%% Math 351 Midterm 1, Fall 2018
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\markright{MATH 351, Midterm 2 Solutions}

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\begin{center}

\textbf{College of the Holy Cross, Fall Semester, 2018} \\
\textbf{MATH 351, Solutions for Midterm 2} \\
\textbf{Friday, November 16}
\end{center}


\begin{enumerate}

\item[I.] Let $G = U(20)$ (where the operation is multiplication mod $20$), and $N = \langle 9\rangle$ in $G$.
\begin{enumerate}
\item[(A)] (5) How do you know is $N$ a normal subgroup of $G$?  

\item[] {\it Solution:} $G$ is an abelian group, so \emph{every} subgroup $H$ in $G$ is normal.  
This follows since if $g \in G$, $h \in H$, then by commutativity $g h g^{-1} = (gg^{-1})h = eh = h \in H$.  
Therefore $H$ is normal.

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

\item[] {\it Solution:}  The subgroup $N = \{1,9\}$.  The distinct left cosets are
$N$, $3N = \{3,27\}$, $11N = \{11,9\}$, and $13N = \{13,17\}$.  The group table is

\[
\begin{array}{cccccc}
   & \vline &  N & 3N & 11N & 13N\\ \hline 
  N & \vline &   N &  3N & 11N & 13N\\ 
 3N & \vline &  3N &   N & 13N & 11N\\
11N & \vline & 11N & 13N &   N &  3N\\
13N & \vline & 13N & 11N &  3N &   N
\end{array} 
\]

For instance, by the definition of the coset product, 
$13N \cdot 13N = 169 N = 9N$ since $169 \equiv 9 \bmod 20$.  However, $9 \in N$, so 
$9N = N$.  

From the form of the table, the group $G/N = U(20)/\langle 9\rangle$
is non-cyclic of order 4, hence isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$. 
This relies on Lemma 4.1, or the Fundamental Theorem for finite abelian groups.   
\end{enumerate}

\item[II.]
\begin{enumerate}
\item[(A)] (10)  Let $\alpha : G \to H$ be a group homomorphism.  Show that $\ker(\alpha)$ is a 
normal subgroup of $G$.

\item[] {\it Solution:}  By definition, $\ker(\alpha) = \{g \in G \mid \alpha(g) = e\}$ (the identity
in $H$).  We always have $\alpha(e) = e$ for a group homomorphism, so $e \in \ker(\alpha)$.  Moreover,
if $a,b \in \ker(\alpha)$, then 
$$\alpha(a^{-1}b) = (\alpha(a))^{-1}\alpha(b) = e^{-1} e = e.$$
Hence $a^{-1}b \in \ker(\alpha)$.  Since this is true for all $a,b\in \ker(\alpha)$, we have shown
$\ker(\alpha)$ is a subgroup of $G$.  Finally, to show that $\ker(\alpha)$ is normal in $G$, 
let $g \in G$ and $a \in \ker(\alpha)$.  Then 
$$\alpha(gag^{-1}) = \alpha(g)\alpha(a)(\alpha(g))^{-1} = \alpha(g)e(\alpha(g))^{-1} = e.$$
Hence $gag^{-1} \in \ker(\alpha)$ whenever $g \in G$ and $a \in \ker(a)$.  By part 2 of Theorem 4.3, 
this shows $\ker(\alpha)$ is normal in $G$.  

\item[(B)] (10)  State the First Isomorphism Theorem for groups.

\item[] {\it Solution:}  Let $\alpha : G \to H$ be a group homomorphism.  Then $G/\ker(\alpha) \cong \alpha(G)$.
In words, the image of $\alpha$ (that is, the subgroup  $\alpha(G) \subseteq H$) is isomorphic as a group
to the factor group $G/\ker(\alpha)$).  

\item[(C)] (10) Let $G = \mathbb{Z} \times \mathbb{Z}$ and $N = \{(a,2a) \mid a \in \mathbb{Z}\}$.  
Using the First Isomorphism Theorem, determine a group isomorphic to $G/N$.

\item[] {\it Solution:}  Consider the mapping $\alpha : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$
defined by $\alpha(x,y) = y - 2x$.  Then $\alpha$ is a group homomorphism since 
$$\alpha(x+x',y+y') = y+y' - 2(x+x') = (y - 2x) + (y' - 2x') = \alpha(x,y) + \alpha(x',y').$$
The subgroup $N$ is the kernel of this $\alpha$ since $y - 2x = 0$ if and only if $(x,y) = (x,2x) \in N$.  
Moreover $\alpha$ is clearly surjective since given any $z \in \mathbb{Z}$, $\alpha(0,z) = z - 2\cdot 0 = z$.
Therefore, the First Isomorphism Theorem says 
$$(\mathbb{Z} \times \mathbb{Z})/N \cong \alpha(\mathbb{Z}\times \mathbb{Z}) = \mathbb{Z}.$$  
\end{enumerate}

\item[III.]  Let $G$ be a group of order $14$.   
\begin{enumerate}
\item[(A)] (15)  Show that $G$ contains elements of order $2$ and elements of order $7$.  
You may use without proof any general facts we know that apply here.  

\item[] {\it Solution 1:}  If you recall Theorem 4.15 in the text, then you can use the
fact that every group of order $14$ is isomorphic to either $\mathbb{Z}_{14}$ or $D_{14}$ (the
symmetries of a regular heptagon).  In $\mathbb{Z}_{14}$, $|2| = 7$ and $|7| = 2$, so we have
elements of both order $2$ and order $7$.  In $D_{14}$, the rotation $R_{370/7}$ has order
$7$ and the ``flip'' across any symmetry line has order $2$.  

\item[] {\it Solution 2:}  If you didn't recall Theorem 4.15, you could still derive this,
essentially by repeating a portion of the proof of that theorem in this special case (possibly taking
things we showed later into account).  If $G$ is cyclic
of order $14$ with generator $a$ and $|a| = 14$, then $G$ also has elements of both order
$2$ and order $7$, since  $|a^7| = 2$  and $|a^2| = 7$.  If $G$ is not cyclic, then by 
Lagrange's theorem, the orders of the non-identity elements of $G$ can only be $2$ or $7$.
If $G$ has only elements of order $1$ and $2$, then $G$ must be abelian (Exercise 3.32, which we
did earlier).  But then $G$ would be an abelian $2$-group and the order would be a power of 
$2$.  Since $14$ is not a power of $2$, this case is impossible.  Similarly, if $G$ has only 
elements of order $1$ and $7$, then since $7$ is prime, any element of order $7$ generates
a subgroup of order $7$ which contains the identity and 6 elements of order 7.  The intersection
of any two distinct subgroups of order 7 can contain only the identity.  Therefore, $|G|$ would 
be congruent to $1$ modulo $6$.  But $14$ is not congruent to $1$ mod 6.  So this case also 
is impossible.  $G$ must contain both elements of order $2$ and order $7$.
  
\item[(B)] (10)  Still assuming $G$ has order 14, any element $a$ of order $7$ generates
a normal subgroup.  If $b$ has order $2$, determine all possibilities for $bab = bab^{-1}$. 

\item[] {\it Solution:} (This is one of the steps in the proof of Theorem 4.15 mentioned in
Solution 1 of the previous part.)  We have $\langle a\rangle$ is normal in $G$ since it 
has order equal to half the order of $G$ (Theorem 4.1).  Hence $bab = bab^{-1}$ must be 
an element of $\langle a\rangle$, hence $bab = a^i$ for some $i$.
Now we ``do it'' (i.e. conjugate by $b$) again.  Since $b$ has order $2$ in $G$, 
$$a = b^2ab^2 = b(bab)b = ba^ib = (bab) (bab) \cdots (bab) = (bab)^i = a^{i^2}.$$
This implies $i^2 \equiv 1 \bmod 7$, so $i = 1$ or $i = 6$.   
\end{enumerate}

\item[IV.]  
\begin{enumerate}
\item[(A)] (10) Using the Fundamental Theorem, give a complete list of abelian groups of 
order $72$ up to isomorphism.
  
\item[] {\it Solution:} We have $72 = 2^3\cdot 3^2$.
 Every abelian group of order $72$ is isomorphic to one of the following:
\begin{align*}
\mathbb{Z}_8 \times \mathbb{Z}_9, &\quad \mathbb{Z}_8 \times \mathbb{Z}_3 \times \mathbb{Z}_3,\\
\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_9, &\quad \mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_3\\
\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_9, &\quad \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_3.
\end{align*}

\item[(B)] (5) Let $G = \mathbb{Z}_4 \times \mathbb{Z}_{18}$.  To which group in your list
from part (A) is $G$ isomorphic?

\item[] {\it Solution:}  Since $18 = 2 \cdot 9$ with $\gcd(2,9) = 1$, 
$\mathbb{Z}_{18} \cong \mathbb{Z}_2 \times \mathbb{Z}_9$.  This says
$$\mathbb{Z}_4 \times \mathbb{Z}_{18} \cong \mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_9.$$

\item[(C)] (5) Let $G = \langle a\rangle$ be a cyclic group of order $72$.  Write
$a = b\cdot c$, where $b$ is a $2$-element of $G$ and $c$ is a $3$-element of $G$.  

\item[] {\it Solution:}  $a$ has order $72 = 8\cdot 9$, so $a^9$ has order $8$ and $a^8$ has order $9$.
The $2$-subgroup is generated by $a^9$ and the $3$-subgroup is generated by $a^8$.  
We get $a = (a^9)^k \cdot (a^8)^\ell$
when $9k + 8\ell \equiv 1 \bmod 72$.   This holds when $k = 1$ and $\ell = 8$.  So $b = a^9$ and 
$c = a^{64} = (a^8)^8$. 
\end{enumerate}

\end{enumerate}

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