
\end{eqnarray*}
The right cosets are {\it different}
\begin{eqnarray*}
G() = G &=& \{(), (1234), (13)(24), (1432), (24), (12)(34), (13), (14)(23)\}\\
G(12) &=& \{(12), (134), (1423), (243), (142), (34), (123), (1324) \}\\
G(14) &=& \{(14), (234), (1243), (132), (124), (1342), (143), (23) \}
\end{eqnarray*}

This shows $G$ is {\it not a normal subgroup of $S_4$}.
\vskip 10pt
\noindent
{\it Comment:}  $G$ is isomorphic to $D_4$, the symmetries of the square, here!
\end{itemize}
\vskip 10pt
\centerline{\it Have a peaceful and joyous holiday season!}
\end{document}
