\documentclass[12pt]{article}
\usepackage{latexsym,amsmath,amssymb,epsfig,float,graphicx,subfig}

%\usepackage{my_teach}

%\pagestyle{myheadings}
\markright{MATH 351, Final Exam Review}

\setlength{\topmargin}{-0.5in}
\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\textheight}{8.5in}

\newcommand{\Blank}{$\underline{\phantom{Please write your name in this
        space}}$} 
\newcommand{\Strut}{$\vphantom{\Big|}$}

\begin{document}

\begin{center}

\textbf{College of the Holy Cross, Fall Semester, 2018} \\
\textbf{MATH 351, Modern Algebra I -- Solution for Review Problem for Final Exam} \\
\textbf{November 29}
\end{center}

Consider the following operation table:
\[
\begin{array}{cccccccccccccc}
\cdot & \vline & g_1 & g_2 & g_3 & g_4 & g_5 & g_6 & g_7 & g_8 & g_9 & g_{10} & g_{11} & g_{12} \\ \hline
g_1 & \vline & g_1 & g_2 & g_3 & g_4 & g_5 & g_6 & g_7 & g_8 & g_9 & g_{10} & g_{11} & g_{12} \\
g_2 & \vline & g_2 & g_1 & g_4 & g_3 & g_6 & g_5 & g_8 & g_7 & g_{10} & g_9 & g_{12} & g_{11} \\
g_3 & \vline & g_3 & g_4 & g_7 & g_8 & g_{11} & g_{12} & g_1 & g_2 & g_5 & g_6 & g_9 & g_{10} \\
g_4 & \vline & g_4 & g_3 & g_8 & g_7 & g_{12} & g_{11} & g_2 & g_1 & g_6 & g_5 & g_{10} & g_9 \\
g_5 & \vline & g_5 & g_6 & g_9 & g_{10} & g_1 & g_2 & g_{11} & g_{12} & g_3 & g_4 & g_7 & g_8 \\
g_6 & \vline & g_6 & g_5 & g_{10} & g_9 & g_2 & g_1 & g_{12} & g_{11} & g_4 & g_3 & g_8 & g_7 \\
g_7 & \vline & g_7 & g_8 & g_1 & g_2 & g_9 & g_{10} & g_3 & g_4 & g_{11} & g_{12} & g_5 & g_6 \\
g_8 & \vline & g_8 & g_7 & g_2 & g_1 & g_{10} & g_9 & g_4 & g_3 & g_{12} & g_{11} & g_6 & g_5 \\
g_9 & \vline & g_9 & g_{10} & g_{11} & g_{12} & g_7 & g_8 & g_5 & g_6 & g_1 & g_2 & g_3 & g_4 \\
g_{10} & \vline & g_{10} & g_9 & g_{12} & g_{11} & g_8 & g_7 & g_6 & g_5 & g_2 & g_1 & g_4 & g_3 \\
g_{11} & \vline & g_{11} & g_{12} & g_5 & g_6 & g_3 & g_4 & g_9 & g_{10} & g_7 & g_8 & g_1 & g_2 \\
g_{12} & \vline & g_{12} & g_{11} & g_6 & g_5 & g_4 & g_3 & g_{10} & g_9 & g_8 & g_7 & g_2 & g_1 \\
\end{array}
\]
\noindent
You may assume without proof that this is the operation table for a group $G$ of order 12. 
\begin{enumerate}
\item[a.] (*) Which is the identity element in this group?  Which element is the inverse of each element? Is
$G$ abelian?  
\item[] {\it Solution:}  $g_1$ is the identity element since $g_1\cdot g_j = g_j \cdot g_1 = g_j$ for all 
$j = 1,\ldots, 12$.  We have 
\begin{align*}
g_1^{-1} = g_1, \quad g_2^{-1} = g_2, &\quad g_3^{-1} = g_7, \quad g_4^{-1} = g_8\\
g_5^{-1} = g_5, \quad g_6^{-1} = g_6, &\quad g_7^{-1} = g_3, \quad g_8^{-1} = g_4\\
g_9^{-1} = g_9, \quad g_{10}^{-1} = g_{10}, &\quad g_{11}^{-1} = g_{11}, \quad g_{12}^{-1} = g_{12}.
\end{align*}
$G$ is \emph{not abelian}, since for example $g_4 \cdot g_9 = g_6$, but $g_9 \cdot g_4 = g_{12}$.  
\item[b.] (*) What are the orders of each element in this group?
\item[] {\it Solution:} We have
\begin{align*}
|g_1| = 1, \quad |g_2| = 2, &\quad |g_3| = 3, \quad |g_4| = 6\\
|g_5| = 2, \quad |g_6| = 2, &\quad |g_7| = 3, \quad |g_8| = 6\\
|g_9| = 2, \quad |g_{10}| = 2, &\quad |g_{11}| = 2, \quad |g_{12}| = 2.
\end{align*}
For example, to see $|g_4| = 6$, compute from the table: 
$$g_4^2 = g_7, g_4^3 = g_4\cdot g_7 = g_2, g_4^4 = g_4\cdot g_2 = g_3,$$
then 
$$g_4^5 = g_4\cdot g_3 = g_8,\quad {\rm and}\quad g_4^6 = g_4 \cdot g_8 = g_1 = e.$$
Hence the cyclic subgroup $\langle g_4\rangle = \{g_1,g_4,g_7,g_2,g_3,g_8\}$, which has order $6$.
\item[c.] (*) Why is $H = \{g_1,g_2,g_{11},g_{12}\}$ a subgroup of $G$?  
\item[] {\it Solution:}  Probably the best way to see this is to pick out the 
entries from the table for $G$ in the rows and columns corresponding to these four elements.
We find:
\[
\begin{array}{cccccc}
\cdot & \vline & g_1 & g_2 & g_{11} & g_{12} \\ \hline
g_1   & \vline & g_1 & g_2 & g_{11} & g_{12} \\
g_2   & \vline & g_2 & g_1 & g_{12} & g_{11} \\
g_{11} & \vline & g_{11} & g_{12} & g_1 & g_2 \\
g_{12} & \vline & g_{12} & g_{11} & g_2 & g_1
\end{array}
\]
This shows that $e = g_1 \in H$, $H$ is closed under products (only elements from $H$ appear in these
entries in the table), and $H$ is closed under taking inverses (since every element in $H$ is its
own inverse).  Hence $H$ is a subgroup of $G$.  

\item[d.] To which ``standard'' group of order $4$ is $H$ isomorphic?
\item[] {\it Solution:}  $H$ is a non-cyclic group of order $4$, hence isomorphic to 
$\mathbb{Z}_2 \times \mathbb{Z}_2$.
\item[e.] (*) Why is $J = \{g_1,g_3,g_5,g_7,g_9,g_{11}\}$ a subgroup of $G$?  
\item[] {\it Solution:}  Proceed as in part c.  Selecting the rows and columns corresponding
to the elements of $J$, we obtain:
\[
\begin{array}{cccccccc}
\cdot & \vline & g_1 & g_3 & g_5 & g_7 & g_9 & g_{11}\\ \hline
g_1   & \vline & g_1 & g_3 & g_5 & g_7 & g_9 & g_{11}\\
g_3   & \vline & g_3 & g_7 & g_{11} & g_1 & g_5 & g_9\\
g_5   & \vline & g_5 & g_9 & g_1 & g_{11} & g_3 & g_7\\
g_7   & \vline & g_7 & g_1 & g_9 & g_3 & g_{11} & g_5\\
g_9   & \vline & g_9 & g_{11} & g_7 & g_5 & g_1 & g_3\\
g_{11} & \vline & g_{11} & g_5 & g_3 & g_9 & g_7 & g_1\\
\end{array}
\]
As in part c, the form of the table shows that $J$ is a subgroup of $G$.  
\item[f.] To which ``standard'' group of order 6 is $J$ isomorphic?
\item[] {\it Solution:}  $J$ is nonabelian of order $6$, since, for instance $g_5\cdot g_{11} = g_7$
but $g_{11} \cdot g_5 = g_3$.  Hence $J$ is isomorphic to $D_6$ (which is also isomorphic to $S_3$).
Recall that any group of order $2p$, where $p$ is an odd prime, is isomorphic to either $\mathbb{Z}_{2p}$ (abelian), 
or to $D_{2p}$ (nonabelian).  That applies here since $6 = 2 \cdot 3$.  
\item[g.] (*) Why is $K = \{g_1,g_3,g_7\}$ a subgroup of $G$? 
\item[] {\it Solution:}  Same idea as parts c and e.  Alternately, by the calculations done above, 
you can see that $K = \langle g_3\rangle = \langle g_7\rangle$, so it's the cyclic subgroup 
generated by an element.  
\item[h.] (*) Which of the subgroups $H,J,K$ from the previous parts is normal in $G$?
\item[] {\it Solution:}  $J$ \emph{must be normal} in $G$, since it has $|J| = 6 = \frac{1}{2} |G|$.
$H$ is \emph{not normal} in $G$.  We can see that because, for instance 
$$g_5\cdot g_{11}\cdot g_5^{-1} = g_7 \cdot g_5 = g_9 \notin H.$$
Finally, we claim $K$ is \emph{normal} in $G$.  Here's another way to check that, 
by looking at the distinct left and right cosets.
\begin{align*}
g_1 K = g_3 K = g_7 K &= K = K g_1 = K g_3 = K g_7\\
g_2 K = g_4 K = g_8 K &= \{g_2, g_4, g_8\} = K g_2 = K g_4 = K g_8\\
g_5 K = g_9 K = g_{11} K &= \{g_5, g_9, g_{11}\} = K g_5 = K g_9 = K g_{11}\\
g_6 K = g_{10} K = g_{12} K &= \{g_6, g_{10}, g_{12}\} = K g_6 = K g_{10} = K g_{12}.
\end{align*}
\item[i.] For each subgroup that is normal in $G$ from the previous part, construct the factor group.
(That is construct $G/H$ if $H$ is normal, $G/J$ if $J$ is normal in $G$, and $G/K$ if $K$ is normal).
\item[] {\it Solution:}  The normal subgroups are $J$ and $K$.  We have $G/J$ is a group of order
$2$ consisting of the cosets $J, g_2 J$.  The table looks like
\[
\begin{array}{cccc}
\cdot & \vline & J & g_2 J\\ \hline
J & \vline & J & g_2 J\\
g_2 J & \vline & g_2 J & J
\end{array}
\]
Note that because of the way the coset product is defined, 
$$(g_2 J)\cdot (g_2 J) = (g_2\cdot g_2) J = g_1 J = J.$$
For the factor group $G/K$, let's label the left cosets as $K, g_2 K, g_5 K, g_6 K$.  Then the 
table for the factor group looks like this:
\[
\begin{array}{cccccc}
\cdot & \vline & K & g_2 K & g_5 K & g_6 K\\ \hline
K & \vline & K & g_2 K & g_5 K & g_6 K\\
g_2 K & \vline & g_2 K & K & g_6 K & g_5 K\\
g_5 K & \vline & g_5 K & g_6 K & K & g_2 K\\
g_6 K & \vline & g_6 K & g_5 K & g_2 K & K
\end{array}
\]
(This is the non-cyclic group of order $4$.)
\item[j.] Construct a group homomorphism $\alpha : G \to L$ for some group $L$ to make 
$\ker(\alpha) = \{g_1,g_2\}$.  To which ``standard'' group of order $6$ is $L$ isomorphic?  (Hint: To
see how to define $\alpha$, you might note the way 
the whole table for $G$ breaks up into $2\times 2$ blocks according to the cosets of the subgroup
$\{g_1,g_2\}$!)
\item[] {\it Solution:}  This can be phrased in multiple ways.  The ``slickest'' is to note
how the table for $G$ breaks up into $2\times 2$ blocks according to the cosets of the 
subgroup $N = \{g_1, g_2\}$.  Those cosets are:
\begin{align*} 
N,\ &g_3 N = \{g_3,g_4\} = N g_3,\ g_5 N = \{g_5, g_6\} = N g_5,\\
g_7 N = \{g_7, g_8\} = N g_7,\ &g_9 N = \{g_9, g_{10}\} = N g_9,\ g_{11} N = \{g_{11}, g_{12}\} = N g_{11}.
\end{align*}
This shows that $N$ is a normal subgroup of $G$.  Hence the factor group $L = G/N$ is defined 
and the map $\alpha : G \to G/N$ taking $g_i \in G$ to $g_i N \in L$ is a mapping as required with 
$\ker(\alpha) = N$.  (Note that if $i$ is even, then $g_i N$ is equal to one of the cosets above.
For instance $g_8 N = \{g_8, g_7\} = g_7 N$.)
\item[k.] Show that $G$ is the internal direct product of its subgroups $L = \{g_1,g_2\}$
and $J = \{g_1,g_3,g_5,g_7,g_9,g_{11}\}$.
\item[] {\it Solution:}  We know $J$ is normal in $G$ from part h above and we saw in part j 
that $N$ is normal in $G$.  Moreover $J \cap N = \{g_1\} = \{e\}$.  Therefore the 12 products
of elements of $J$ and $N$ are all distinct and give all the elements in $G$:  $J\cdot N = G$.  Therefore,
by definition $G$ is the internal direct product of $J$ and $N$.  Recall that this means $G$
is also isomorphic to the external direct product $J \times N \cong D_6 \times \mathbb{Z}_2$.  
\item[l.] What is $Z(G)$, the center of $G$?  What is the centralizer of the element $g_2$?  
\item[] {\it Solution:}  The center of $G$ is the collection of elements commuting with all elements
in $G$.  This means that in the table we're looking for $g_i$ such that if you read across
the row for $g_i$, then the elements are listed \emph{in the same order} as they appear in
the column for $g_i$.  The elements of $G$ for which this is true are $g_1,g_2$.  Hence 
$Z(G) = \{g_1,g_2\}$.  This implies that the centralizer of $g_2$ (the subgroup
$$C(g_2) = \{g \in G \mid g\cdot g_2 = g_2 \cdot g\})$$
is all of $G$:  $C(g_2) = G$.    
\item[m.] What are the conjugacy classes in $G$?
\item[] {\it Solution:}  The elements of the center appear alone in separate conjugacy classes.
Elements in any conjugacy class must have the same order, so the best way to organize this is
to look one order at a time.  The two elements of order $6$ ($g_4$ and $g_8$) 
are conjugate, since for instance
$$g_{11}\cdot g_4 \cdot g_{11}^{-1} = g_6 \cdot g_{11} = g_8.$$ 
The two elements of order 3 ($g_3$ and $g_7$) are also conjugate, since for instance
$$g_{11} \cdot g_3 \cdot g_{11}^{-1} = g_5 \cdot g_{11} = g_7.$$
The remaining 6 elements of the group all have order 2.  But they are not all conjugate since
for instance if we look at the element $g_{12}$, then its centralizer is
$C(g_{12}) = \{g_1,g_2,g_{11},g_{12}\}$ which has order $4$.  This means that the conjugacy
class of $g_{12}$ contains only $[G : C(g_{12})] = 12/4 = 3$ distinct elements:
$$\{g_{12}, g_6, g_{10}\}$$
Similarly the class of $g_{11}$ contains only 3 distinct elements ($g_{11}$ has the same centralizer
as $g_{12}$):
$$\{g_{11}, g_5, g_9\}.$$
To summarize, $G$ has 
\begin{itemize}
\item Two classes of size 1 from the elements of the center: $\{g_1\}, \{g_2\}$
\item One class consisting of the elements of order 6: $\{g_4, g_8\}$
\item One class consisting of the elements of order 3: $\{g_3, g_7\}$, and 
\item Two classes of noncentral elements of order 2: $\{g_5,g_9,g_{11}\}$ and $\{g_6, g_{10}, g_{12}\}$.
\end{itemize}
\item[n.] How many different Sylow $2$-subgroups are there in $G$?  How many different Sylow $3$-subgroups?
Verify that the statement of Sylow III holds here.  
\item[] {\it Solution:} Since $12 = 2^2 \cdot 3$, any Sylow $3$-subgroup must have order 3, hence
be isomorphic to $\mathbb{Z}_3$.  Since $G$ contains only two elements of order 3 ($g_3$ and $g_7$),
there is exactly one Sylow $3$-subgroup, namely the subgroup $K = \{g_1,g_3,g_7\}$ studied above.
(Note that it must be normal since there is only one!)  The Sylow $2$-subgroups must be isomorphic 
to $\mathbb{Z}_2 \times \mathbb{Z}_2$ since $G$ does not contain any elements of order 4.  Each
such subgroup must contain 3 of the elements of order $2$ and the ones in each subgroup 
must commute among themselves.  
Note that the only pairs of the noncentral elements of order 2 that do commute with each other 
are $g_5$ and $g_6$, $g_9$ and $g_{10}$, and $g_{11}$ and $g_{12}$.  Hence we get 3 Sylow $2$-subgroups
$$\{g_1,g_2,g_5,g_6\}, \{g_1,g_2,g_9,g_{10}\}, \{g_1,g_2,g_{11},g_{12}\}.$$
This all agrees with Sylow III since that theorem says the number of Sylow $3$-subgroups 
must be $\equiv 1 \bmod 3$ and divide $4$, while the number of Sylow $2$-subgroups must be 
$\equiv 1 \bmod 2$ and divide $3$.
\end{enumerate}

\end{document}


