\documentclass{article} \usepackage{amssymb} \usepackage{latexsym} \title{Group Theory: Math30038, Sheet 1} \author{GCS: Solutions} \date{} \begin{document} \maketitle {\it The course web site is available via {\tt http://www.bath.ac.uk/$\sim$masgcs/}} \begin{enumerate} \item Let $G$ be a group. Suppose that $x \in G$. Let $C_G(x) = \{ g \in G \mid gx = xg \} \subseteq G.$ Prove that $C_G(x) \leq G$.\newline \noindent {\bf Solution: } $1 \in C_G(x) \not = \emptyset.$ Moreover if $g \in G$ and $gx = xg$, then $xg^{-1} = g^{-1}x$ so if $a, b \in C_G(x)$, then $ab^{-1}x = axb^{-1} = xab^{-1}$ and therefore $ab^{-1} \in C_G(x)$. \item {\em Let $G$ be a group. Suppose that $S \subseteq G$. Let $C_G(S) = \{ g \in G \mid gs = sg \forall s \in S\}.$ Prove that $C_G(S) \leq G$.} \newline \noindent {\bf Solution: } We have $C_G(S) = \cap_{s \in S} C_G(s)$ and since the intersection of subgroups is a subgroup, we are done. \item Let $G$ be a group. Suppose that $S \subseteq G$. Let $N_G(S) = \{ g \in G \mid gS = Sg \}$ where $gS = \{ gs \mid s \in S\}$ and $Sg = \{ sg \mid s \in S \}$. Prove that $C_G(S) \leq N_G(S) \leq G$. \newline \noindent {\bf Solution: } The fact that $N_G(S)$ is a group follows the outline of the proof that $C_G(x)$ is a group. The condition to be in $C_G(S)$ is stronger than that to be in $N_G(S)$ so $C_G(S) \leq N_G(S) \leq G$. \item {\em Let $G$ be a group. Suppose that $S \subseteq G$ and that $x \in N_G(S)$. Prove that $C_G(S)x = xC_G(S)$.} \newline \noindent {\bf Solution: } We will show that $C_G(S) = x^{-1}C_G(S)x$ which will suffice. Suppose that $s \in S$, then $xs = s'x$ for some $s' \in S$. Now shoose any $c \in C_G(S)$, then $x^{-1}cxs = x^{-1}cs'x = x^{-1}s'cx = x^{-1}(xsx^{-1})cx = sx^{-1}cx$ and therefore $sx^{-1}c \in C_G(S)$ and so $x^{-1}C_G(S)x \leq C_G(S).$ Replacing $x$ by $x^{-1}$ in this argument yields that $x C_G(S)x^{-1} \leq C_G(S).$ Premultiplying by $x^{-1}$ and postmultiplying by $x$ gives $C_G(S) \leq x^{-1}C_G(S)x$. Now we have two mutually reverse inclusions so $C_G(S) = x^{-1}C_G(S)x$ for all $x \in N_G(S).$ \item {\em Let $G$ be a finite group. Suppose that $\emptyset \not = H \subseteq G$ has the property that if $a,b \in H$, then $ab \in H$. Does it follow that $H \leq G$? What happens if we relax the condition that $G$ is finite?} \newline \noindent {\bf Solution: } If $h \in H$, then $\langle h \rangle \leq H$ and so $o(h)$ is a natural number $n$. If $h = 1$ then $h^{-1} = h \in H$. Otherwise $n > 1$ and then $h^{-1} = h^{n-1} \in H$ so $H$ is a subgroup of $G$. In the event that $G$ is infinite, things fall apart. For example perhaps $G = \mathbb Z$ under addition, the $\mathbb N$ is a non-empty additively closed subset which is not a subgroup since $-1 \not \in \mathbb N$. \item Suppose that $G$ is a group and that $x \in G$. Prove that $(x^{-1})^{-1} = x$. \newline \noindent {\bf Solution: } $x^{-1} x = 1 = x^{-1} (x^{-1})^{-1}.$ Premultiplying by $x$ gives the result. \item {\em Does there exist a group $G$ containing elements $a,b$ such that $a^2 = b^2 = (ab)^3 = 1$?} \newline \noindent {\bf Solution: } Yes, the trivial group will do it. More interestingly, let $G$ be the group $S_3$, and put $a = (1,2)$ and $b = (2,3)$. Now the orders of $a, b$ and $ab$% are exactly as advertized''. \item {\em Suppose that $G$ is a group with the property that $x^2 = 1$ whenever $x$ is an element of $G$. Show that $G$ must be abelian.} \newline \noindent {\bf Solution: } Suppose that $a, b \in G$, then $abab = 1 = baab$. Postmultiplying by $ba$ yields $ab = ba$. \item {\em (Challenge) Suppose that $G$ is a group with the property that $x^3 = 1$ whenever $x$ is an element of $G$. Show that $G$ need not be abelian.} \newline \noindent {\bf Solution: } Consider the set of 3 by 3 upper triangular matrices with entries in $F = \mathbb Z_3$, the field of integers modulo 3 which have 1s on the leading diagonal. It is easy to verify that these 27 matrices form a group. Each matrix is of the form $I + \Delta$ wheer $I$ is the 3 by 3 identity matrix and $\Delta$ is strictly upper triangular. Note that $\Delta^3$ is the zero matrix, so the inverse of $I + \Delta$ is $I - \Delta + \Delta^2$ and moreover $(I + \Delta)^3 = I + 3\Delta + 3\Delta^2 = I$ so ev ery element of the group has order dividing 3. The matrices $I + E_{12}$ and $I + E_{23}$ do not commute as may be verified by direct calculation; here $E_{ab}$ is the 3 by 3 matrix with entries in $F$ where the entrie in the $i$-th row and $j$-th column is $\delta_{ia}\delta_{jb}$ where $\delta$ is the Kronecker delta. \end{enumerate} \vfill \end{document}