%% Math 351 Final Solutions, Fall 2018
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\textbf{College of the Holy Cross, Fall Semester, 2018} \\
\textbf{MATH 351, Final Examination Solutions} \\
\textbf{Friday, December 14}
\end{center}

\vskip 10pt

\begin{enumerate}

\item[I.]  Both parts of this
question deal with  $SL(2,\mathbb{Z})$, the set of $2\times 2$ integer matrices
of determinant 1, a group under matrix multiplication.  Let
$$H = \left\{A = \begin{pmatrix} a & b\\ c & d\end{pmatrix} \in SL(2,\mathbb{Z}) : 
c \equiv 0 \bmod 5\right\}.$$
\begin{enumerate}
\item[(A)] (*) (15)  Is $H$ a subgroup of $SL(2,\mathbb{Z})$?  Why or why not?
\item[] {\it Solution:}  The answer is \emph{yes}.  First, the identity matrix
$\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix}$ (the identity element in $SL(2,\mathbb{Z})$
is in $H$ since the element in the second row and first column is
$0 \equiv 0 \bmod 5$.  Next, if $A = \begin{pmatrix} a & b\\ 5k & d\end{pmatrix}$
and $B = \begin{pmatrix} a' & b'\\ 5k' & d'\end{pmatrix}$ are elements of $H$, then
the matrix product
$$AB = \begin{pmatrix} a & b\\ 5k & d\end{pmatrix}\cdot 
\begin{pmatrix} a' & b'\\ 5k' & d'\end{pmatrix} 
= \begin{pmatrix} aa' + 5bk' & ab' + bd'\\ 5(ka' + dk') & 5kb' + dd'\end{pmatrix}$$
has lower left entry divisible by $5$.  Finally, if 
$A = \begin{pmatrix} a & b\\ 5k & d\end{pmatrix}$ is in $H$, then
$$A^{-1} = \begin{pmatrix} d & -b\\ -5k & a\end{pmatrix}$$
since $\det(A) = 1$.  This matrix is also in $H$, so $H$ is a subgroup of 
$SL(2,\mathbb{Z})$.  

\item[(B)] (*) (15)  Is the cyclic subgroup generated by 
$$A = \begin{pmatrix} 1 & 0 \\ 5 & 1\end{pmatrix}$$
in $G$ finite or infinite?  Explain.

\item[] {\it Solution:} It is easy to see, and easy to prove by induction,
that 
$$A^k = \begin{pmatrix} 1 & 0 \\ 5k & 1\end{pmatrix}$$
for all $k \in \mathbb{Z}$.  Since $A^k = \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$
only when $k = 0$, this shows that the matrix $A$ is an element of infinite order, and 
the cyclic subgroup generated by $A$ is infinite. 
\end{enumerate}

\vskip 10pt

\item[II.] Let $G = \langle a\rangle$ be a cyclic group of order $100$.
\begin{enumerate}
\item[(A)]  (*) (10)  How many different generators does $G$ have?

\item[] {\it Solution:}  The number is $\phi(100)$ (the Euler phi-function).
Since $100 = 2^2\cdot 5^2$, $\phi(100) = 2^1 \cdot (2 - 1)\cdot 5^1\cdot (5 - 1) = 40$.  
The generators themselves are the powers $a^i$ where $\gcd(i,100) = 1$.  

\item[(B)]  (*) (10)  What is the order of the element $a^{30}$ in $G$?

\item[] {\it Solution:}  The order is 
$$|a^{30}| = \frac{100}{\gcd(100,30)} = \frac{100}{10} = 10.$$
(Alternately, one could compute powers of $a^{30}$ until an exponent 
divisible by $100$ is obtained.)

\item[(C)]  (*)  (10)  Suppose you know that a subgroup $H$ of $G$ contains both 
$a^{30}$ and $a^{56}$. What can you say about the order of $H$?  

\item[] {\it Solution:}  We see $\gcd(30,56) = 2$.  Moreover, because $H$
is a subgroup of $G$, repeating the steps of the Euclidean algorithm in
the exponents, we find:
\begin{align}
56 &= 1\cdot 30 + 26 \Rightarrow a^{56}\cdot a^{-30} = a^{26} \in H\\
30 &= 1\cdot 26 + 4 \Rightarrow a^{30} \cdot a^{-26} = a^4 \in H\\
26 &= 4\cdot 6 + 2 \Rightarrow a^{26} \cdot a^{-24} = a^2 \in H.
\end{align}
Hence there are two possibilities:  Either $H$ is all of $G$ and 
$|H| = 100$, or else $H = \langle a^2\rangle$, which says $|H| = 50$.  
\end{enumerate}
\vskip 10pt
\item[III.]
\begin{enumerate}
\item[(A)] (10)  Let $\alpha : G \to H$ be a group homomorphism.  Show that $\alpha(G)$ is a 
subgroup of $H$.
\item[] {\it Solution:}  For all group homomorphisms $\alpha(e_G) = e_H$.  Hence
$e_H \in \alpha(G)$.  If $c,d \in \alpha(G)$, then $c = \alpha(a)$ and $d = \alpha(b)$
for some $a,b \in G$.  Therefore, since $\alpha$ is a group homomorphism,
$$c\cdot d = \alpha(a)\cdot \alpha(b) = \alpha(a \cdot b).$$
This shows $c\cdot d \in \alpha(G)$.  Finally, if $c = \alpha(a)$, then 
$$c^{-1} = (\alpha(a))^{-1} = \alpha(a^{-1}).$$
It follows that $c^{-1} \in \alpha(G)$.  Therefore $\alpha(G)$ is a subgroup of $H$.
\item[(B)] (20)  {\it State and prove} the First Isomorphism Theorem for groups.
\item[] {\it Solution:}  The First Isomorphism Theorem states that if 
$\alpha : G \to H$ is a group homomorphism, then the image $\alpha(G)$ is isomorphic
as a group to $G/\ker(\alpha)$.  To prove this we will simplify the notation
by writing $\ker(\alpha) = N$ and consider the mapping
\begin{align*}
\phi : G/N &\longrightarrow \alpha(G)\\
           g N &\longmapsto \alpha(g)
\end{align*}
Since this mapping is defined with domain a factor group, we need to start
by showing that it is well-defined.  If the cosets $g N$ and
$g' N$ are equal, though, $g^{-1}g' \in N$ and this
implies $\alpha(g^{-1}g') = (\alpha(g))^{-1} \alpha(g') = e_H$.  It
follows that $\alpha(g) = \alpha(g')$, so the mapping $\phi$ is well-defined.
Next, we claim that $\phi$ is a group homomorphism.  This follows from the
way the coset product is defined in the factor group:
$$\phi(g N\cdot g' N) = \phi((gg')N) = 
\alpha(gg') = \alpha(g)\cdot \alpha(g') = \phi(gN)\cdot \phi(g'N).$$
This shows $\phi$ is a gorup homomorphism.  Since every element $g$ in $G$ yields
some coset of the kernel, every $\alpha(g)$ for $g \in G$ is in the image
of $\phi$, so the mapping $\phi$ is surjective.  So, it remains
to show that $\phi$ is injective.  Suppose 
$$\phi(gN) = \alpha(g) = \alpha(g') = \phi(g'N).$$
This shows that 
$$(\alpha(g))^{-1}\cdot \alpha(g') = \alpha(g^{-1}g') = e_H,$$ 
so $g^{-1}g' \in N$, and we know that that implies the cosets of $g$ and 
$g'$ are equal: $g N = g' N$.  Therefore $\phi$ is also injective,
and we have shown that $\phi$ is an isomorphism of groups, which is what we had
to do.  
\end{enumerate}

\item[IV.]  All parts of this question refer to the group $G$ of
order $8$ whose (corrected!) operation table is given below:
\[
\begin{array}{cccccccccc}
\cdot & \vline & g_1 & g_2 & g_3 & g_4 & g_5 & g_6 & g_7 & g_8 \\ \hline
g_1 & \vline & g_1 & g_2 & g_3 & g_4 & g_5 & g_6 & g_7 & g_8 \\
g_2 & \vline & g_2 & g_5 & g_4 & g_7 & g_6 & g_1 & g_8 & g_3 \\
g_3 & \vline & g_3 & g_8 & g_5 & g_2 & g_7 & g_4 & g_1 & g_6 \\
g_4 & \vline & g_4 & g_3 & g_6 & g_5 & g_8 & g_7 & g_2 & g_1 \\
g_5 & \vline & g_5 & g_6 & g_7 & g_8 & g_1 & g_2 & g_3 & g_4 \\
g_6 & \vline & g_6 & g_1 & g_8 & g_3 & g_2 & g_5 & g_4 & g_7 \\
g_7 & \vline & g_7 & g_4 & g_1 & g_6 & g_3 & g_8 & g_5 & g_2 \\
g_8 & \vline & g_8 & g_7 & g_2 & g_1 & g_4 & g_3 & g_6 & g_5
\end{array}
\]

\begin{enumerate}
\item[(A)] (*) (5) What is the inverse of the element $g_2$?

\item[] {\it Solution:} By inspection of the table, we
see $g_1$ is the identity element.  Since $g_2 \cdot g_6 = g_6 \cdot g_2 = g_1$,
the inverse of $g_2$ is $g_6$.

\item[(B)] (*) (5) What elements are in the subgroup $\langle g_3\rangle$?

\item[] {\it Solution:}  We see $g_3^2 = g_5$, $g_3 \cdot g_5 = g_7$ and 
$g_3 \cdot g_7 = g_1$.  Therefore, $g_3$ has order 4 and 
$$\langle g_3 \rangle = \{g_1, g_3, g_5, g_7\}.$$

\item[(C)] (*) (5) Is the subgroup $\langle g_3\rangle$ normal in $G$?  Why or why not?

\item[] {\it Solution:}  Since $|\langle g_3\rangle| = 4 = \frac{1}{2}|G|$, this 
subgroup is normal in $G$.  

\item[(D)] (*) (5) What is the center of $G$, that is, the subgroup $Z(G)$?

\item[] {\it Solution:}  By inspection of the table, the elements that commute
with all elements of $G$ are $g_1$ and $g_5$.  Therefore 
$$Z(G) = \{g_1, g_5\}.$$

\item[(E)] (20) Construct the group table for the factor group $G/Z(G)$.
To which ``standard'' group is this isomorphic?

\item[] {\it Solution:}  The distinct left cosets of $N = Z(G)$ are 
$$N, g_2N = \{g_2, g_6\}, g_3N = \{g_3, g_7\}, g_4N = \{g_4, g_8\}$$
The group table for the factor group is found using the usual coset product

\[
\begin{array}{cccccc}
\cdot & \vline & N & g_2N & g_3N & g_4N \\ \hline
N & \vline & N & g_2N & g_3N & g_4N \\
g_2N & \vline & g_2N & N & g_4N & g_3N \\
g_3N & \vline & g_3N & g_4N & N & g_2N \\
g_4N & \vline & g_4N & g_3N & g_2N & N
\end{array}
\]
This is a non-cyclic group of order $4$, hence isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$. 
\item[] {\it Note:}  This group of order 8 is isomorphic to the \emph{quaternion group}, $Q$, 
the last of the order 8 groups that we just ``met'' the last day of class.  You didn't
need to know that to do any part of this problem, though!
\end{enumerate}

\vskip 10pt

\item[V.]   
\begin{enumerate}
\item[(A)] (15)  Up to isomorphism, how many different \emph{abelian} groups 
of order $600$ are there?  List one group from each isomorphism class.
\item[] {\it Solution:}  Since $600 = 2^3 \times 3 \times 5^2$, there are
three possible structures for the $2$-subgroup, one for the $3$-subgroup, and two
for the $5$-subgroup.  By the fundamental theorem of finite abelian groups, 
the following list includes all the possibilities, up to isomorphism:
\begin{align*}
&\mathbb{Z}_8 \times \mathbb{Z}_3 \times \mathbb{Z}_{25}\\
&\mathbb{Z}_8 \times \mathbb{Z}_3 \times \mathbb{Z}_5 \times \mathbb{Z}_5\\
&\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_{25}\\
&\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_5 \times \mathbb{Z}_5\\
&\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_{25}\\
&\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_5 \times \mathbb{Z}_5
\end{align*}
\item[(B)] (15)  Up to isomorphism, how many different groups of order $2018$ are
there?  List one group from each isomorphism class.  (Hint:  This is \emph{not} a long list!) 
\item[] {\it Solution:}  Factoring we find $2018 = 2\cdot 1009$, and $1009$ is an odd prime.
(This takes some checking, but it is routine; you just have to check that none of the primes $3,5,7,11,13,17,19,23,29,31$ divides $1009$.  Those are the primes $\le \sqrt{1009}$.)
In this situation we know that there are only two isomorphism classes of groups 
of order $2p$, namely, every group of order 2018 is isomorphic to either
$$\mathbb{Z}_{2018} \quad {\rm or} \quad D_{2018}$$
($D_{2018}$ is the dihedral group of rotational and reflection symmetries of a regular $1009$-gon, which would be hard to distinguish from a circle if you drew it -- unless the edges
were made very long, of course!).  
\end{enumerate}

\vskip 10pt

\item[VI.] 
\begin{enumerate}
\item[(A)] (20) Use the Sylow Theorems to show that there are no simple groups
of order $100$.  
\item[] {\it Solution:}  We have $100 = 2^2 \cdot 5^2$.  By Sylow III, the 
number of Sylow $5$-subgroups must be congruent to 1 mod 5 and it must divide
$4$.  The only possible number is $1$, and Sylow II implies that subgroup must
be a normal subgroup of order $25$.  Hence if $|G| = 100$, then $G$ is not a simple group.
\item[(B)] (20)  How many different Sylow $5$-subgroups does the alternating group 
$A_5$ have?
\item[] {\it First Solution:}  The alternating group $A_5$ has order $\frac{1}{2}\cdot 5! = 60$.
This factors as $60 = 2^2 \cdot 3 \cdot 5$.  The Sylow $5$-subgroups must 
have order $5$, and hence are cyclic since $5$ is prime.  
The only elements of order $5$ in $A_5$ (or $S_5$)
are the $5$-cycles $(abcde)$, where $\{a,b,c,d,e\} = \{1,2,3,4,5\}$.  There are 
$\frac{5!}{5} = 24$ distinct $5$-cycles, but groups of $4$ of them generate the 
same subgroup since if $\sigma$ is one of the $5$-cycles, $\sigma^2, \sigma^3, \sigma^4$
all generate the same subgroup as $\sigma$.  
Hence the number of distinct Sylow $5$-subgroups is $\frac{24}{4} = 6$.  
Note that this agrees with the statement of Sylow III.  The number of Sylow
$5$-subgroups in a group of order 60 is congruent to 1 mod 5 and divides 12, hence
is either 1 or 6. 
\item[] {\it Second Solution:} (``sneaky'')   By Sylow III, the number of Sylow
$5$-subgroups in a group of order 60 is congruent to 1 mod 5 and divides 12, hence
is either 1 or 6.  If there is just one Sylow $5$-subgroups, then Sylow II implies
that subgroup is normal, of order $5$.  However, we know that $A_5$ is a simple, nonabelian
group, so it has no normal subgroups other than $\{e\}$ and $A_5$.  Therefore, the 
number must be $6$(!)
\end{enumerate}

\end{enumerate}

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