%% Math 351 Midterm 1, Fall 2018
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\begin{center}

\textbf{College of the Holy Cross, Fall Semester, 2018} \\
\textbf{MATH 351, Midterm 1 Solutions} \\
\textbf{Thursday, October 4}
\end{center}

\begin{enumerate}

\item[I.] Let $G = {\rm SL}(2,\mathbb{Z})$, the set of $2\times 2$ integer matrices with determinant 1, 
which is a group under the operation of matrix multiplication.
Let $$H = \left\{A = \begin{pmatrix} a & b \\ c & d\end{pmatrix} \in G \mid b \equiv 0 \bmod 4\right\}.$$ 
\begin{enumerate}
\item[(A)] (15) Is $H$ a subgroup of $G$?  Why or why not?
\item[] {\it Solution:}  $H$ \emph{is a subgroup of} $G$.  Here's a proof.  First, the identity element in 
$G$, namely the $2\times 2$ identity matrix: 
$$I = \begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix}$$
is in $H$ since the upper right entry is $0 \equiv 0 \bmod 4$.  Next if 
$$A = \begin{pmatrix} a & 4k \\ c & d\end{pmatrix}, \quad {\rm and} \quad 
B = \begin{pmatrix} e & 4\ell \\ f & g\end{pmatrix}$$
are elements of $H$, then the product 
\begin{align*}
A^{-1}B &= \begin{pmatrix} d & -4k \\ -c & a\end{pmatrix} \begin{pmatrix} e & 4\ell \\ f & g\end{pmatrix}\\
&= \begin{pmatrix} de - 4kf & 4ld - 4kg\\ -ce + af & -4\ell c + ag \end{pmatrix}.
\end{align*}
The upper right entry in the product is $4(ld - kg)$ and $l,d,k,g$ are all integers, so this 
element is $\equiv 0 \bmod 4$.  Hence $A^{-1}B \in H$ and $H$ is a subgroup of $G$.  
\item[(B)] (10)  Compute this matrix product, noting that the left and right matrices are inverses of each other
and the middle matrix is in $H$:
$$\begin{pmatrix} 2 & 1\\
                  3 & 2\end{pmatrix} \begin{pmatrix} 1 & 8 \\
                                                     1 & 9\end{pmatrix} \begin{pmatrix} 2 & -1\\
                                                                                       -3 & 2\end{pmatrix}$$
What does the result tell you about the subgroup $H$?
\item[] {\it Solution:}  All the matrices are in $G$, since they are all integer matrices
with determinant equal to $1$. The product is 
$$\begin{pmatrix} -69 & 47\\ -116 & 79\end{pmatrix}$$
Note that the upper right entry is $47 \equiv 3 \bmod 4$, not $0 \bmod 4$.  
Hence this product is \emph{not an element
of} $H$.  It follows that $H$ is \emph{not a normal subgroup of} $G$.

\end{enumerate}

\item[II.]
\begin{enumerate}
\item[(A)] (15)  Let $G = \langle a\rangle$ be a cyclic group.  Prove that every subgroup of $G$ is 
also cyclic.
\item[] {\it Solution:}  Let $H$ be any subgroup of $G$.  If $H = \{e = a^0\}$ then $H = \langle e\rangle$,
so $H$ is cyclic.  Otherwise, let $a^k \in H$, where $k$ is the smallest strictly positive power of $a$
for which this holds.  Since $H$ is a subgroup, this means that every power of $a^k$ is also in $H$, and 
hence $\langle a^k \rangle \subseteq H$.  Now to show the other containment, 
every other element of $H$ is in $G$, so it has the form $a^n$ for some $n$.
Use division in $\mathbb{Z}$ to divide $k$ into $n$.  This means that we can write $n = qk + r$, where
$0 \le r < n$.  Since $a^n \in H$ and $a^k \in H$, $a^r = a^n \cdot (a^k)^{-q} \in H$.  But by the choice
of $k$, this implies $r = 0$ and $a^r = e$.  Hence $a^n = (a^k)^q \in \langle a^k\rangle$.  Hence
since $a^n$ was an arbitrary element of $H$, we also have $H \subseteq \langle a^k\rangle$.  Therefore,
$H = \langle a^k\rangle$ is cyclic.  This is what we had to show.  
\item[(B)] (5)  In part (A), suppose that $a$ has order 120.  
List all the integers that are orders of elements of $G$. 
\item[] {\it Solution:}  The orders are the divisors of 120:
$$1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120.$$
\item[(C)] (10) Still assuming $a$ has order 120, how many elements of $G$ have order $24$? 
What are they? 
\item[] {\it Solution:}  There are $\phi(24)$ of them.  Since $24 = 2^3\cdot 3$, this number 
is $\phi(24) = 2^2\cdot (2-1)\cdot (3 - 1) = 8$.  To find them, recall that we want the $a^i$
for which $\frac{120}{\gcd(120,i)} = 24$, so $\gcd(120,i) = 5$.  The $i$ that work 
are $i = 5, 25, 35, 55, 65, 85, 95, 115$.  
\end{enumerate}

\item[III.]  
\begin{enumerate}
\item[(A)] (10) Let $H$ be a subgroup of a group $G$.  Show that $aH = bH$ if and only if 
$a^{-1}b \in H$.
\item[] {\it Solution:}  $\Rightarrow:$  Suppose that $aH = bH$.  Then for every $h \in H$
there exists $h' \in H$ such that $ah = bh'$.  Multiplying by $a^{-1}$ on the left and $h'^{-1}$
on the right we get $a^{-1}b = h(h')^{-1}$.  Since $H$ is a subgroup and $h,h' \in H$, 
$h(h')^{-1} \in H$.  Hence $a^{-1}b \in H$.  

\item[] $\Leftarrow:$  Suppose that $a^{-1}b \in H$. Then for all $h \in H$, we have 
$a^{-1}bh = h' \in H$.  Hence $bh = ah'$.  This implies $bH \subseteq aH$.  Since the left
cosets of $H$ in $G$ partition $G$, this implies that $bH = aH$.  (Recall any two left cosets are 
either identical or disjoint.)

\item[(B)] (15) Let $G = S_3$ and let 
$$H = \left\langle \begin{pmatrix} 1 & 2 & 3\\  3 & 2 & 1\end{pmatrix}\right\rangle.$$
Find all of the left and right cosets of $H$ in $G$.  

\item[] {\it Solution:}  The generator for $H$ is an element of order $2$ in $S_3$.  Hence
$$H = \left\{ \begin{pmatrix} 1 & 2 & 3\\  1 & 2 & 3\end{pmatrix},\begin{pmatrix} 1 & 2 & 3\\  3 & 2 & 1\end{pmatrix}\right\}.$$
Since $|G| = 6$ and $|H| = 2$, there are 3 left cosets and 3 right cosets.  The distinct left cosets are 
\begin{align*}
\begin{pmatrix} 1 & 2 & 3\\  1 & 2 & 3\end{pmatrix}H &= H\\
\begin{pmatrix} 1 & 2 & 3\\  3 & 1 & 2\end{pmatrix}H &= 
\left\{ \begin{pmatrix} 1 & 2 & 3\\  3 & 1 & 2\end{pmatrix},\begin{pmatrix} 1 & 2 & 3\\  2 & 1 & 3\end{pmatrix}\right\}\\
\begin{pmatrix} 1 & 2 & 3\\  2 & 3 & 1\end{pmatrix}H &= 
\left\{ \begin{pmatrix} 1 & 2 & 3\\  2 & 3 & 1\end{pmatrix},\begin{pmatrix} 1 & 2 & 3\\  1 & 3 & 2\end{pmatrix}\right\}
\end{align*}
The right cosets are computed similarly:
\begin{align*}
H \begin{pmatrix} 1 & 2 & 3\\  1 & 2 & 3\end{pmatrix} &= H\\
H \begin{pmatrix} 1 & 2 & 3\\  3 & 1 & 2\end{pmatrix} &= 
\left\{ \begin{pmatrix} 1 & 2 & 3\\  3 & 1 & 2\end{pmatrix},\begin{pmatrix} 1 & 2 & 3\\  1 & 3 & 2\end{pmatrix}\right\}\\
H \begin{pmatrix} 1 & 2 & 3\\  2 & 3 & 1\end{pmatrix} &= 
\left\{ \begin{pmatrix} 1 & 2 & 3\\  2 & 3 & 1\end{pmatrix},\begin{pmatrix} 1 & 2 & 3\\  2 & 1 & 3\end{pmatrix}\right\}
\end{align*}
\end{enumerate}

\item[IV.]  Let $G$ be a group containing subgroups $H$ and $K$ with $|H| = 28$ and $|K| = 65$.
\begin{enumerate}
\item[(A)] (10)  What is the smallest possible value for $|G|$?
\item[] {\it Solution:}  By Lagrange's Theorem, $|G|$ must be divisible by both $28 = 2^2\cdot 7$ and 
$65 = 5 \cdot 13$.  Since these two integers are relatively prime, ${\rm lcm}(28,65) = 28\cdot 65 = 1820$.  
\item[(B)] (10) Show that $H \cap K = \{e\}$.   
\item[] {\it Solution:}  By Lagrange's Theorem again, the intersection $H\cap K$ is a subgroup 
of both $H$ and $K$, so its order must divide both $28$ and $65$.  Since $\gcd(28,65) = 1$, the
intersection can contain only the identity element:  $H \cap K = \{e\}$.  
\end{enumerate}


\end{enumerate}

\end{document}


