\documentclass{article} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{enumerate} \usepackage{ifthen} \usepackage{graphicx} \setlength{\unitlength}{0.1in} \setlength{\oddsidemargin}{0.0in} \setlength{\evensidemargin}{0.0in} \setlength{\topmargin}{0.0in} \setlength{\headheight}{0.0in} \setlength{\headsep}{0.0in} \setlength{\topskip}{0.0in} \setlength{\textheight}{9.5in} \setlength{\textwidth}{6.5in} \newcommand{\comp}{ \,{\scriptstyle \stackrel{\circ}{}}\, } \newcommand{\nullset}{\mathrm{O}\!\!\!\!\big/\,} \newcommand{\divides}{\,\Big|\,} \newcommand{\varspace}[2]{\mbox{\parbox[c][#1]{#2}{\hfill}}} \newcommand{\myspace}{\varspace{0.35in}{0.35in}} \begin{document} \pagestyle{empty} \noindent \parbox{2in}{\bf Math 3110} \hfill {\Large \bf Final Exam} \hfill \parbox{2in}{\bf \hfill May 4$^{th}$, 2009} \vspace{-0.1in} \begin{center} \Large ANSWER KEY \end{center} \noindent \noindent {\bf \large 1. (12 points)} Tables for Sarah. Let $R = \{ 0,3,6,9 \} \subset \mathbb{Z}_{12}$. \begin{enumerate}[(a)] \item Finish filling out the following addition and multiplication tables for $R$: \vspace{0.1in} \noindent \mbox{} \hspace{0.1in} \begin{tabular}{|c||c|c|c|c|} \hline $+$ & 0 & 3 & 6 & 9 \\ \hline \hline 0 & 0 & 3 & 6 & 9 \\ \hline 3 & 3 & 6 & 9 & 0 \\ \hline 6 & 6 & 9 & 0 & 3 \\ \hline 9 & 9 & 0 & 3 & 6 \\ \hline \end{tabular} \hspace{0.2in} \begin{tabular}{|c||c|c|c|c|} \hline $\times$ & 0 & 3 & 6 & 9 \\ \hline \hline 0 & 0 & 0 & 0 & 0 \\ \hline 3 & 0 & 9 & 6 & 3 \\ \hline 6 & 0 & 6 & 0 & 6 \\ \hline 9 & 0 & 3 & 6 & 9 \\ \hline \end{tabular} \vspace{0.1in} \item Fill out the following table of information about $R$. \vspace{0.1in} \noindent \begin{tabular}{|c|c|c|} \hline Property & Is this true about $R$? & Briefly Explain \\ \hline \hline $R$ is a subring of $\mathbb{Z}_{12}$ & Yes & The tables show closure under all operations. \\ \hline $R$ has a ``unity'' & Yes & Notice that $9x=x=x9$ for all $x$. \\ \hline $R$ is commutative & Yes & Because $\mathbb{Z}_{12}$ is commutative. \\ \hline $R$ is an integral domain & No & 6 is a zero divisor. \\ \hline $R$ is a field & No & 6 is not a unit. \\ \hline \end{tabular} \vspace{0.1in} \item Fill out the following table concerning $R$. {\it Hint:} These tables are longer than they need to be. \vspace{0.1in} \mbox{} \hspace{-0.25in} \begin{tabular}{|c|c|} \hline Zero Divisors & I am a zero divisor because... \\ \hline \hline 6 & $6 \cdot 6 = 0$ \\ \hline \end{tabular} \hspace{0.5in} \begin{tabular}{|c|c|} \hline Units & My multiplicative inverse is... \\ \hline \hline 3 & $3^{-1}=3$ since $3 \cdot 3 = 9$ (the identity). \\ \hline 9 & $9^{-1}=9$ since $9 \cdot 9 = 9$ (the identity). \\ \hline \end{tabular} \end{enumerate} \vspace{0.2in} \noindent {\bf \large 2. (13 points)} Recall $A_4 = \{ (1), (123), (132), (124), (142), (134), (143), (234), (243),$\\ $(12)(34), (13)(24), (14)(23) \}$. Let $H = \{ (1), (12)(34), (13)(24), (14)(23) \}$. \begin{enumerate}[(a)] \item Quickly explain why $H$ is a subgroup of $A_4$ (I am thinking of a certain one word answer). \vspace{0.1in} ``Closure'' --- since $H$ is a finite non-empty subset of the group $A_4$, we only need to check closure under function composition (closure under inverses comes for free since $H$ is finite). \vspace{0.1in} \item Find all of the left and right cosets of $H$ in $A_4$. Is $H$ a {\bf normal} subgroup of $A_4$? \vspace{0.1in} If you think about this for a moment, you should realize that you only need to perform a few permutation compositions. The first coset ``$H$'' comes for free. Next, $|A_4|=12$ and $|H|=4$ so the index of $H$ in $A_4$ is $[A_4:H]=12/4=3$. Since there are only 3 cosets and we already know $H$, after a second coset is computed the third coset must be the ``left-overs''. \noindent {\bf Left Cosets:} \begin{itemize} \item $H = \{ (1), (12)(34), (13)(24), (14)(23) \}$ \item $(123)H = \{ (123)(1), (123)(12)(34), (123)(13)(24), (123)(14)(23) \}$ $= \{ (123), (134), (243), (142) \}$ \item $(132)H = \{ (132), (143), (234), (124) \}$ \end{itemize} \noindent {\bf Right Cosets:} \begin{itemize} \item $H = \{ (1), (12)(34), (13)(24), (14)(23) \}$ \item $H(123) = \{ (1)(123), (12)(34)(123), (13)(24)(123), (14)(23)(123) \}$ $= \{ (123), (243), (142), (134) \}$ \item $H(132) = \{ (132), (143), (234), (124) \}$ \end{itemize} Since the left and right cosets match, $H$ is {\bf normal} in $A_4$. \vspace{0.1in} \item Construct a Cayley table for $\displaystyle{{A_4 \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop H}}$. \vspace{0.1in} \begin{tabular}{c||c|c|c|} $\displaystyle{{A_4 \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop H}}$ & $H$ & $(123)H$ & $(132)H$ \\ \hline \hline $H$ & $H$ & $(123)H$ & $(132)H$ \\ \hline $(123)H$ & $(123)H$ & $(132)H$ & $H$ \\ \hline $(132)H$ & $(132)H$ & $H$ & $(123)H$ \\ \hline \end{tabular} \vspace{0.1in} \item Is $\displaystyle{{A_4 \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop H}}$ Abelian? Cyclic? Why or why not? \vspace{0.1in} $\displaystyle{{A_4 \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop H}}$ is both abelian and cyclic. The order of this quotient group is 3 and since 3 is prime and all groups of prime order are cyclic, this quotient group must be cyclic. It's abelian since cyclic implies abelian. Alternatively, we can see that this group is cyclic directly since $((123)H)^2 = (123)^2H = (132)H$ and thus $\displaystyle{{A_4 \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop H} = \left\langle (123)H \right\rangle}$. \end{enumerate} \vspace{0.2in} \noindent {\bf \large 3. (13 points)} An ``ideal'' exam question. \begin{enumerate}[(a)] \item Find all of the principle ideals of $\mathbb{Z}_{20}$. {\it Hint:} Principle ideals of $\mathbb{Z}_n$ = Cyclic subgroups of $\mathbb{Z}_n$. \vspace{0.1in} Remember that for the rings $\mathbb{Z}_n$ and $\mathbb{Z}$ we have: cyclic subgroups = subgroups = subrings = ideals = principle ideals. [We know that all subgroups of cyclic groups are automatically cyclic. Then the rest of the equivalences follow because in these rings multiplication is just repeated addition, so closure under addition guarantees closure under multiplication too.] We know that since $\mathbb{Z}_{20}$ is cyclic, there is exactly one (cyclic) subgroup of each possible order dividing 20. So there are unique subgroups of orders 1, 2, 4, 5, 10, and 20. \begin{description} \item[Order 1] $\displaystyle{ \left\langle \frac{20}{1} \right\rangle = \langle 20 \rangle = \langle 0 \rangle = \{ 0 \} }$ \item[Order 2] $\displaystyle{ \left\langle \frac{20}{2} \right\rangle = \langle 10 \rangle = \{ 0, 10 \} }$ \item[Order 4] $\displaystyle{ \left\langle \frac{20}{4} \right\rangle = \langle 5 \rangle = \{ 0,5,10,15 \} }$ \item[Order 5] $\displaystyle{ \left\langle \frac{20}{5} \right\rangle = \langle 4 \rangle = \{ 0,4,8,12,16 \} }$ \item[Order 10] $\displaystyle{ \left\langle \frac{20}{10} \right\rangle = \langle 2 \rangle = \{ 0, 2,4,6, 8,10,12, 14,16,18 \} }$ \item[Order 20] $\displaystyle{ \left\langle \frac{20}{20} \right\rangle = \langle 1 \rangle = \mathbb{Z}_{20} }$ \end{description} This is a complete list of all (cyclic) subgroups / subrings / (principle) ideals of $\mathbb{Z}_{20}$. \vspace{0.1in} \item It turns out that $\mathbb{Z}_{15}$ is a principle ideal of itself. Find all $k \in \mathbb{Z}_{15}$ such that $(k)=\mathbb{Z}_{15}$.\\ In general, what do we call the set of all $k \in \mathbb{Z}_n$ such that $(k)=\mathbb{Z}_n$?\\ {\it Hint:} Principle ideals of $\mathbb{Z}_n$ = cyclic subgroups in $\mathbb{Z}_n$, so I'm really asking you about the generators of the group $\mathbb{Z}_n$. \vspace{0.1in} If you quickly check all $(k) = \langle k \rangle = \{ mk \in \mathbb{Z}_{15} \,|\, m \in \mathbb{Z}_{15} \}$, you will find that $\mathbb{Z}_{15} = (1) = (2) = (4) = (7) = (8) = (11) = (13) = (14)$. In fact, these are just the units of $\mathbb{Z}_{15}$: $U(\mathbb{Z}_{15}) = U(15) = \{ 1,2,4,7,8,11,13,14 \}$. In general, $(k) = \mathbb{Z}_n$ if and only if $k \in U(n)$. [proof: Suppose $(k) = \mathbb{Z}_n$. Then $1 \in (k)$, so $mk=1$ for some $m \in \mathbb{Z}_n$. Thus $k$ is a unit. Conversely, if $k$ is a unit, then $k \in (k)$ and any ideal containing a unit is equal to the whole ring (see the next part of this problem).] \vspace{0.1in} \item Let $R$ be a ring with $1$, let $I$ be an ideal of $R$, and suppose $u \in R$ is a unit.\\ Prove that $u \in I$ implies that $I=R$. \vspace{0.1in} Since $u$ is a unit, $u^{-1}$ exists in $R$. But $1 = u^{-1}u \in I$ since $u \in I$ and $I$ absorbs multiplications from outside itself. Let $r \in R$. Notice that $r = r1 \in I$ because $1 \in I$ (and absorbtion again). Therefore, $R \subseteq I$ and so $I=R$. \end{enumerate} \vspace{0.2in} \noindent {\bf \large 4. (14 points)} Random Questions. \begin{enumerate}[(a)] \item Consider $22 \in \mathbb{Z}_{125}$. Is $22$ a unit, zero divisor, both, or neither?\\ If $22$ is a unit, find its inverse. If $22$ is a zero divisor find $b \not=0$ such that $22b = 0 \mbox{ mod } 125$. \vspace{0.1in} We need to run the Euclidean algorithm on 22 and 125 to determine if it's a unit or not. $125 = 22(5)+15$, \qquad $22 = 15(1) + 7$, \qquad $15 = 7(2) + 1$, \quad and \quad $7 = 1(7) + 0$ The last non-zero remainder is 1. So 22 and 125 are relatively prime and thus 22 is a {\bf unit} in $\mathbb{Z}_{125}$. We need to run the algorithm backwards to find the inverse of 22 modulo 125. \begin{itemize} \item $(-2)7 + (1)15 = 1$ \item $(-1)15 + (1)22 = 7$ $\Rightarrow$ $(-2)[(-1)15+(1)22] + (1)15 = 1$ $\Rightarrow$ $(3)15 + (-2)22 = 1$ \item $(-5)22 + (1)125 = 15$ $\Rightarrow$ $(3)[(-5)22 + (1)125] + (-2)22 = 1$ $\Rightarrow$ $(-17)22 + (3)125 = 1$ \end{itemize} Thus $22^{-1} = -17 = 108$ (mod 125). \vspace{0.1in} \item Let $G$ be an Abelian group and $H$ a subgroup of $G$. Prove that $H$ is a normal subgroup of $G$. \vspace{0.1in} Let $x \in G$. $xH = \{xh\,|\,h\in H\} = \{hx\,|\,h\in H\} = Hx$ since $xh=hx$ because $G$ is abelian. Thus $xH=Hx$ for all $x \in G$, so $H$ is normal in $G$. \vspace{0.1in} \item Let $G$ be a cyclic group and let $H$ be a subgroup of $G$. Quickly explain why $\displaystyle{{G \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop H}}$ is a group.\\ (i.e. Why can we quotient by $H$?) Then prove that $\displaystyle{{G \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop H}}$ is cyclic. \vspace{0.1in} Since $G$ is cyclic, $G$ is abelian and thus $H$ is normal (by the last part of this problem). Thus it makes sense to quotient $G$ by $H$. Now $G$ cyclic implies there exists some $g \in G$ such that $G = \langle g \rangle$. Suppose that $xH \in G/H$. Then $x\in G=\langle g \rangle$ so there exists some $k \in \mathbb{Z}$ so that $x = g^k$. Next, notice that $xH = g^kH = (gH)^k$. Thus $xH \in \langle gH \rangle$. Therefore, $\langle gH \rangle = \displaystyle{{G \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop H}}$. \end{enumerate} \vspace{0.2in} \noindent {\bf \large 5. (12 points)} Isomorphic or not? \begin{enumerate}[(a)] \item $\mathbb{R}_{>0}$ is the group of positive real numbers under multiplication. $\mathrm{GL}_2(\mathbb{R})$ is the group of invertible $2 \times 2$ real matrices. Is $\mathbb{R}_{>0} \cong \mathrm{GL}_2(\mathbb{R})$? Why or why not? \vspace{0.1in} These are both groups of infinite order, but ``No'' these groups are not isomorphic since $\mathbb{R}_{>0}$ is abelian and $\mathrm{GL}_2(\mathbb{R})$ is not abelian. Another (more difficult) way to see that they are not isomorphic, is that all non-identity elements of $\mathbb{R}_{>0}$ are of infinite order whereas $\mathrm{GL}_2(\mathbb{R})$ has elements of finite orders such as $\displaystyle{ A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}}$ (the order of $A$ is 2 since $A^2=I_2$). \vspace{0.1in} \item Is $U(10) \cong \mathbb{Z}_4$? Why or why not? \vspace{0.1in} Recall that $U(10)$ is the group of units of $\mathbb{Z}_{10}$. It consists of the elements of $\mathbb{Z}_{10}$ which are relatively prime to 10. $U(10) = \{ 1,3,7,9 \}$. Let's look at the orders of the elements of $U(10)$. $9^2=81=1$ (mod 10) so the order of 9 is 2. $3^2=9$, $3^3=27=7$ (mod 10), $3^4=81=1$ (mod 10) so the order of 3 is 4. Similarly the order of 7 is also 4. Thus $U(10)=\langle 3 \rangle$ (and also $U(10)=\langle 7 \rangle$). Thus $U(10)$ is a {\bf cyclic group of order 4}. Since any two cyclic groups of the same order are isomorphic, we can conclude that $U(10) \cong \mathbb{Z}_4$. \vspace{0.1in} \item Is $S_4 \cong D_{12}$? Why or why not? \vspace{0.1in} Both of these are non-abelian groups of order 24 ($|S_4|=4! = 4\cdot 3\cdot 2\cdot 1 = 24$, $|D_{12}| = 2\cdot 12 = 24$). So we need to look deeper. Consider the order of elements. Remember that we use lcm's of lengths of (disjoint) cycles to compute orders in $S_4$. $S_4$ has elements which are 1, 2, 3, and 4 cycles along with elements made up of 2 disjoint 2-cycles. So the elements of $S_4$ have orders 1, 2, 3, and 4. On the other hand, $D_{12}$ has rotations of orders 1, 2, 3, 4, 6, and 12 (as well as reflections of order 2). Since $D_{12}$ has elements of order 12 and $S_4$ does not, they cannot be isomorphic: $D_{12} \not\cong S_4$. {}[We could also see that these groups are not isomorphic by counting the number of elements of each order, and seeing that these counts don't match.] \end{enumerate} \vspace{0.2in} \noindent {\bf \large 6. (13 points)} Subgroups \begin{enumerate}[(a)] \item Show that $K = \{ \ell \in \mathbb{Z} \,|\, 4 \mbox{ divides } \ell \}$ is a subgroup of $\mathbb{Z}$. \vspace{0.1in} Remember that $\mathbb{Z}$ is a group under addition. So to check that $K$ is a subgroup we need to check that $K$ is closed under addition and negation (it's easy to see that $K$ is a subset of $\mathbb{Z}$ and is obviously non-empty because 4 divides 4 and thus $4 \in K$). Let $x,y \in K$. Then $4$ divides $x$ and $4$ divides $y$. Hence there exists some $k,\ell \in \mathbb{Z}$ such that $4k=x$ and $4\ell=y$. Notice that $x+y=4k+4\ell=4(k+\ell)$ thus 4 divides $x+y$ so $x+y \in K$. Also, $-x = -4k = 4(-k)$ so 4 divides $-x$ and thus $-x \in K$. Therefore, $K$ is a subgroup of $\mathbb{Z}$. \vspace{0.1in} \item Let $G$ be an Abelian group with identity $e$. Let $H = \{ g \in G \,|\, g^2=e \}$.\\ Show that $H$ is a subgroup of $G$. \vspace{0.1in} First, obviously $H \subseteq G$. Next, $e^2=e$ so $e \in H$ thus $H$ is non-empty. Suppose $x,y \in H$ so $x^2=e$ and $y^2=e$. Notice that $(xy)^2=xyxy=xxyy=x^2y^2=ee=e$ (since $G$ is abelian) so $xy \in H$. Also, $(x^{-1})^2 = x^{-2} = (x^2)^{-1} = e^{-1} = e$ so $x^{-1} \in H$. Thus, $H$ is a subgroup of $G$. \vspace{0.1in} \item Let $N = \{ x \in \mathbb{Z}_{99} \,|\, 5x \equiv 2 \mbox{ mod } 99 \}$. Is $N$ a subgroup of $\mathbb{Z}_{99}$? Why or why not? \vspace{0.1in} This is {\bf not} a subgroup of $\mathbb{Z}_{99}$ since $N$ does not contain the identity because $5(0) \equiv 0 \not\equiv 2$ (mod 99). Another way to see that $N$ is not a subgroup is that $N$ is not closed under either addition or negation. Since $5(x+y) = 5x+5y \equiv 2 + 2 = 4 \not\equiv 2$ (mod 99) if $x,y \in N$ or $5(-x) = -5x \equiv -2 \equiv 97 \not\equiv 2$ (mod 99) if $x \in N$. \end{enumerate} \vspace{0.2in} \noindent {\bf \large 7. (14 points)} Subrings \begin{enumerate}[(a)] \item Let $\displaystyle{ S = \left\{ \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \,\Big|\, a,b,c \in \mathbb{R} \right\}}$ is a subring of $M_2(\mathbb{R}) = \mathbb{R}^{2 \times 2}$ (the ring of $2 \times 2$ real matrices).\\ Show $S$ is a subring of $M_2(\mathbb{R})$. \vspace{0.1in} $S$ is obviously a non-empty subset of $M_2(\mathbb{R})$. We need to check closure under subtraction and closure under multiplication. Let $\displaystyle{ \begin{bmatrix} a_1 & b_1 \\ 0 & c_1 \end{bmatrix}, \begin{bmatrix} a_2 & b_2 \\ 0 & c_2 \end{bmatrix} \in S}$. $\displaystyle{\begin{bmatrix} a_1 & b_1 \\ 0 & c_1 \end{bmatrix} - \begin{bmatrix} a_2 & b_2 \\ 0 & c_2 \end{bmatrix} = \begin{bmatrix} a_1-a_2 & b_1-b_2 \\ 0 & c_1-c_2 \end{bmatrix} \in S}$ and \\ $\displaystyle{\begin{bmatrix} a_1 & b_1 \\ 0 & c_1 \end{bmatrix} \begin{bmatrix} a_2 & b_2 \\ 0 & c_2 \end{bmatrix} = \begin{bmatrix} a_1a_2 & a_1b_2+b_1c_2 \\ 0 & c_1c_2 \end{bmatrix} \in S}$. Thus $S$ is a subring of $M_2(\mathbb{R})$. \vspace{0.1in} \item Let $R$ be a ring and $a \in R$. Show that $T = \{ r \in R \,|\, ar = 0\}$ is a subring of $R$. \vspace{0.1in} First, notice that $a0=0$ so $0 \in T$ thus $T$ is a non-empty subset of $R$. Now we need to check closure under subtraction and closure under multiplication. Let $x,y \in T$. Thus $ax=0$ and $ay=0$. Hence, $a(x-y)=ax-ay=0-0=0$ and thus $x-y \in T$. Also, $a(xy) = (ax)y = 0y = 0$ thus $xy \in T$. Therefore, $T$ is a subring of $R$. \vspace{0.1in} \item Let $T$ be the subring of $R$ from part (b). If $a=0$, what is $T$? \vspace{0.1in} If $a=0$, then $T = \{ r \in R \,|\, 0r=0 \} = \{ r \in R \,|\, 0=0 \} = R$ (since $0=0$ is always true). \vspace{0.1in} \item Let $T$ be the subring of $R$ from part (b). If $R$ is an integral domain and $a \not=0$, what is $T$? \vspace{0.1in} Suppose that $a \not=0$, $r \in R$, and $ar=0$. Now $R$ is an integral domain so $r=0$ (otherwise $r$ would be a zero divisor). Thus the only element in $T$ is 0 that is $T = \{ 0 \}$ (the trivial subring). \end{enumerate} \vspace{0.2in} \noindent {\bf \large 8. (15 points)} Homomorphisms and Kernels. \vspace{0.1in} Let $\mathbb{R}^{\#}$ be the group of non-zero real numbers under multiplication. \noindent \begin{enumerate}[(a)] \item Let $f : \mathbb{R}^{\#} \rightarrow \mathbb{R}$ be defined by $f(x) = \ln|x|$ (notice the absolute value sign around $x$).\\ Show that $f$ is a homomorphism. Find $\mathrm{Ker}(f)$. Is $f$ 1-1? onto? an isomorphism? \vspace{0.1in} $f(xy)=\ln|xy| = \ln|x|+\ln|y| = f(x)+f(y)$ (using a law of logarithms). Thus $f$ is a (group) homomorphism. For $x \in \mathbb{R}^{\#}$ to be in the kernel of $f$, we need $f(x)=0$ (0 is the identity of $\mathbb{R}$). Thus $\ln|x|=0$ so $|x|=1$. Thus $x=\pm 1$. Hence $\mathrm{Ker}(f) = \{ \pm 1 \}$. $f$ is 2-to-1 not 1-to-1. On the hand $f$ is onto, since $f(e^x) = \ln|e^x| = \ln(e^x) = x$ for all $x \in \mathbb{R}$. Since $f$ is not 1-to-1 is cannot be an isomorphism. \vspace{0.1in} \item Let $G$ be a group and $x,g \in G$. Let $\varphi_g(x) = gxg^{-1}$. Recall that $\varphi_g$ is an automorphism of $G$ (that is $\varphi_g \in \mbox{Aut}(G)$). Also, recall that $\mbox{Aut}(G)$ (automorphisms of $G$) is a group (under function composition). Consider the map $\psi:G \rightarrow \mbox{Aut}(G)$ defined by $\psi(g) = \varphi_g$. \begin{enumerate}[(i)] \item Show that $\varphi_g \circ \varphi_h = \varphi_{gh}$.\\ (This shows that $\psi$ is a homomorphism.) \vspace{0.1in} Let $g,h,x \in G$. $\varphi_g \circ \varphi_h(x) = \varphi_g(hxh^{-1}) = g(hxh^{-1})g^{-1} = (gh)x(gh)^{-1}$ $= \varphi_{gh}(x)$. Therefore, $\psi(g) \circ \psi(h) = \varphi_g \circ \varphi_h = \varphi_{gh} = \psi(gh)$ so $\psi$ is a (group) homomorphism. \vspace{0.1in} \item Show that $\mbox{Ker}(\psi) = Z(G) = \{ g \in G \,|\, gx=xg \; \mbox{ for all } x \in G\}$\\ (Therefore, the center of $G$ is a normal subgroup). \vspace{0.1in} $g \in \mbox{Ker}(\psi)$ if and only if $\psi(g)=\varphi_g = \mbox{id}_G$ (the identity map is the identity element of the group of automorphisms) if and only if $gxg^{-1} = \varphi_g(x) = \mbox{id}_G(x) = x$ for all $x \in G$ if and only if $gx=xg$ for all $x \in G$ if and only if $g \in Z(G)$. Therefore, $\mbox{Ker}(\psi) = Z(G)$ (and hence normal since kernels are normal subgroups). \vspace{0.1in} \item Let $\mbox{Inn}(G) = \{ \varphi_g \,|\, g \in G\}$ (``inner automorphisms'' of $G$).\\ Use part (ii) to explain why $\displaystyle{{G \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop Z(G)}} \cong \mbox{Inn}(G)$. \vspace{0.1in} This is just the first isomorphism theorem applied to $\psi$: $G$ mod kernel is isomorphic to the image where the kernel of $\psi$ is $Z(G)$ and the image is $\mbox{Inn}(G)$. \end{enumerate} \end{enumerate} \end{document}