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\begin{frame}
\begin{center}
\large Illinois State University Undergraduate Colloquium in Mathematics -- March 23, 2022

%\vspace{0.15in}


\LARGE \textcolor{highlight}{Fuchs' Problem for Dicyclic Groups}

%\force\hfill
%\includegraphics[width=3in]{2017-09-08-OCSA_report_frames-figure00.jpg}
%\hfill\force

%\vspace{0.5in}

\normalsize

\begin{tabular}{rcl}
Bill Cook & $\Longleftrightarrow$ &
\texttt{cookwj@appstate.edu}
\end{tabular}\\
\textcolor{appstate}{Appalachian State University}\\
Boone, North Carolina

\vspace{0.25in}       

Joint work with \hu{Joshua Carr} and \hu{Lindsey Wise}.

\vspace{0.25in}
       
\textcolor{highlight}{\large Thursday, March $23^\mathrm{rd}$\!, 2021}
\end{center}

\small
Every ring with identity has an associated group of units. Fuchs' problem seeks to determine which group structures can appear as groups of units. In this talk, we will discuss Fuchs' problem in general and then focus on groups of small orders and Dicyclic groups. 

\end{frame}

\begin{frame}
\frametitle{Background:}

A \hu{ring} $R$ is an Abelian group under addition equipped with an associative multiplication and obeys distributive laws.\\
\fbox{We assume that all of our rings are \hlr{rings with $1$}.}

\begin{itemize}\setlength\itemsep{0mm}
\item \hu{Closure:} If $a,b \in R$, then $a+b,ab \in R$
\item \hu{Associativity:} If $a,b,c \in R$, then $(a+b)+c=a+(b+c)$ and $a(bc)=(ab)c$.
\item \hu{Identity:} There are $0,1 \in R$ such that if $a \in R$, then $a+0=a=0+a$ and $1a=a=a1$.
\item \hu{(Additive) Inverses:} For all $a \in R$ there is some $-a \in R$ such that $a+(-a)=0=(-a)+a$.
\item \hu{Commutativity (Addition):} If $a,b \in R$, then $a+b=b+a$.
\item \hu{Distributivity:} If $a,b,c \in R$, then $a(b+c)=ab+ac$ and $(a+b)c=ac+bc$.
\end{itemize}

\end{frame}

\begin{frame}
\frametitle{Background:}


Let $R$ be a ring. We say $u \in R$ is a \hu{unit} if it has a multiplicative inverse: there exists some $b \in R$ such that \fbox{\hlr{$uu^{-1}=1=u^{-1}u$}}. 


\force \hfill
Let \fbox{\hlr{$R^\times$}} $= \{  u \in R \;|\; u \text{ is a unit in }R \}$ 
be the \hu{group of units}. 
\hfill \force

This is a group under \hlr{multiplication} since\dots 

\begin{itemize}\setlength\itemsep{0mm}
\item \hu{Closure} -- {\small A unit times a unit is a unit:} $(ab)^{-1}=b^{-1}a^{-1}$.
\item \hu{Associativity} -- {\small Multiplication is associative:} $a(bc)=(ab)c$.
\item \hu{Identity} -- {\small The multiplicative identity, $1$, is a unit:} $1^{-1}=1$.
\item \hu{Inverses} -- {\small The inverse of a unit is a unit:} $(u^{-1})^{-1}=u$.
\end{itemize}

\pause

\hu{Examples:} \hlr{$\mathbb{Z}^\times = \{\pm 1\}$}, $\mathbb{R}[x]^\times = \mathbb{R}^\times = \mathbb{R}_{\not= 0}$, $(\mathbb{R}^{n\times n})^\times = \mathrm{GL}_n(\mathbb{R})$,\\ \pause \force\qquad\qquad\  \hlr{$\mathbb{Z}_n^\times = U(\mathbb{Z}_n)=U(n) = \{ k \in \mathbb{Z}_n \;|\; \mathrm{gcd}(k,n)=1\}$}


\vspace{0.1in}

\pause$U(10)=\{1,3,7,9\}$\\ \force \qquad \qquad where $1^{-1}=1$, $3^{-1}=7$, $7^{-1}=3$, and $9^{-1}=9$.


\end{frame}


\begin{frame}
\frametitle{Fuchs' Problem:}

Which group structures can appear as unit groups? \hl{(Fuchs $\approx$1960)}

\vspace{0.1in}

\force\hfill
\fbox{\hlr{Given a group $G$, is there a ring $R$ such that $R^\times=G$?}}
\hfill \force

\vspace{0.1in}

We say the $G$ is \hu{realizable} as a unit group if there is some ring $R$ such that $R^\times=G$.

\vspace{0.2in}

\pause

\hu{Careful!} This is not asking if $G$ can appear among the units.\\
For example, let $R$ be any commutative ring. We can form the group ring, $R[G]$, and we have that $G \subseteq R[G]^\times$.

\vspace{1.25in}

\force

\end{frame}

\begin{frame}
\frametitle{Fuchs' Problem:}

Which group structures can appear as unit groups?

\vspace{0.1in}

\force\hfill
\fbox{\hlr{Given a group $G$, is there a ring $R$ such that $R^\times=G$?}}
\hfill \force

\vspace{0.1in}

We say the $G$ is \hu{realizable} as a unit group if there is some ring $R$ such that $R^\times=G$.

\vspace{0.1in}

\hu{Examples:} Notice \hlr{$\{0\}^\times=\{0\}$} and \hlr{$\mathbb{Z}_2^\times = \{1\}$} both realize the trivial group. Also, \hlr{$\mathbb{Z}_3^\times = \{1,2\} = \langle 2 \rangle \cong C_2$} realizes the cyclic group of order 2. Likewise, \hlr{$\mathbb{Z}_5^\times =\{1,2,3,4\} = \langle 2 \rangle \cong C_4$}, \hlr{$\mathbb{Z}_7^\times \cong C_6$}, and \hlr{$\mathbb{Z}_{8}^\times = \{1,3,5,7\} \cong C_2 \times C_2$}.

\vspace{0.1in}

\pause

\hu{Finite Fields:} There is a unique (up to isomorphism) field of order $q=p^k$ where $p$ is prime and $k$ is a positive integer. Call such a field $\mathbb{F}_q$.\\ Any finite subgroup of the multiplicative group of a field is cyclic, so \hlr{$\mathbb{F}_q^\times = \mathbb{F}_q-\{0\}\cong C_{q-1}$}. For example, \hlr{$\mathbb{F}_4^\times \cong C_3$}, \hlr{$\mathbb{F}_8^\times \cong C_7$}, and \hlr{$\mathbb{F}_9^\times \cong C_8$}.


\end{frame}


\begin{frame}
\frametitle{Characteristic of a Ring:}

For convenience, we let \hl{$\mathbb{Z}_0=\mathbb{Z}$}.

\vspace{0.1in}

For any ring $R$, there exists some $n \in \mathbb{Z}_{\geq 0}$ such that $R$ contains an isomorphic \hl{copy of $\mathbb{Z}_n$} called the \hu{prime subring} of $R$. 

\vspace{0.1in}

\force \hfill
\fbox{\hlr{$\mathrm{char}(R)=n$}} is the \hu{characteristic} of $R$
\hfill \force

\vspace{0.1in}

\hu{Examples:} $\mathrm{char}(\mathbb{Z})=0$ and $\mathrm{char}(\mathbb{Z}_n)=n$

\vspace{0.1in}

\pause

Important Observation: Since $1$ commutes with all $R$, so do additive powers of $1$. Thus\\ \force \hfill \hu{the prime subring of $R$ lies in the center of $R$}. \hfill \force

\vspace{0.1in}

\pause

\force \hspace*{0.5in} This implies that \fbox{\hlr{$Z(R^{\times})$ must contain a copy of $\mathbb{Z}_n^\times$}}.

\vspace{0.1in}

\pause

\hu{Example:} If $Z(R^\times)=\{1\}$, then either $R=\{0\}$ or $\mathrm{char}(R)=2$

\force \hfill since $|\mathbb{Z}_n^\times|>1$ for $n>2$.

\end{frame}

\begin{frame}
\frametitle{We Cannot Realize Cyclic Order 5:}

\hu{\large\bf Theorem:} Non-trivial groups of odd order can only be realized in characteristic 2.

{\bf Proof \#1:} $\mathbb{Z}_n^\times$ only has odd order for $n=1$ ($R=\{0\}$) and $n=2$.\\ \pause
{\bf Proof \#2:} If $u \not= -u$, then units come in pairs. 

\pause

\vspace{0.2in}

\hu{Example:} We cannot realize $C_5$. 

\vspace{0.1in}

Suppose $R$ realizes $C_5$ and say $R^\times = \langle u \rangle \cong C_5$. Since $|R^\times|=5$ is odd, the characteristic of $R$ must be $2$ so that $x=-x$ for all $x \in R$. 
\pause
Notice that $(1+u^2+u^3)^3 = u^9 + 3 u^8 + 3 u^7 + 4 u^6 + 6 u^5 + 3 u^4 + 3 u^3 + 3 u^2 + 1$ \pause $= u^9+u^8+u^7+u^4+u^3+u^2+1$ \pause $=2u^4+2u^3+2u^2+1=1$. \pause
Therefore, if $a=1+u^2+u^3$, then $a^3=1$ so that $a^{-1}=a^2$. \pause This means $a$ is a unit of order 1 or 3. \pause But $R^\times$ has order 5. Therefore, $a$'s order is 1. \pause Thus $a=1$ so $u^2+u^3=0$ \pause so $u^2=u^3$ so $1=u$ (contradiction). 

\end{frame}


\begin{frame}
\frametitle{Group Rings:}

Let $R$ be a commutative ring (with $1 \not=0$). 
$$R[G] = \{ r_1g_1+\cdots+r_\ell g_\ell \;|\; r_1,\dots,r_\ell \in R \text{ and } g_1,\dots,g_\ell \in G\}$$
Add coordinatewise and extend $G$'s multiplication ``linearly''. 

Then $R[G]$ is the \hu{group ring} of $G$ with coefficients in $R$.

\vspace{0.1in}

\hu{Example:} Let $D_3 = \langle a,x \;|\; a^3=1,x^2=1,xa=a^{-1}x \rangle$ $= \{1,a,a^2,x,ax,a^2x\}$ be the Dihedral group of order 6. The group ring of $D_3$ with coefficients in $\mathbb{Z}_4$ is a ring with $4^6=4096$ elements.
$$\mathbb{Z}_4[D_3] = \{ r_1+r_2a+r_3a^2+r_4x+r_5ax+r_6a^2x \;|\; r_1,\dots,r_6 \in \mathbb{Z}_4 \}$$
\small
$$(2+a+3x)+(3+2a^2+x)\pause = (2+3)+a+2a^2+(3+1)x\pause = 1+a+2a^2\pause$$

$(1+2a^2+x)(3a+2ax)\pause = (1+2a^2+x)3a+(1+2a^2+x)2ax$\pause $= 3a+6a^3+3xa+2ax+4a^3x+2xax\pause= 3a+2+3a^2x+2ax+0+2a^2$\pause
$=2+3a+2a^2+2ax+3a^2x$

\end{frame}


\begin{frame}
\frametitle{Summary Basic Results:}

\hu{\bf Theorem:} Let $G$ be a group and $n$ a non-negative integer. Suppose $G$ can be realized by a ring $R$ of characteristic $n$.
\begin{itemize}
\item There is an ideal $I$ of $\Z_n[G]$ such that $(\Z_n[G]/I)^\times \cong G$. Moreover, $G$ can be realized by a finite ring if and only if $G$ is finite and realizable in a positive characteristic.
\item Let $S$ be the subring of $R$ generated by the units $G$. Then $S^\times=G$.
\item If a ring realizes an Abelian group, one of its commutative subrings realizes that group. 
\item If $G$ is cyclic, then there is some ideal $I$ of $\Z_n[x]$ such that $(\Z_n[x]/I)^\times \cong G$.
\item Units of the prime subring are central units: $\Z_n^\times \subseteq Z(R^\times)=Z(G)$. 
\item We have $-1$ is a unit of order 1 or 2. Moreover, $-1$ is a unit of order 2 if and only if the characteristic of $R$ is not 1 or 2. 
\item Non-trivial realizable groups of odd order must be realized in characteristic 2.
\end{itemize}

\end{frame}

\begin{frame}
\frametitle{Using our Tools:}


\vspace{0.1in}

\hu{Example:} {\small Cyclic groups $C_n$ where $n=5,11,$ or $13$ cannot be realized.}

\pause

\vspace{0.1in}

Suppose $R$ realizes $C_n$. These groups have odd order so we must work in characteristic 2.\pause

Note that in $\mathbb{Z}_2[x]$ for $n=5,11,$ and $13$, the factorization of $x^n-1=x^n+1$ into irreducibles is  $(x+1)(x^{n-1}+\cdots+x+1)$.\pause

Thus can assume $R$ is a quotient of $$\mathbb{Z}_2[x]/(x^n-1) \cong \mathbb{Z}_2[x]/(x+1) \times \mathbb{Z}_2[x]/(x^{n-1}+\cdots+x+1)\pause \cong \mathbb{F}_2 \times \mathbb{F}_{2^{n-1}}\pause$$ Thus $R$ must be isomorphic to $\{0\}$, $\mathbb{F}_2$, $\mathbb{F}_{2^{n-1}}$, or $\mathbb{F}_2 \times \mathbb{F}_{2^{n-1}}$.\pause 

But none of these realize $C_n$. 

\end{frame}

\begin{frame}
\frametitle{What is known?}

\begin{description}
\item[{\Large \hl{Cyclic Groups:}}]  {\small The work of Gilmer (1963) and then Pearson \& Schneider (1970) settled which cyclic groups can and cannot be realized.}
\end{description}

\hu{\bf Theorem:} A finite cyclic group is realizable if its order is the product of a set of pairwise relatively prime integers drawn from the following list:
\begin{enumerate}[(a)]
\item $p^k-1$ where $p$ is prime and $k \geq 1$;
\item $(p-1)p^k$ where $p>2$ and $k \geq 1$;
\item $2k$ where $k$ is odd;
\item $4k$ where any prime dividing $k$ is congruent to 1 modulo 4.
\end{enumerate}

\end{frame}

\begin{frame}
\frametitle{What is known?}

\begin{description}
\item[{\Large \hl{Symmetric and Alternating Groups:}}] {\small  Davis \& Occhipinti (2014) gave a complete solution for which symmetric ($S_n$) and alternating ($A_n$) groups are realizable.}

The only realizable symmetric and alternating groups are\dots \quad
$S_1, S_2, S_3, S_4, A_1, A_2, A_3, A_4, \text{ and } A_8$

\vspace{0.1in}

\pause

\item[{\Large \hl{Finite Simple Groups:}}] {\small In a different paper, Davis \& Occhipinti (2014) settled which finite simple groups can be realized.}

The only realizable finite simple groups are\dots\\
\force \hspace*{-0.5in}
$C_2, C_p \text{ where $p$ is a Mersenne prime}, \mathrm{PSL}_n(\mathbb{F}_2) \text{ where }n \geq 3$

\end{description}

\end{frame}

\begin{frame}
\frametitle{What  is known?}

\begin{description}
\item[{\Large \hl{Dihedral Groups:}}] {\small Chebolu \& Lockridge (2016) gave a complete solution for which dihedral groups can be realized.}
\end{description}

Recall $D_n = \langle a,x \;|\; a^n=1, x^2=1, ax=xa^{-1} \rangle = \{ 1,a,\dots,a^{n-1},x,ax,\dots,a^{n-1}x\}$ is the \hu{dihedral group} of order $2n$.

The only realizable dihedral groups are\dots \quad
\[ D_2, D_4, D_6, \text{ and }D_{k} \text{ where $k$ is odd.} \]

\end{frame}

\begin{frame}
\frametitle{Generalized Quaternion = Dicyclic Groups:}

Let $\mathrm{Dic}_n = \langle a,x \;|\; a^{2n}=1, a^n=x^2, ax=xa^{-1} \rangle$\\ \force \hspace{0.5in} $= \{ 1,a,\dots,a^{2n-1},x,ax,\dots,a^{2n-1}x \}$\\ be the \hu{dicyclic group} of order $4n$.

\begin{itemize}
\item These groups can be constructed inside the quaternion algebra letting $a$ be a primitive $2n$-th root of unity and $x=j$. \pause
\item $\mathrm{Dic}_1 \cong C_4$ is the only Abelian dicyclic group.  \pause
\item $\mathrm{Dic}_2 \cong Q = \{ \pm 1, \pm i, \pm j, \pm k\}$ is the \hl{quaternion group} of order $8$. \pause
\item $\langle a \rangle = \{1,a,\dots,a^{2n-1}\}$ has order $2n$. 
\item The elements $x,ax,\dots,a^{2n-1}x$ all have order $4$. \pause
\item $a^n=x^2$ is the only element of order 2. \pause
\item $Z(\mathrm{Dic}_n) = \{1,a^n \}$ when $n>1$. 
\end{itemize}

\end{frame}

\begin{frame}
\frametitle{Generalized Quaternion = Dicyclic Groups:}


Let $n>1$. Then since $|Z(\mathrm{Dic}_n)|=2$, if $R$ realizes $\mathrm{Dic}_n$, we have

\vspace{0.1in}

\force \hfill
\fbox{\hlr{$\mathrm{char}(R)=0,2,3,4, \text{ or }6$}.}
\hfill \force

\pause

\vspace{0.2in}

Note that $\mathrm{Dic}_1 = Z(\mathrm{Dic}_1) \cong C_4$ allows for more possibilities:

\vspace{0.1in}

\force \hfill $\mathrm{char}(R)=0,2,3,4,5,6, \text{ or }10$. \hfill \force

\end{frame}




\begin{frame}
\frametitle{Generalized Quaternion = Dicyclic Groups:}

\hu{\bf Proposition:} Dicyclic groups cannot be realized in char. $3$ or $6$.\pause

\vspace{0.05in}

{\bf Proof:} Suppose $R$ realizes $\mathrm{Dic}_n$. If $R$'s characteristic is not 2, then $1 \not= -1$ so $-1$ has order $2$. Thus $a^n=x^2=-1$. \pause

Suppose $\mathrm{char}(R)=3$. Then $(1+x)^4$\pause $=1+4x+6x^2+4x^3+x^4$\pause $=1+x+0+(-1)x+1=2=-1$. \pause So $(1+x)^8=1$ and thus $1+x$ is a unit of order 8.\pause Therefore, $1+x=a^k$ for some $k$ since $x,ax,\dots,a^{2n-1}x$ have order 4. \pause Then $a(1+x)=a^{k+1}=(1+x)a$ \pause so $a+ax=a+xa$ and so $ax=xa$. \pause But $xa^{-1}=ax=xa$ so that $a^{-1}=a$.\pause Thus $a^2=1$ and so $n=1$.\pause But $\mathrm{Dic}_1 \cong C_4$ does not have an element of order 8! (contradiction)\pause

Suppose $\mathrm{char}(R)=6$. Then $(2+x)^4=2^4+4\cdot 2^3x+6\cdot 2^2x^2+4 \cdot 2x^3+x^4=4+2x+2(-1)x+1=5=-1$. Thus $(2+x)^8=1$ and thus $2+x$ is a unit of order 8. Again, we must have $2+x=a^k$ for some $k$ and again $a(2+x)=a^{k+1}=(2+x)a$ implies $ax=xa$. This leads to the same contradiction. $\square$

\end{frame}

\begin{frame}
\frametitle{Generalized Quaternion = Dicyclic Groups:}

\hu{\bf Proposition:} $\mathrm{Dic}_n$ can be realized in char. $2$ or $4$ if and only if $n=1$ or $n=2$.

\vspace{0.05in}

{\bf Proof:} This is similar in flavor to the characteristic 3 and 6 proof, but it is a bit more involved. $\square$\pause

\vspace{0.1in}

\hu{\bf Example:} $\mathbb{Z}[-1,j]^\times \cong C_4 \cong \mathrm{Dic}_1$, $\mathbb{Z}[i,j]^\times \cong Q \cong \mathrm{Dic}_2$, and\dots\\ \force \hfill $\mathbb{Z}[e^{\pi i/3}, j]^\times \cong \mathrm{Dic}_3$.\pause

\vspace{0.05in}

\hu{\bf Proposition:} We have a complete picture for small Dicyclic groups\dots
\begin{itemize}
\item $\mathrm{Dic}_1$ can only be realized in char. 0, 2, 4, 5, and 10. \pause
\item $\mathrm{Dic}_2$ can only be realized in char. 0, 2, and 4.\pause
\item $\mathrm{Dic}_3$ can only be realized in char. 0.\pause\\ \force \hfill \hlr{$\mathrm{Dic}_3$ cannot be realized by a finite ring!}
\end{itemize}

\end{frame}

\begin{frame}
\frametitle{Some Partial Results:}

\vspace{0.1in}

Sadly, $\mathbb{Z}[e^{\pi i/(2n)}]^\times$ is infinite for $n>3$, so we cannot use $\mathbb{Z}[e^{\pi i/(2n)},j]$ to realize $\mathrm{Dic}_n$ for $n>3$.\pause

\vspace{0.1in}

\hu{\bf Proposition:} If every prime factor of $n$ is congruent to 1 modulo 4 (i.e., $\mathbb{Z}_n^\times$ has an element of  order 4 that squares to be $-1$), then we can realize $\mathrm{Dic}_n$ in characteristic $0$.

\vspace{0.1in}

Let $R=\Z_n \rtimes \Z[i]$ ($=\Z_n \times \Z[i]$ as a group under addition). Define $f(x+yi)=x+y\tau$ and $g(x+yi)=x-y\tau$ where $\tau^2=-1$ in $\mathbb{Z}_n$. We multiply as follows: $(a,b)(x,y)=(af(y)+xg(b),by)$. Then $R=\Z_n \rtimes \mathbb{Z}[i]$ realizes $\mathrm{Dic}_n$ (in characteristic $0$).\pause

\vspace{0.2in}

\hu{\bf Proposition:} If $G$ is realizable in characteristic $n$, then so is every maximal Abelian subgroup of $G$.\pause

\vspace{0.1in}

\hu{\bf Corollary:} If we can realize $\mathrm{Dic}_n$ in characteristic 0, then we must be able to realize $C_{2n}$ as well. As a consequence, for example, $\mathrm{Dic}_{4n}$ cannot be realized for any $n \geq 1$.

\end{frame}

\begin{frame}
\frametitle{Groups of Small Order:}

\force  \hspace*{-0.25in}
\small
\begin{tabular}{|c|c|} \hline
Order & Isomorphism Classes \large\strut \\ \hline
1,2,3 & $C_1 \cong \mathbb{Z}_2^\times$, $C_2 \cong \mathbb{Z}_3^\times$, $C_3 \cong \mathbb{F}_4^\times$ \large\strut \\ \hline
4 & $C_4 \cong \mathbb{Z}_5^\times$ and $C_2 \times C_2 \cong (\mathbb{Z}_3 \times \mathbb{Z}_3)^\times$ \large\strut \\ \hline
5 & $C_5$ is not realizable \large\strut \\ \hline
6 & $C_6 \cong \mathbb{Z}_7^\times$ and $S_3 \cong (\mathbb{Z}_2^{2 \times 2})^\times$ \Large\strut \\ \hline
7 & $C_7 \cong \mathbb{F}_8^\times$ \large\strut \\ \hline
\multirow{2}{*}{8} & $C_8 \cong \mathbb{F}_9^\times$, $C_2 \times C_4 \cong (\mathbb{Z}_3 \times \mathbb{Z}_5)^\times$, $C_2\times C_2\times C_2 \cong (\mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3)^\times$, \Large\strut \\ %\cline{2-2}  % usage \cline{col1-col2}
   & $D_4 \cong U_3(\mathbb{Z}_2)^\times$, and $Q \cong (\mathbb{Z}_2[\mathrm{Dic}_2]/(1+x+a+ax))^\times$ \large\strut \\ \hline
9 & $C_9$ is not realizable and $C_3 \times C_3 \cong (\mathbb{F}_4 \times \mathbb{F}_4)^\times$ \large\strut \\ \hline
10 & $C_{10} \cong \mathbb{Z}_{11}^\times$ and $D_5$ is not realizable \large\strut \\ \hline
11 & $C_{11}$ is not realizable \large\strut \\ \hline
\multirow{2}{*}{12} & $C_{12} \cong \mathbb{Z}_{13}^\times$, $C_2 \times C_6 \cong (\mathbb{Z}_3 \times \mathbb{Z}_7)^\times$, $D_6 \cong U_2(\mathbb{Z}_3)^\times$, \large\strut \\ %\cline{2-2}
    & $A_4 \cong (\mathcal{O}/2\mathcal{O})^\times$, and $\mathrm{Dic}_3 \cong (\mathbb{Z}[e^{\pi i/3},j])^\times$ \large\strut \\ \hline
13 & $C_{13}$ is not realizable \large\strut \\ \hline
14 & $C_{14} \cong (\mathbb{Z}_3 \times \mathbb{F}_8)^\times$ and $D_7$ is not realizable \large\strut \\ \hline
15 & $C_{15} \cong \mathbb{F}_{16}^\times$ \large\strut \\ \hline
\end{tabular}



\end{frame}



\begin{frame}
\frametitle{References:}

\small
\begin{itemize}\setlength\itemsep{0mm}
\item S. Chebolu, K. Lockridge, ``Fuchs’ problem for dihedral groups,'' {\it Journal of Pure and Applied Algebra}, 2016.
\item C. Davis and T. Occhipinti, ``Which alternating and symmetric groups are unit groups?'' {\it Journal of Algebra and Its Applications}. Vol. 13, No. 3 (2014).
\item C. Davis and T. Occhipinti, ``Which finite simple groups are unit groups?'' {\it Journal of Pure and Applied Algebra}. Vol. 13, No. 3 (2014).
\item R. Gilmer, ``Finite rings having a cyclic multiplicative group of units,'' {\it American Journal of Mathematics}, 1963, 447--452.
\item K. R. Pearson and J. E. Schneider, ``Rings with a cyclic group of units,'' {\it Journal of Algebra}. Vol. 16 (1970).
\end{itemize}

\hu{Theses:}
\begin{itemize}
\item Joshua Carr, ``Realizable Unit Groups'' (May 2017)
\item Lindsey Wise, ``Realizable Dicyclic Groups'' (December 2020)
\end{itemize}

\end{frame}


\begin{frame}

\begin{center}
\Huge \hl{Thank you} for listening!

 \hl{Questions?}
\end{center}
\end{frame}


\end{document}

\begin{frame}
\frametitle{Characteristic of a Ring:}

Let $R$ be a ring. We always have the homomorphism $\varphi:\mathbb{Z} \to R$ where \hl{$\varphi(k)=k1_R$} where $k1_R$ is the $k^\mathrm{th}$ additive power of the multiplicative identity of $R$.

\vspace{0.05in}

{\small For example, $\varphi(3)=1_R+1_R+1_R$ and $\varphi(-2)=(-1_R)+(-1_R)$.}

\vspace{0.1in}

Since $\mathbb{Z}$ is a principal ideal domain, $\mathrm{ker}(\varphi) = n\mathbb{Z}$ for some $n \in \mathbb{Z}$. In fact, since $-n\mathbb{Z}=n\mathbb{Z}$, we can require $n \geq 0$. The isomorphism theorem then implies $\mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z} \cong \varphi(\mathbb{Z}) \subseteq R$. Note that we identify \hl{$\mathbb{Z}_0=\mathbb{Z}$}.

Call \fbox{\hlr{$n$}} the \hu{characteristic} and \fbox{\hlr{$\varphi(\mathbb{Z})$}} the \hu{prime subring} of $R$.

\vspace{0.1in}


Notice that since $1$ commutes with all $R$, everything in $\varphi(\mathbb{Z})$ does too: \hl{the prime subring lies in the center of the ring}. Next, since homomorphisms send units to units, we have that the center of the unit group \fbox{\hlr{$Z(R^{\times})$ must contain a copy of $\mathbb{Z}_n^\times$}}.

\end{frame}

