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\noindent
\parbox{2in}{\bf Math 3110} 
\hfill {\Large \bf Test \#3} \hfill
\parbox{2in}{\bf \hfill November $18^{\mathrm{th}}$, 2011}

\vspace{0.3in}

\noindent {\large\bf Name:} \underline{\color{red}\Large\sc Answer Key} \hfill {\bf Be sure to show your work!}

\vspace{0.2in}

\noindent {\large 1. (25 points)} 3-2-1...Go!

\begin{enumerate}[(a)]
   \item How many elements of order 3 are in $\mathbb{Z}_{9000} \oplus \mathbb{Z}_{3333333}$?

\vspace{0.1in}

The order of an element of $\mathbb{Z}_{9000} \oplus \mathbb{Z}_{3333333}$ is computed by finding the least
common multiple of the orders of each of its components: $|(a,b)| = \mathrm{lcm}(|a|,|b|)$. If we have to have an
element of order 3, then $3=|(a,b)|=\mathrm{lcm}(|a|,|b|)$. The only ways to get an lcm of 3 are $\mathrm{lcm}(3,1)=\mathrm{lcm}(1,3)=\mathrm{lcm}(3,3)=3$.

Now both $\mathbb{Z}_{9000}$ and $\mathbb{Z}_{3333333}$ are cyclic groups of orders divisible by 3. So they
both have exactly 1 element of order 1 (the identity) and 2 elements of order 3 (the two generators of the unique
subgroup of order 3). Thus there are $2\cdot 1 + 1\cdot 2+ 2 \cdot 2 = 8$ elements of order 3.

Alternatively, taking all elements of order 1 or 3, we have $3 \cdot 3 = 9$ choices. Only the identity has order 1, so
that leaves exactly $9-1=8$ elements of order 3.

{\bf Unnecessary Extra:}
By the way, if $3$ divides $n$, then $n/3$ and $n/3+n/3=2n/3$ are the elements of order 3 in $\mathbb{Z}_n$. So since $0$ has order 1 in both groups, $3000$ and $6000$ have order 3 in $\mathbb{Z}_{9000}$, and $1111111$ and $2222222$ have order 3 in $\mathbb{Z}_{3333333}$. Thus the elements of order 3 are:
$(3000,0)$, $(6000,0)$, $(0,1111111)$, $(0,2222222)$, $(3000,1111111)$, $(3000,2222222)$, $(6000,1111111)$, and $(6000,2222222)$.

\vspace{0.1in}

   \item Let $H$ be a subgroup of $G$ where $G$ is abelian. Quickly explain why $H$ is a normal subgroup of $G$.
   
\vspace{0.1in}
 
    Let $a \in G$. Then $aH = \{ ah \;|\; h \in H \} = \{ ha \;|\; h \in H \} = Ha$ (the middle equality holds because $G$ is abelian).
    
    Alternate proof: Let $a \in G$ and $h \in H$. Then $aha^{-1} = haa^{-1} = h \in H$ (the first equality again follows from the fact that $G$ is abelian). Thus $aHa^{-1} \subseteq H$ so $H$ is normal.

\vspace{0.1in}
   
  
   \item Consider for example: $\diagquotient{\mathbb{Z}_{12}}{\langle 4 \rangle} \cong \mathbb{Z}_4$. Now let $k\ell=n$ (where $k,\ell,n$ are positive integers) and let $G=\diagquotient{\mathbb{Z}_n}{\langle k \rangle}$. Briefly, explain why the $G$ has order $k$, why $G$ is cyclic, and why $G \cong \mathbb{Z}_k$.

\vspace{0.1in}

Recall that $|\langle k \rangle| = n/k$ if $k$ divides $n$. Therefore, $|G| = \diagquotient{|\mathbb{Z}_n|}{|\langle k \rangle|} = \frac{n}{n/k} = \frac{n}{\ell} = k$. $G$ is cyclic because $G$ is a quotient of $\mathbb{Z}_n$ and any quotient of a cyclic group is cyclic. Finally, $G$ and $\mathbb{Z}_k$ are both cyclic groups of order $k$. Thus they must be isomorphic (any two cyclic groups of the same order are isomorphic).

\vspace{0.1in}

   \item Let $G$ and $H$ be group. Prove that $\pi:G \oplus H \rightarrow G$ defined by $\pi((g,h))=g$ is a homomorphism which is onto. Find the kernel of $\pi$. Finally, show that $\diagquotient{G \oplus H}{\{e\}\oplus H} \cong G$.

\vspace{0.1in}

Let $(g,h),(x,y) \in G \oplus H$. $\pi((g,h)(x,y))=\pi((gx,hy))=gx=\pi((g,h))\pi((x,y))$. Thus $\pi$ is a homomorphism. Next, let $g \in G$. Since $H$ is a group, it contains at least an identity element, say $e \in H$. Then $(g,e)\in G \oplus H$ and $\pi((g,e))=g$. Therefore, $\pi$ is onto. $\mathrm{ker}(\pi) = \{ (g,h) \in G \oplus H \;|\; \pi((g,h))=e \} = \{ (g,h) \in G \oplus H \;|\; g=e \} = \{ (e,h) \;|\; h \in H \} = \{ e \} \oplus H$. Finally, noting that $\mathrm{ker}(\pi) = \{e\}\oplus H$ and $\mathrm{image}(\pi)=G$ (since $\pi$ is onto), we have (by the first isomorphism theorem) that $\diagquotient{G \oplus H}{\{e\}\oplus H} = \diagquotient{G \oplus H}{\mathrm{ker}(\pi)} \cong \mathrm{image}(\pi) = G$.

\vspace{0.1in}


\end{enumerate}

\noindent {\large 2. (25 points)} Quotients

\begin{enumerate}[(a)]
   \item Given: $K=\{R_0, R_{180} \}$ is a normal subgroup of $D_6$. 
   
\vspace{0.1in}   

   The order of $\diagquotient{D_6}{K}$ is \underline{\quad $|D_6|/|K|=12/2=6$ \quad}.

\vspace{0.1in}   

   The identity of $\diagquotient{D_6}{K}$ is \underline{\quad $K$ (or equivalently $R_0K=R_{180}K=\{R_0,R_{180}\}$) \quad}.
   
\vspace{0.1in}

   $(R_{60}K)^{-1}=$ \underline{\quad $R_{60}^{-1}K = R_{-60}K=R_{300}K=R_{120}K=\{R_{120},R_{300}\}$ \quad}.

\vspace{0.1in}

  The order of $R_{60}K$ in $\diagquotient{D_6}{K}$ is \underline{\quad 3 \quad}. \hfill The size of the set $R_{60}K$ is \underline{\quad 2 \quad}.

\vspace{0.05in}

Scratch work:

$R_{60}K$ contains two elements because all left cosets have the same size (i.e. that of the subgroup $K$). 

$R_{60}K \not= K$, $(R_{60}K)^2 = R_{60}^2K=R_{120}K\not=K$, $(R_{60}K)^3=R_{60}^3K=R_{180}K=K$. Thus the order of $R_{60}K$ is 3.

\vspace{0.1in}   

   \item Let $H$ be a (normal) subgroup of $G$ where $G$ is abelian. Prove that $\diagquotient{G}{H}$ is abelian.
   
\vspace{0.1in}   

   Let $aH,bH \in \diagquotient{G}{H}$. Then $aHbH=abH=baH=bHaH$ where $ab=ba$ because $G$ is abelian.

\vspace{0.1in}   

 \item Consider $\diagquotient{\mathbb{Z}_{12}}{H}$ where $H=\langle 4 \rangle = \{0,4,8\}$. List all of the cosets (and their contents). Then make a Cayley table for this quotient group.

\vspace{0.1in}

There are $|\mathbb{Z}_{12}|/|H|= 12/3=4$ cosets.

\begin{itemize}
   \item $H = \{ 0,4,8 \}$
   \item $1+H = \{ 1,5,9 \}$
   \item $2+H = \{ 2,6,10\}$
   \item $3+H = \{ 3,7,11\}$
\end{itemize}

\vspace{0.1in} 

Example: $(2+H)+(3+H)=(2+3)+H=5+H=1+H$ (since $1$ and $5$ belong to the same coset).

\vspace{0.1in} 

\begin{tabular}{c||c|c|c|c} 
             & $    H$ & $1+H$ & $2+H$ & $3+H$ \\ \hline\hline
$    H$   & $    H$ & $1+H$ & $2+H$ & $3+H$ \\ \hline
$1+H$  & $1+H$ & $2+H$ & $3+H$ & $    H$ \\ \hline
$2+H$  & $2+H$ & $3+H$ & $    H$ & $1+H$ \\ \hline
$3+H$  & $3+H$ & $     H$ & $1+H$ & $2+H$ 
\end{tabular} 

\vspace{0.1in}

{\bf Bonus Problem:} I almost put this on the test and then got rid of it because of length.

\item Let $G$ be a group. A group $H$ is called a {\it homomorphic image} of $G$ if there exists a homomorphism $\varphi : G \rightarrow H$ which is onto (i.e. the image of $\varphi$ is $H$). Explain why homomorphic images and quotients of $G$ share the same group properties. Specifically why is it that if $G$ has an abelian quotient of order 123 if and only if $G$ has an abelian homomorphic image of order 123).

\vspace{0.1in} 

Suppose that $H$ is a homomorphic image of $G$. Then there is a homomorphism $\varphi:G \to H$ which is onto. Therefore, by the first isomorphism theorem, $\diagquotient{G}{\mathrm{ker}(\varphi)}\cong H$. Thus $H$ is (isomorphic to) a quotient of $G$. Conversely, any quotient of $G$ is the image of the corresponding projection homomorphism: $\pi : G \to \diagquotient{G}{K}$ defined by $\pi(g)=gK$ ($\pi$ is an onto homomorphism and its kernel is $K$).

Therefore if, for example, $G$ has an abelian quotient of order 123, say $G/K$ is abelian of order 123. Then $\pi:G \to G/K$ is a homomorphism from $G$ onto an abelian group of order 123. Thus $G$ has a homomorphic image which is abelian of order 123.

\vspace{0.1in} 


\end{enumerate}

\noindent {\large 3. (25 points)} Oh no! I've made a mistake. $2^2 \cdot 5 \not= 50$

\vspace{0.1in}

\noindent
In each of the following situations, explain \underline{why} we know {\bf a mistake has been made}.\\ (Why are these statements wrong?)

\begin{enumerate}[(a)]
    \item We found a homomorphism $\varphi: \mathbb{Z}_{50} \oplus \mathbb{Z}_{10} \rightarrow D_{5}$  which is onto.

\vspace{0.1in}

If there was an onto homomorphism from $\mathbb{Z}_{50} \oplus \mathbb{Z}_{10}$ to $D_5$. Then (by the first isomorphism theorem) we would have that
there is a quotient of $\mathbb{Z}_{50} \oplus \mathbb{Z}_{10}$ which is isomorphic to $D_5$. But this is impossible, quotients of abelian groups are abelian and $D_5$ is not abelian.

OR, briefly, an abelian group cannot have a non-abelian homomorphic image.

\vspace{0.1in}

    \item Let $H = Z(D_6) = \{ R_0, R_{180} \}$ (the center of $D_6$). After some shoddy computations, I've found that $\diagquotient{D_6}{H} \cong \mathbb{Z}_6$. \hfill [{\it Hint:} No computations needed to shoot this down.]

\vspace{0.1in}

The $G/Z$ theorem states that if a group quotiented by its center is cyclic, then the group must be abelian. $\mathbb{Z}_6$ is cyclic. But $D_6$ is not abelian, so 
this is impossible.

\vspace{0.1in}

   \item I just found a normal subgroup $H \triangleleft S_{4}$ such that $\diagquotient{S_4}{H}$ is an abelian group of order $18$.

\vspace{0.1in}

$|S_4|=4!=24$. We are told that $|S_4/H|=|S_4|/|H|=24/|H|=18$. There is no way to divide 24 by a postive integer and get 18. So this is impossible.
OR, briefly, there is no quotient of order 18 because 18 is not a divisor of $|S_4|=4!=24$.

\vspace{0.1in}

   \item My friend Herbert found an element $(x,y) \in D_6 \oplus \mathbb{Z}_{10}$ whose order is $60$.

\vspace{0.1in}

In a direct product of groups we have $|(x,y)|=\mathrm{lcm}(|x|,|y|)$. $x \in D_6$ so its order is $1,2,3,$ or $6$. $y \in \mathbb{Z}_{10}$ so
its order is $1,2,5,$ or $10$. The largest possible order (coming from these choices) is $\mathrm{lcm}(6,10)=30$. There are no elements of order 60 in this group.

\vspace{0.1in}

\end{enumerate}

\noindent {\large 4. (25 points)} Finite Abelian Groups

\begin{enumerate}[(a)]
   \item List all of the non-isomorphic abelian groups of order $100=2^25^2$. ({\it Note}: $4 \cdot 25 = 100$ $\leftarrow$ I can multiply!)

\vspace{0.1in}

    There are 2 partitions of 2: $2=2$ and $2=1+1$. Thus there are 2 abelian groups of order $2^2=4$ and 2 abelian groups of order $5^2=25$. Therefore,
there are a total of $2 \cdot 2 = 4$ non-isomorphic abelian groups of order $100$.

\begin{itemize}
   \item $\mathbb{Z}_{4} \oplus \mathbb{Z}_{25}$ ($\cong \mathbb{Z}_{100}$)
   \item $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_{25}$ ($\cong \mathbb{Z}_2 \oplus \mathbb{Z}_{50}$)
   \item $\mathbb{Z}_4 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5$ ($\cong \mathbb{Z}_{5} \oplus \mathbb{Z}_{20}$)
   \item $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5$ ($\cong \mathbb{Z}_{10} \oplus \mathbb{Z}_{10}$) 
\end{itemize}

   \item How many non-isomorphic abelian groups of order 449,878,000 are there?

   {\it Note:} 449,878,000 $=2^45^311^313^2$ and there are 5 non-isomorphic abelian groups of order $2^4=16$. \smiley

\vspace{0.1in}

There are 5 abelian groups of order $2^4$, 3 abelian groups of order $5^3$, 3 abelian groups of order $11^3$, and 2 abelian groups of order $13^2$.
Multiplying these (independent choices) together we get $5 \cdot 3 \cdot 3 \cdot 2 = 90$ non-isomorphic abelian groups of order $2^45^311^313^2$ [I think 
I'll skip listing these \smiley].

\vspace{0.1in}

   \item Are the groups $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12} \oplus \mathbb{Z}_{25}$ and $\mathbb{Z}_{180} \oplus \mathbb{Z}_{30}$ isomorphic? Explain your answer.

\vspace{0.1in}

Note that both groups are of the same order: $18 \cdot 12 \cdot 25 = 5400 = 180 \cdot 30$.
There are many ways to go about answering this question. I will break up both groups into cyclic groups of prime power orders and then compare.

For the first group, $18=2\cdot 3^2$, $12=2^2\cdot 3$, and $25=5^2$. Therefore,
$\mathbb{Z}_{18} \oplus \mathbb{Z}_{12} \oplus \mathbb{Z}_{25} \cong
  \mathbb{Z}_9 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_{25}$

For the second group, $180=2^2\cdot 3^2\cdot 5$ and $30=2\cdot 3\cdot 5$. Therefore,
$\mathbb{Z}_{180} \oplus \mathbb{Z}_{30} \cong
  \mathbb{Z}_4 \oplus \mathbb{Z}_9 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5$

Both groups have $\mathbb{Z}_2 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_9$ in common. However, $\mathbb{Z}_{25}$ cannot be split into $\mathbb{Z}_5 \oplus \mathbb{Z}_5$. Therefore, these groups are not isomorphic [for example, the first group has elements of order 25 and the second does not.]

\vspace{0.1in}

   \item Is the group $\mathbb{Z}_{15} \oplus \mathbb{Z}_{7} \oplus \mathbb{Z}_{121}$ cyclic? (Note: $121=11^2$) Explain you answer.

\vspace{0.1in}

Yes. Notice that $15=3\cdot 5$, $7$, and $121=11^2$ are relatively prime (they don't share any prime factors). Therefore, $\mathbb{Z}_{15} \oplus \mathbb{Z}_{7} \oplus \mathbb{Z}_{121} \cong \mathbb{Z}_{3\cdot 5\cdot 7\cdot 11^2}=\mathbb{Z}_{12705}$ (which is cyclic).

\vspace{0.1in}
\end{enumerate}

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