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\noindent
\parbox{2in}{\bf Math 3110} 
\hfill {\Large \bf Final Exam} \hfill
\parbox{2in}{\bf \hfill December $7^{\mathrm{th}}$, 2015}

\vspace{0.1in}

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

\vspace{0.1in}

\noindent
{\large 1. (10 points)} Working in $\mathbb{Z}_{18}$.

\begin{enumerate}[(a)]

\item $I = (6)=\langle 6 \rangle = \{ 0,6, 12 \}$ and $\diagquotient{\mathbb{Z}_{18}}{I} = \{I,1+I,2+I,3+I,4+I,5+I \}$.

\item For each element in $\diagquotient{\mathbb{Z}_{18}}{I}$, state whether that element is zero, a zero divisor, a unit, or none of the above. If it is a unit, give its inverse. If it is an zero divisor, show that this is the case.

\vspace{0.05in}

$I$ is the zero of $\diagquotient{\mathbb{Z}_{18}}{I}$. $1+I$ is a unit (it is {\it the} unit) and is its own inverse. $2+I$ is a zero divisor: $(2+I)(3+I)=2\cdot 3+I=6+I=I$ (since $6 \in I$). By the last calculation, $3+I$ is also a zero divisor. $4+I$ is a zero divisor: $(4+I)(3+I)=4 \cdot 3 +I=12+I=I$ (since $12 \in I$). $5+I$ is a unit: $(5+I)(5+I)=5\cdot 5+I = 25+I = 7+I =(1+6)+I=1+I$. So $5+I$ is also its own inverse.

\vspace{0.05in}

\item It turns out that there is a ring homomorphism $\varphi: \mathbb{Z}_{18} \to \mathbb{Z}_6$ which is onto and has $I=(6)=\mathrm{Ker}(\varphi)$. Is $\varphi$ one-to-one? Explain why or why not. What does the first isomorphism theorem say in this case?

\vspace{0.05in}

No. $\varphi$ isn't one-to-one since $\mathrm{Ker}(\varphi)=\{0,6,12\} \not= \{0\}$. In fact, it must be a 3-to-1 map. The first isormorphism theorem says that $\diagquotient{\mathbb{Z}_{18}}{\mathrm{Ker}(\varphi)} \cong \mathrm{Im}(\varphi)$ so that $\diagquotient{\mathbb{Z}_{18}}{I} \cong \mathbb{Z}_6$.

{\it Note:} I claim that there is such a homomorphism. Specifically, $\varphi(x)=x$ works. Notice that if $x=y$ (mod 18), then there is some $k \in \mathbb{Z}$ such that $x=y+18k$ so that $x=y+6(3k)$. Thus $x=y$ (mod 6). This means that $\varphi$ is well-defined. Also, $\varphi(x+y)=x+y=\varphi(x)+\varphi(y)$ so $\varphi$ is a homomorphism. Obviously $\varphi$ is onto. Finally, $\mathrm{Ker}(\varphi) = \{ x \in \mathbb{Z}_{18} \;|\; \varphi(x)=0 \} = \{ x \in \mathbb{Z}_{18} \;|\; x=0 \mbox{ (mod 6)} \}$. This is exactly the set of multiples of 6 (i.e. $\mathbb{Z}_{18} = (6) = \{ 0,6,12\}$).

\vspace{0.05in}

\item Is $I=(6)$ a prime or maximal ideal in $\mathbb{Z}_{18}$? Why or why not?

\vspace{0.05in}

No. Notice that $\diagquotient{\mathbb{Z}_{18}}{I} \cong \mathbb{Z}_6$ is neither a field nor an integral domain (it has zero divisors). Thus $I$ is neither a maximal nor a prime ideal.

\vspace{0.05in}

\end{enumerate}

\noindent
 {\large 2. (12 points)} Groups: Isomorphic or not.
   
   \begin{enumerate}[(a)]

    \item Explain why $A_4 \not\cong \mathbb{Z}_2 \oplus \mathbb{Z}_6$ [not isomorphic].

\vspace{0.05in}

Both are groups of order 12: $|A_4| = \dfrac{4!}{2} = 12$ and $|\mathbb{Z}_2 \oplus \mathbb{Z}_6| = 2 \cdot 6 =12$. However, $A_4$ is not abelian while $\mathbb{Z}_2 \oplus \mathbb{Z}_6$ is abelian. Therefore, they cannot be isomorphic. [Of course, there are other reasons these groups aren't isomorphic, for example, $A_4$ has elements of orders 1, 2, and 3 whereas $\mathbb{Z}_2 \oplus \mathbb{Z}_6$ has elements of orders 1, 2, 3, and 6.]


\vspace{0.05in}
   
     \item Explain why $Q = \{ \pm 1, \pm i, \pm j, \pm k \} \not\cong D_4$ [not isomorphic].

\vspace{0.05in}

Both are non-abelian groups of order 8 (no help here). Even more, they both have elements of orders 1, 2, and 4. Even their centers are isomorphic (still no help)! However, $Q$ has 6 elements of order 4 (i.e. $\pm i, \pm j, \pm k$) while $D_4$ has only 2 elements of order 4 (i.e. the $90^\circ$ and $270^\circ$ rotations), so they cannot be isomorphic. Alternatively, we could notice that $Q$ only has 1 element of order 2 (i.e. $-1$) while $D_4$ has 5 elements of order 2 (i.e. the $180^\circ$ rotation and 4 reflections).

\vspace{0.05in}

   \item Explain why $U(10) \cong \mathbb{Z}_4$ [are isomorphic].
   
\vspace{0.05in}

In $U(10)$: $3^1=3$, $3^2=9$, $3^3=27=7$, and $3^4=1$. Therefore, $U(10)=\{1,3,7,9\}=\langle 3 \rangle$. Thus $U(10)$ is a cyclic group of order 4. Since any two cyclic groups of the same order are isomorphic, $U(10) \cong \mathbb{Z}_4$.

\vspace{0.05in}
   
   \end{enumerate}

\newpage

\noindent
 {\large 3. (8 points)} Rings: Explain why each pair of {\bf rings} are not isomorphic.
   
   \begin{enumerate}[(a)]
   
     \item $\mathbb{Z} \not\cong \mathbb{E}$ \qquad where \quad $\mathbb{E} = 2\mathbb{Z} = \{ n \in \mathbb{Z} \;|\; n \mbox{ is even}\}$.

\vspace{0.05in}

These are both (countably) infinite commutative rings of characteristic 0 and neither have any zero divisors (no help here). However, $\mathbb{Z}$ has an identity: $1$ whereas $\mathbb{E}$ has no identity: $xy=x$ for all $x \in \mathbb{E}$ implies $y=1$ but $1 \not\in \mathbb{E}$. Therefore, these rings cannot be isomorphic.

\vspace{0.05in}

     \item $\mathbb{R}^{3 \times 3} \not\cong \mathbb{C}$ \qquad where \quad $\mathbb{R}^{3 \times 3}$ is the ring of $3 \times 3$ real matrices \quad and \quad $\mathbb{C}$ is the complex numbers.

\vspace{0.05in}

Both are infinite rings with unity of characteristic zero and have infinite groups of units (no help here). However, $\mathbb{R}^{3 \times 3}$ is not commutative while $\mathbb{C}$ is commutative. Therefore, they cannot be isomorphic. Alternatively, $\mathbb{C}$ is a field and an integral domain (has no zero divisors). On the other hand, $\mathbb{R}^{3 \times 3}$ is not a field and is not an integral domain and does have many many zero divisors.

\vspace{0.05in}
   
   \end{enumerate}

\noindent
{\large 4. (9 points)} Workin' in $\mathbb{Z}_{88}$. \qquad [Note: $88 = 2^3 \cdot 11$]

\begin{enumerate}[(a)]

\item Fill out the following table (for $\mathbb{Z}_{88}$):
         
         \begin{tabular}{|r|c|c|c|c|c|c|c|c|} \hline
         order = & 1 & 2 & 4 & 8 & 11 & 22 & 44 & 88 \Large\strut \\ \hline
         \parbox[c][0.5in]{1.25in}{number of elements with this order = } & 1 & 1 & 2 & 4 & 10 & 10 & 20 & 40 \Large\strut
         \\ \hline 
         \end{tabular}

\item Draw $\mathbb{Z}_{88}$'s lattice of ideals.

\vspace{-1.75in}

\force \hfill
\begin{tikzpicture}[node distance=1.5cm]
\node(88){$88$};
\node(44)[below left of = 88]{$44$};
\node(8)[below right of = 88]{$8$};
\node(22)[below of = 44]{$22$};
\node(4)[below of = 8]{$4$};
\node(11)[below of = 22]{$11$};
\node(2)[below of = 4]{$2$};
\node(1)[below left of = 2]{$1$};

\draw(88)--(44);
\draw(44)--(22);
\draw(22)--(11);
\draw(11)--(1);
\draw(88)--(8);
\draw(8)--(4);
\draw(4)--(2);
\draw(2)--(1);
\draw(44)--(4);
\draw(22)--(2);
\end{tikzpicture}
\qquad \qquad
\begin{tikzpicture}[node distance=1.5cm]
\node(88){$\mathbb{Z}_{88}=(1)$};
\node(44)[below left of = 88]{$(2)$};
\node(8)[below right of = 88]{$(11)$};
\node(22)[below of = 44]{$(4)$};
\node(4)[below of = 8]{$(22)$};
\node(11)[below of = 22]{$(8)$};
\node(2)[below of = 4]{$(44)$};
\node(1)[below left of = 2]{$(0)$};

\draw(88)--(44);
\draw(44)--(22);
\draw(22)--(11);
\draw(11)--(1);
\draw(88)--(8);
\draw(8)--(4);
\draw(4)--(2);
\draw(2)--(1);
\draw(44)--(4);
\draw(22)--(2);
\end{tikzpicture}

\force \hfill Divisibility Lattice \hspace{0.65in} Ideal Lattice \hspace{0.05in} \force

\vspace{-0.75in}

Remember that in $\mathbb{Z}$ and $\mathbb{Z}_n$ we have:\\ \force \quad subgroup = cyclic subgroup = subring = ideal = principal ideal.

So finding the ideal lattice is the same as finding the subgroup lattice.

\vspace{0.05in}


\item Which ideals are prime? maximal?

\vspace{0.05in}

$\mathbb{Z}_{88}$ is a {\bf finite} commutative ring with $1$. Thus a quotient is a finite commutative ring with $1$. This means that any quotient is an integral domain if and only if it is a field (since finite integral domains are fields and fields are always integral domains). Therefore, a quotient of $\mathbb{Z}_{88}$ is a field iff it is an integral domain. Also, a quotient is a field iff the ideal we are quotienting by is a maximal ideal. Likewise, integral domain corresponds to prime. Since quotients are integrals domains iff they are fields, we have that prime and maximal are the same thing for $\mathbb{Z}_{88}$ (this is true for any finite commutative ring with 1). 

Maximal ideals are proper ideals not contained in any other ideal. Looking at our lattice we can see that these are exactly: \fbox{$(2)$ and $(11)$} (these ideals are both prime and maximal). 

\vspace{0.05in}

\end{enumerate}

\noindent
{\large 5. (6 points)} Workin' in $\mathbb{Z}_{323}$. 

\begin{enumerate}[(a)]

\item Is $221$ zero, a unit, a zero divisor, or none of the above in $\mathbb{Z}_{323}$? If $221$ is a zero divisor, prove it.
          If $221$ is a unit, find it's inverse.

\vspace{0.05in}

We need to run the Euclidean algorithm. Dividing we get: $323=(1)221+102$. Next, $221=(2)102+17$. Finally, $102=(6)17+0$. The last non-zero remainder was 17. Therefore, $\mathrm{gcd}(221,323)=17 \not= 1$, so $221$ isn't a unit. We know that in $\mathbb{Z}_n$, non-zero non-units are zero divisors. Therefore, $17$ must be a \fbox{zero divisor}. To see this note that $323/17 = 19$ and $221/17=13$. So $221 \cdot 19 = 13 \cdot 17 \cdot 19 = 13 \cdot 323 = 0$ (mod $323$). Thus $221$ is a zero divisor.

\vspace{0.05in}
   
\item Is $20$ zero, a unit, a zero divisor, or none of the above in $\mathbb{Z}_{323}$? If $20$ is a zero divisor, prove it.
          If $20$ is a unit, find it's inverse.
          
\vspace{0.05in}

Same as in part (a), $323=(16)20+3$ then $20=(6)3+2$ then $3=(1)2+1$ and finally $2=(2)1+0$. Thus $20$ and $323$ are relatively prime. This means that $20 \in U(323)$ (i.e. it's a unit). 

We need to run the algorithm backwards to find the inverse of $20$: $(-1)2+(1)3=1$. Then using $2=(-6)3+(1)20$ we get $1=(-1)[(-6)3+(1)20]+(1)3 = (7)3+(-1)20$. Finally, using $3=(-16)20+(1)323$ we get $1 = (7)[(-16)20+(1)323]+(-1)20 = (-113)20+(7)323$. Therefore, \fbox{$20^{-1}=-113=210$} (in $\mathbb{Z}_{323}$). Double check: $20 \cdot 210 = 4200 = 1+13 \cdot 323 =1$ (mod 323). 

\vspace{0.05in}

\end{enumerate}

\newpage
\noindent
{\large 6. (18 points)} Sub-things

\begin{enumerate}[(a)]

\item Let $H = \{ 1, y \}$. Explain why $H$ is a subgroup of $D_5 =  \{1,x,x^2,x^3,x^4,y,xy,x^2y,x^3y,x^4y\}$ (of course $x^5=1$, $y^2=1$, and $xyxy=1$), then show $H$ is {\bf not} a {\bf normal} subgroup of $D_5$.

\vspace{0.05in}

Notice that $y^2=1$ and $1y=y1=y$ and $1^2=1$. Thus $H$ is closed under the operation. Therefore, by the finite subgroup test, $H$ is a subgroup. Alternatively, we could just notice that $H=\langle y \rangle$ (a cyclic subgroup and so a subgroup). 

Why isn't $H$ normal? Let's compute a left and right coset pair: $xH = \{x1,xy\}=\{x,xy\}$ and $Hx=\{1x,yx\}=\{x,x^4y\}$ (since $yx=x^{-1}y=x^4y$ in $D_5$). Therefore, $xH \not= Hx$. Therefore, the left and right cosets don't match and so $H$ is not a normal subgroup. Alternatively, notice that $xyx^{-1} = xxy=x^2y \not\in H$. Thus $H$ isn't invariant under conjugation. Again, not normal.

\vspace{0.05in}

\item Let $\displaystyle S = 2\mathbb{Z} \oplus 3\mathbb{Z} = \left\{ (2x,3y) \;|\; x,y \in \mathbb{Z} \right\}$. Show $S$ is a {\bf subring} of $\mathbb{R} \oplus \mathbb{Q}$.

\vspace{0.05in}

This is a non-empty set whose elements are ordered pairs of integers. Thus it is a non-empty subset of $\mathbb{R} \oplus \mathbb{Q}$. 

Suppose $(2k,3\ell), (2x,3y) \in S$ (where $k,\ell,x,y \in \mathbb{Z}$). Then $(2k,3\ell)-(2x,3y) = (2k-2x,3\ell-3y)=(2(k-x),3(\ell-y)) \in S$ since $k-x,\ell-y \in \mathbb{Z}$. Therefore, $S$ is closed under subtraction. Also, $(2k,3\ell)(2x,3y) = (4kx,9\ell y) = (2(2kx),3(3\ell y)) \in S$ since $2kx, 3\ell y \in \mathbb{Z}$. Therefore, $S$ is closed under multiplication. This means $S$ is a subring of $\mathbb{R} \oplus \mathbb{Q}$.  

\vspace{0.05in}

\item Let $\mathbb{Q}[i] = \{ a+bi \;|\; a,b \in \mathbb{Q} \}$ where $i=\sqrt{-1}$. Show that $\mathbb{Q}[i]$ is a sub\underline{\bf field} of the complex numbers $\mathbb{C}$.

\vspace{0.05in}

Clearly $\mathbb{Q}[i]$ is a non-empty subset of $\mathbb{C}$. Let $a+bi,c+di \in \mathbb{Q}[i]$ (where $a,b,c,d \in \mathbb{Q}$). Then $(a+bi)-(c+di) = (a-c)+(b-d)i \in \mathbb{Q}[i]$ since $a-c,b-d \in \mathbb{Q}$. Likewise, $(a+bi)(c+di) = ac+adi+bci+bdi^2 = (ac-bd)+(ad+bc)i \in \mathbb{Q}[i]$ since $ac-bd,ad+bc \in \mathbb{Q}[i]$. Thus $\mathbb{Q}[i]$ is a subring of $\mathbb{C}$.

Now we have $1 =1+0i \in \mathbb{Q}[i]$, so it is $\mathbb{Q}[i]$ is a commutative ring with $1 \not=0$ (all subrings of $\mathbb{C}$ are automatically commutative). To see that $\mathbb{Q}[i]$ is a field, we need to check to see if all non-zero elements are units.

Let $a+bi \in \mathbb{Q}[i]$ (where $a,b \in \mathbb{Q}$) such that $a+bi \not=0$. Then $(a+bi)^{-1}$ exists (all non-zero complex numbers have inverses because $\mathbb{C}$ is a field). Notice that $\dfrac{1}{a+bi}  = \dfrac{1}{a+bi} \cdot \dfrac{a-bi}{a-bi} = \dfrac{a-bi}{(a+bi)(a-bi)} = \dfrac{a-bi}{a^2+b^2} = \dfrac{a}{a^2+b^2}+\dfrac{-b}{a^2+b^2}i \in \mathbb{Q}[i]$ since $\dfrac{a}{a^2+b^2},\dfrac{-b}{a^2+b^2} \in \mathbb{Q}$.
Therefore every non-zero element of $\mathbb{Q}[i]$ has an inverse in $\mathbb{Q}[i]$. This means that $\mathbb{Q}[i]$ is a field.

\vspace{0.05in}

\item Can a subring of an integral domain fail to be an integral domain? What about a quotient of an integral domain?
          
\vspace{0.05in}

First, yes, a subring of an integral domain can fail to be an integral domain. Let $R$ be an integral domain. Then $\{0\}$ is a subring. But $\{0\}$ is not an integral domain (because $0 \not= 1$). Another example, from problem \#3(a). $\mathbb{E}$ is not an integral domain (it has no identity) but it is a subring of an integral domain $\mathbb{Z}$. 

However, if $R$ is an integral domain and $S$ is a subring with $1$, then $S$ will also be an integral domain (since $1 \not=0$ in $R$ this is true in $S$, $S$ is automatically commutative, and $S$ has no zero divisors -- otherwise $R$ would have zero divisors). 

How about quotients? Quotients of integral domains don't have to be integral domains either. A silly example: If $R$ is an integral domain, then $R/R \cong \{0\}$ is not an integral domain (since $1=0$). A more interesting example might be: $\mathbb{Z}/(4) \cong \mathbb{Z}_4$ is not an integral domain (since $2$ is a zero divisor in $\mathbb{Z}_4$). Of course, some quotients can be: If $R$ is an integral domain, then $R/\{0\} \cong R$ is an integral domain. Also, $\mathbb{Z}/(3) \cong \mathbb{Z}_3$ is an integral domain (in fact it's a field).

\vspace{0.05in}

\end{enumerate}

\noindent
{\large 7. (15 points)} An ideal question.

\begin{enumerate}[(a)]
   \item Let $\varphi:R \to S$ be a ring homomorphism. Show that $\mathrm{Ker}(\varphi)$ is an ideal of $R$.

\vspace{0.05in}

Recall that $\mathrm{Ker}(\varphi) = \{ x \in R \;|\; \varphi(x)=0 \}$. Notice that this is a subset of $R$. Also, $\varphi(0)=0$ so $0 \in \mathrm{Ker}(\varphi)$ so the kernel is always non-empty.

Let $a,b \in \mathrm{Ker}(\varphi)$. Then $\varphi(a)=\varphi(b)=0$ (by definition). Therefore, $\varphi(a-b)=\varphi(a)-\varphi(b)=0-0=0$ so that $a-b \in \mathrm{Ker}(\varphi)$ (the kernel is closed under subtraction).

Let $a \in \mathrm{Ker}(\varphi)$ and $r \in R$. Then $\varphi(a)=0$ (by definition). Notice that $\varphi(ra)=\varphi(r)\varphi(a)=\varphi(r)0=0$ and $\varphi(ar)=\varphi(a)\varphi(r)=0\varphi(r)=0$. Therefore, $ra,ar \in \mathrm{Ker}(\varphi)$ (the kernel absorbs multiplication on the left and right).

The kernel thus passes the ideal test, so $\mathrm{Ker}(\varphi) \triangleleft R$.


\vspace{0.05in}
             
   \item Let $R$ be a commutative ring with $1$ and $I$ an ideal of $R$. Show that $I=R$ if and only if $I$ contains a unit of $R$.
   
\vspace{0.05in}

Let $R$ be a commutative ring with $1$ and suppose $I \triangleleft R$.

Suppose $I=R$. Then $1 \in R=I$ so $I$ contains a unit (in fact, it contains {\it the} unit).

Conversely, suppose that $I$ contains a unit, say $u \in I$ (where $u \in U(R)$). Then $u^{-1}$ exists in $R$. By the absorption property we have $1=u^{-1}u \in I$ (since $u^{-1}\in R$ and $u \in I$). Let $r \in R$. Again by the absorption property,  $r=r1 \in I$ (since $r \in R$ and $1 \in I$). Thus $R \subseteq I$. However, $I \subseteq R$ (because $I$ is an ideal of $R$). Thus $I=R$.

\vspace{0.05in}
   
   \item Let $\varphi:R \to S$ be a ring homomorphism where $\mathrm{Ker}(\varphi)=\{0\}$. Show that $\varphi$ is one-to-one.
   
\vspace{0.05in}

Suppose that $\varphi$ is a ring homomorphism with trivial kernel: $\mathrm{Ker}(\varphi) = \{0\}$. Suppose $\varphi(x)=\varphi(y)$ for some $x,y \in R$. Then $\varphi(x-y)=\varphi(x)-\varphi(y)=0$ since $\varphi(x)=\varphi(y)$. This means that $x-y \in \mathrm{Ker}(\varphi)$ (since $x-y$ maps to $0$). But $\mathrm{Ker}(\varphi)=\{0\}$ so $x-y=0$ and thus $x=y$. This means $\varphi$ is one-to-one.

\vspace{0.05in}
   
\end{enumerate}

\noindent
{\large 8. (8 points)} The Fundamental Theorem of Finite Abelian Groups.

\begin{enumerate}[(a)]

   \item List all of the non-isomorphic abelian groups of order $100 = 2^2 \cdot 5^2$. Circle any that are cyclic.
   
\vspace{0.05in}

There are $p(2)p(2)=2\cdot 2=4$ non-isomorphic abelian groups of order 100. Specifically, they are: 
\begin{itemize}
\item \fbox{$\mathbb{Z}_{4} \oplus \mathbb{Z}_{25} \cong \mathbb{Z}_{100}$} \qquad [These are cyclic.]
\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 Which of the abelian groups of order 100 contain elements of order 25?

\vspace{0.05in}

The first two have elements of order 25: $\mathbb{Z}_{4} \oplus \mathbb{Z}_{25} \cong \mathbb{Z}_{100}$ and $\mathbb{Z}_2 \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{25} \cong \mathbb{Z}_{2} \oplus \mathbb{Z}_{50}$. Notice that $4$ has order $100/4=25$ in $\mathbb{Z}_{100}$ and $(0,2)$ has order $\mathrm{lcm}(|0| \mbox{ in $\mathbb{Z}_2$}, |2| \mbox{ in $\mathbb{Z}_{50}$}) = \mathrm{lcm}(1,25) = 25$.

On the other hand $\mathbb{Z}_4 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5 \cong \mathbb{Z}_5 \oplus \mathbb{Z}_{20}$ and $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_{5} \cong \mathbb{Z}_{10} \oplus \mathbb{Z}_{10}$ do not have elements of order 25. This is more or less because we can't get 25 from taking an least common multiple of divisors of the various orders involved.

\vspace{0.05in}

\end{enumerate}

\noindent
{\large 9. (7 points)} Let $H = \{ (1), (12)(34), (13)(24), (14)(23) \}$. It can be shown that $H \triangleleft A_4$. Write down a Cayley table for $\diagquotient{A_4}{H}$. Is this quotient a cyclic group?

\vspace{0.05in}

$\left|\diagquotient{A_4}{H}\right| = \diagquotient{|A_4|}{|H|} = \dfrac{12}{4} =3$, so we should have 3 cosets in the quotient. 

\vspace{0.05in}

Notice $(123)H= \{ (123)(1), (123)(12)(34), (123)(13)(24), (123)(14)(23) \} = \{ (123), (134), (243), (142) \}$. The only coset left is $(132)H$ which must contain all of the leftovers, $(132)H = \{ (132), (143), (234), (124)\}$.

\force\hfill
\begin{tabular}{c||c|c|c|} 
& $H$ & $(123)H$ & $(132)H$ \Large\strut \\ \hline \hline
$H$ & $H$ & $(123)H$ & $(132)H$ \Large\strut \\ \hline
$(123)H$ & $(123)H$ & $(132)H$ & $H$ \Large\strut \\ \hline
$(132)H$ & $(132)H$ & $H$ & $(123)H$ \Large\strut \\ \hline
\end{tabular}
\hfill \force


\vspace{0.05in}

\noindent
{\large 10. (7 points)} Recall that if $G$ is a group and $g \in G$, then $\varphi_g(x)=gxg^{-1}$ is called an inner automorphism. Prove that $\varphi_g$ is an automorphism.

\vspace{0.05in}

\begin{itemize}
\item For all $x,y \in G$ we have $\varphi_g(xy) = g(xy)g^{-1} = gxg^{-1}gyg^{-1} = \varphi_g(x)\varphi_g(y)$. So $\varphi_g$ is operation preserving.
\item Suppose $\varphi_g(x)=e$ for some $x \in G$ (let $e$ be the identity of $G$). Then $gxg^{-1}=e$ so $g^{-1}gxg^{-1}g=g^{-1}eg$ so $x=e$. Thus $\mathrm{Ker}(\varphi_g)=\{e\}$ so $\varphi_g$ is one-to-one. Alternatively, suppose $\varphi_g(x)=\varphi_g(y)$ so $gxg^{-1}=gyg^{-1}$. Thus $g^{-1}gxg^{-1}g = g^{-1}gyg^{-1}g$ and so $x=y$ (again one-to-one).
\item Finally, suppose $y \in G$. Then $\varphi_g(g^{-1}yg) = g(g^{-1}yg)g^{-1}=y$. Thus $\varphi_g$ is onto.
\end{itemize}

Alternatively, we could should $\varphi_g$ is one-to-one and onto by exhibiting an inverse, namely $\varphi_{g^{-1}}$. Notice that $\varphi_{g^{-1}}(\varphi_g(x)) = \varphi_{g^{-1}}(gxg^{-1})=g^{-1}gxg^{-1}(g^{-1})^{-1} = x$. Likewise, $\varphi_g(\varphi_{g^{-1}}(x))=x$. Therefore, $(\varphi_g)^{-1}=\varphi_{g^{-1}}$ (an inverse exists).

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