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It doesn't stop at `\end{asy}`. ``` if(!settings.multipleView) settings.batchView=false; settings.tex="pdflatex"; defaultfilename="Napkin-1"; if(settings.render < 0) settings.render=4; settings.outformat=""; settings.inlineimage=true; settings.embed=true; settings.toolbar=false; viewportmargin=(2,2); import extras; size(6cm); usepackage("amsmath"); usepackage("amssymb"); defaultpen(fontsize(11pt)); settings.tex = "latex"; settings.outformat = "pdf"; size(12cm); picture aoeu(string a, string b, string c, string d, string e, string x) { draw(dir(0)--dir(72)--dir(144)--dir(216)--dir(288)--cycle); MP(a, dir(0), dir(0)); MP(b, dir(72), dir(72)); MP(c, dir(144), dir(144)); MP(d, dir(216), dir(216)); MP(e, dir(288), dir(288)); MP(x, origin, origin); return CC(); } picture one = aoeu("1", "2", "3", "4", "5", "1"); picture r = aoeu("5", "1", "2", "3", "4", "r"); picture s = aoeu("1", "5", "4", "3", "2", "s"); picture sr = aoeu("5", "4", "3", "2", "1", "sr"); picture rs = aoeu("2", "1", "5", "4", "3", "rs"); add(shift( (0,0) ) * one); add(shift( (3,0) ) * r); add(shift( (6,0) ) * s); add(shift( (9,0) ) * sr); add(shift( (12,0) ) * rs); \end{asy} \end{center} Trivia: the dihedral group $D_{12}$ is my favorite example of a non-abelian group, and is the first group I try for any exam question of the form ``find an example\dots''. More examples: \begin{example} [Products of groups] Let $(G, \star)$ and $(H, \ast)$ be groups. We can define a \vocab{product group} $(G \times H, {\cdot})$, as follows. The elements of the group will be ordered pairs $(g,h) \in G \times H$. Then \[ (g_1, h_1) \cdot (g_2, h_2) = (g_1 \star g_2, h_1 \ast h_2) \in G \times H \] is the group operation. \label{def:product_group} \end{example} \begin{ques} What are the identity and inverses of the product group? \end{ques} \begin{example} [Trivial group] The \vocab{trivial group}, often denoted $0$ or $1$, is the group with only an identity element. I will use the notation $\{1\}$. \end{example} \begin{exercise} Which of these are groups? \begin{enumerate}[(a)] \ii Rational numbers with odd denominators (in simplest form), where the operation is addition. (This includes integers, written as $n/1$, and $0 = 0/1$). \ii The set of rational numbers with denominator at most $2$, where the operation is addition. \ii The set of rational numbers with denominator at most $2$, where the operation is multiplication. \ii The set of nonnegative integers, where the operation is addition. \end{enumerate} \end{exercise} \section{Properties of groups} \prototype{$\Zm p$ is possibly best.} \begin{abuse} From now on, we'll often refer to a group $(G, \star)$ by just $G$. Moreover, we'll abbreviate $a \star b$ to just $ab$. Also, because the operation $\star$ is associative, we will omit unnecessary parentheses: $(ab)c = a(bc) = abc$. \end{abuse} \begin{abuse} From now on, for any $g \in G$ and $n \in \NN$ we abbreviate \[ g^n = \underbrace{g \star \dots \star g}_{\text{$n$ times}}.\] Moreover, we let $g\inv$ denote the inverse of $g$, and $g^{-n} = (g\inv)^n$. \end{abuse} In mathematics, a common theme is to require that objects satisfy certain minimalistic properties, with certain examples in mind, but then ignore the examples on paper and try to deduce as much as you can just from the properties alone. (Math olympiad veterans are likely familiar with ``functional equations'' in which knowing a single property about a function is enough to determine the entire function.) Let's try to do this here, and see what we can conclude just from knowing \Cref{def:group}. It is a law in Guam and 37 other states that I now state the following proposition. \begin{fact} Let $G$ be a group. \begin{enumerate}[(a)] \ii The identity of a group is unique. \ii The inverse of any element is unique. \ii For any $g \in G$, ${(g\inv)}\inv = g$. \end{enumerate} \end{fact} \begin{proof} This is mostly just some formal manipulations, and you needn't feel bad skipping it on a first read. \begin{enumerate}[(a)] \ii If $1$ and $1'$ are identities, then $1 = 1 \star 1' = 1'$. \ii If $h$ and $h'$ are inverses to $g$, then $1_G = g \star h \implies h' = (h' \star g) \star h = 1_G \star h = h$. \ii Trivial; omitted. \qedhere \end{enumerate} \end{proof} Now we state a slightly more useful proposition. \begin{proposition}[Inverse of products] Let $G$ be a group, and $a,b \in G$. Then $(ab)\inv = b\inv a\inv$. \end{proposition} \begin{proof} Direct computation. We have \[ (ab)(b\inv a\inv) = a (bb\inv) a\inv = aa\inv = 1_G. \] Similarly, $(b\inv a\inv)(ab) = 1_G$ as well. Hence $(ab)\inv = b\inv a\inv$. \end{proof} Finally, we state a very important lemma about groups, which highlights why having an inverse is so valuable. \begin{lemma}[Left multiplication is a bijection] Let $G$ be a group, and pick a $g \in G$. Then the map $G \to G$ given by $x \mapsto gx$ is a bijection. \label{lem:group_mult_biject} \end{lemma} \begin{exercise} Check this by showing injectivity and surjectivity directly. (If you don't know what these words mean, consult \Cref{ch:sets_functions}.) \end{exercise} \begin{example} Let $G = \Zm 7$ (as in \Cref{def:mult_mod_p}) and pick $g=3$. The above lemma states that the map $x \mapsto 3 \cdot x$ is a bijection, and we can see this explicitly: \begin{align*} 1 &\overset{\times 3}{\longmapsto} 3 \pmod 7 \\ 2 &\overset{\times 3}{\longmapsto} 6 \pmod 7 \\ 3 &\overset{\times 3}{\longmapsto} 2 \pmod 7 \\ 4 &\overset{\times 3}{\longmapsto} 5 \pmod 7 \\ 5 &\overset{\times 3}{\longmapsto} 1 \pmod 7 \\ 6 &\overset{\times 3}{\longmapsto} 4 \pmod 7. \end{align*} \end{example} The fact that the map is injective is often called the \vocab{cancellation law}. (Why do you think so?) \begin{abuse} [Later on, sometimes the identity is denoted $0$ instead of $1$] You don't need to worry about this for a few chapters, but I'll bring it up now anyways. In most of our examples up until now the operation $\star$ was thought of like multiplication of some sort, which is why $1 = 1_G$ was a natural notation for the identity element. But there are groups like $\ZZ = (\ZZ,+)$ where the operation $\star$ is thought of as addition, in which case the notation $0 = 0_G$ might make more sense instead. (In general, whenever an operation is denoted $+$, the operation is almost certainly commutative.) We will eventually start doing so too when we discuss rings and linear algebra. \end{abuse} \section{Isomorphisms} \prototype{$\ZZ \cong 10\ZZ$.} First, let me talk about what it means for groups to be isomorphic. Consider the two groups \begin{itemize} \ii $\ZZ = (\left\{ \dots,-2,-1,0,1,2,\dots \right\}, +)$. \ii $10\ZZ = (\left\{ \dots, -20, -10, 0, 10, 20, \dots \right\}, +)$. \end{itemize} These groups are ``different'', but only superficially so -- you might even say they only differ in the names of the elements. Think about what this might mean formally for a moment. Specifically the map \[ \phi \colon \ZZ \to 10 \ZZ \text{ by } x \mapsto 10 x \] is a bijection of the underlying sets which respects the group operation. In symbols, \[ \phi(x + y) = \phi(x) + \phi(y). \] In other words, $\phi$ is a way of re-assigning names of the elements without changing the structure of the group. That's all just formalism for capturing the obvious fact that $(\ZZ,+)$ and $(10 \ZZ, +)$ are the same thing. Now, let's do the general definition. \begin{definition} Let $G = (G, \star)$ and $H = (H, \ast)$ be groups. A bijection $\phi \colon G \to H$ is called an \vocab{isomorphism} if \[ \phi(g_1 \star g_2) = \phi(g_1) \ast \phi(g_2) \quad \text{for all $g_1, g_2 \in G$}. \] If there exists an isomorphism from $G$ to $H$, then we say $G$ and $H$ are \vocab{isomorphic} and write $G \cong H$. \end{definition} Note that in this definition, the left-hand side $\phi(g_1 \star g_2)$ uses the operation of $G$ while the right-hand side $\phi(g_1) \ast \phi(g_2)$ uses the operation of $H$. \begin{example} [Examples of isomorphisms] Let $G$ and $H$ be groups. We have the following isomorphisms. \begin{enumerate}[(a)] \ii $\ZZ \cong 10 \ZZ$, as above. \ii There is an isomorphism \[ G \times H \cong H \times G\] by the map $(g,h) \mapsto (h,g)$. \ii The identity map $\id \colon G \to G$ is an isomorphism, hence $G \cong G$. \ii There is another isomorphism of $\ZZ$ to itself: send every $x$ to $-x$. \end{enumerate} \end{example} \begin{example} [Primitive roots modulo $7$] As a nontrivial example, we claim that $\Zc 6 \cong \Zm 7$. The bijection is \[ \phi(\text{$a$ mod $6$}) = \text{$3^a$ mod $7$}. \] \begin{itemize} \ii This map is a bijection by explicit calculation: \[ (3^0, 3^1, 3^2, 3^3, 3^4, 3^5) \equiv (1,3,2,6,4,5) \pmod 7. \] (Technically, I should more properly write $3^{0 \bmod 6} = 1$ and so on to be pedantic.) \ii Finally, we need to verify that this map respects the group operation. In other words, we want to see that $\phi(a+b) = \phi(a) \phi(b)$ since the operation of $\Zc 6$ is addition while the operation of $\Zm 7$ is multiplication. That's just saying that $3^{a+b \bmod 6} \equiv 3^{a \bmod 6} 3^{b \bmod 6} \pmod 7$, which is true. \end{itemize} \end{example} \begin{example} [Primitive roots] More generally, for any prime $p$, there exists an element $g \in \Zm p$ called a \vocab{primitive root} modulo $p$ such that $1, g, g^2, \dots, g^{p-2}$ are all different modulo $p$. One can show by copying the above proof that \[ \Zcc{p-1} \cong \Zm p \text{ for all primes $p$}. \] The example above was the special case $p=7$ and $g=3$. \end{example} \begin{exercise} Assuming the existence of primitive roots, establish the isomorphism $\Zcc{p-1} \cong \Zm p$ as above. \end{exercise} It's not hard to see that $\cong$ is an equivalence relation (why?). Moreover, because we really only care about the structure of groups, we'll usually consider two groups to be the same when they are isomorphic. So phrases such as ``find all groups'' really mean ``find all groups up to isomorphism''. \section{Orders of groups, and Lagrange's theorem} \prototype{$\Zm p$.} As is typical in math, we use the word ``order'' for way too many things. In groups, there are two notions of order. \begin{definition} The \vocab{order of a group} $G$ is the number of elements of $G$. We denote this by $\left\lvert G \right\rvert$. Note that the order may not be finite, as in $\ZZ$. We say $G$ is a \vocab{finite group} just to mean that $\left\lvert G \right\rvert$ is finite. \end{definition} \begin{example}[Orders of groups] For a prime $p$, $\left\lvert \Zm p \right\rvert = p-1$. In other words, the order of $\Zm p$ is $p-1$. As another example, the order of the symmetric group $S_n$ is $n!$ and the order of the dihedral group $D_{2n}$ is $2n$. \end{example} \begin{definition} The \vocab{order of an element} $g \in G$ is the smallest positive integer $n$ such that $g^n = 1_G$, or $\infty$ if no such $n$ exists. We denote this by $\ord g$. \end{definition} \begin{example}[Examples of orders] The order of $-1$ in $\QQ^\times$ is $2$, while the order of $1$ in $\ZZ$ is infinite. \end{example} \begin{ques} Find the order of each of the six elements of $\Zc 6$, the cyclic group on six elements. (See \Cref{def:cyclic_group} if you've forgotten what $\Zc 6$ means.) \end{ques} \begin{example}[Primitive roots] If you know olympiad number theory, this coincides with the definition of an order of a residue mod $p$. That's why we use the term ``order'' there as well. In particular, a primitive root is precisely an element $g \in \Zm p$ such that $\ord g = p-1$. \end{example} You might also know that if $x^n \equiv 1 \pmod p$, then the order of $x \pmod p$ must divide $n$. The same is true in a general group for exactly the same reason. \begin{fact} If $g^n = 1_G$ then $\ord g$ divides $n$. \end{fact} Also, you can show that any element of a finite group has a finite order. The proof is just an olympiad-style pigeonhole argument. Consider the infinite sequence $1_G, g, g^2, \dots$, and find two elements that are the same. \begin{fact} Let $G$ be a finite group. For any $g \in G$, $\ord g$ is finite. \end{fact} What's the last property of $\Zm p$ that you know from olympiad math? We have Fermat's little theorem: for any $a \in \Zm p$, we have $a^{p-1} \equiv 1 \pmod p$. This is no coincidence: exactly the same thing is true in a more general setting. \begin{theorem} [Lagrange's theorem for orders] Let $G$ be any finite group. Then $x^{\left\lvert G \right\rvert} = 1_G$ for any $x \in G$. \end{theorem} Keep this result in mind! We'll prove it later in the generality of \Cref{thm:lagrange_grp}. \section{Subgroups} \prototype{$\SL_n(\RR)$ is a subgroup of $\GL_n(\RR)$.} Earlier we saw that $\GL_n(\RR)$, the $n \times n$ matrices with nonzero determinant, formed a group under matrix multiplication. But we also saw that a subset of $\GL_n(\RR)$, namely $\SL_n(\RR)$, also formed a group with the same operation. For that reason we say that $\SL_n(\RR)$ is a subgroup of $\GL_n(\RR)$. And this definition generalizes in exactly the way you expect. \begin{definition} Let $G = (G, \star)$ be a group. A \vocab{subgroup} of $G$ is exactly what you would expect it to be: a group $H = (H, \star)$ where $H$ is a subset of $G$. It's a \vocab{proper subgroup} if $H \neq G$. \end{definition} \begin{remark} To specify a group $G$, I needed to tell you both what the set $G$ was and the operation $\star$ was. But to specify a subgroup $H$ of a given group $G$, I only need to tell you who its elements are: the operation of $H$ is just inherited from the operation of $G$. \end{remark} \begin{example} [Examples of subgroups] \listhack \begin{enumerate}[(a) ] \ii $2\ZZ$ is a subgroup of $\ZZ$, which is isomorphic to $\ZZ$ itself! \ii Consider again $S_n$, the symmetric group on $n$ elements. Let $T$ be the set of permutations $\tau \colon \{1, \dots, n\} \to \{1, \dots, n\}$ for which $\tau(n) = n$. Then $T$ is a subgroup of $S_n$; in fact, it is isomorphic to $S_{n-1}$. \ii Consider the group $G \times H$ (\Cref{def:product_group}) and the elements $ \left\{ (g, 1_H) \mid g \in G \right\} $. This is a subgroup of $G \times H$ (why?). In fact, it is isomorphic to $G$ by the isomorphism $(g,1_H) \mapsto g$. \end{enumerate} \end{example} \begin{example} [Stupid examples of subgroups] For any group $G$, the trivial group $\{1_G\}$ and the entire group $G$ are subgroups of $G$. \end{example} %\begin{example} % [Center of a Group] % Let $G$ be a group. % Its \vocab{center}, denoted $Z(G)$, is the set $x \in G$ such that % $gx = xg$ for every $g \in G$; in other words, it is the set of % $x \in G$ which commute with every element of $G$. %\end{example} %You can check the center is indeed a group (some boring details\dots). Next is an especially important example that we'll talk about more in later chapters. \begin{example}[Subgroup generated by an element] Let $x$ be an element of a group $G$. Consider the set \[ \left<x\right> = \left\{ \dots, x^{-2}, x^{-1}, 1, x, x^2, \dots \right\}. \] This is also a subgroup of $G$, called the subgroup generated by $x$. \end{example} \begin{exercise} If $\ord x = 2015$, what is the above subgroup equal to? What if $\ord x = \infty$? \end{exercise} Finally, we present some non-examples of subgroups. \begin{example}[Non-examples of subgroups] Consider the group $\ZZ = (\ZZ, +)$. \begin{enumerate}[(a)] \ii The set $\left\{ 0,1,2,\dots \right\}$ is not a subgroup of $\ZZ$ because it does not contain inverses. \ii The set $\{ n^3 \mid n \in \ZZ \} = \{ \dots, -8, -1, 0, 1, 8, \dots \}$ is not a subgroup because it is not closed under addition; the sum of two cubes is not in general a cube. \ii The empty set $\varnothing$ is not a subgroup of $\ZZ$ because it lacks an identity element. \end{enumerate} \end{example} \section{Groups of small orders} Just for fun, here is a list of all groups of order less than or equal to ten (up to isomorphism, of course). \begin{enumerate} \ii The only group of order $1$ is the trivial group. \ii The only group of order $2$ is $\Zc 2$. \ii The only group of order $3$ is $\Zc 3$. \ii The only groups of order $4$ are \begin{itemize} \ii $\Zc 4$, the cyclic group on four elements, \ii $\Zc 2 \times \Zc 2$, called the Klein Four Group. \end{itemize} \ii The only group of order $5$ is $\Zc 5$. \ii The groups of order six are \begin{itemize} \ii $\Zc 6$, the cyclic group on six elements. \ii $S_3$, the permutation group of three elements. This is the first non-abelian group. \end{itemize} Some of you might wonder where $\Zc 2 \times \Zc 3$ is. All I have to say is: Chinese remainder theorem! You might wonder where $D_6$ is in this list. It's actually isomorphic to $S_3$. \ii The only group of order $7$ is $\Zc 7$. \ii The groups of order eight are more numerous. \begin{itemize} \ii $\Zc 8$, the cyclic group on eight elements. \ii $\Zc 4 \times \Zc 2$. \ii $\Zc 2 \times \Zc 2 \times \Zc 2$. \ii $D_8$, the dihedral group with eight elements, which is not abelian. \ii A non-abelian group $Q_8$, called the \emph{quaternion group}. It consists of eight elements $\pm 1$, $\pm i$, $\pm j$, $\pm k$ with $i^2=j^2=k^2=ijk=-1$. \end{itemize} \ii The groups of order nine are \begin{itemize} \ii $\Zc 9$, the cyclic group on nine elements. \ii $\Zc 3 \times \Zc 3$. \end{itemize} \ii The groups of order $10$ are \begin{itemize} \ii $\Zc{10} \cong \Zc5 \times \Zc2$ (again Chinese remainder theorem). \ii $D_{10}$, the dihedral group with $10$ elements. This group is non-abelian. \end{itemize} \end{enumerate} \section{Unimportant long digression} A common question is: why these axioms? For example, why associative but not commutative? This answer will likely not make sense until later, but here are some comments that may help. One general heuristic is: Whenever you define a new type of general object, there's always a balancing act going on. On the one hand, you want to include enough constraints that your objects are ``nice''. On the other hand, if you include too many constraints, then your definition applies to too few objects. So, for example, we include ``associative'' because that makes our lives easier and most operations we run into are associative. In particular, associativity is required for the inverse of an element to necessarily be unique. However we don't include ``commutative'', because examples below show that there are lots of non-abelian groups we care about. (But we introduce another name ``abelian'' because we still want to keep track of it.) Another comment: a good motivation for the inverse axioms is that you get a large amount of \emph{symmetry}. The set of positive integers with addition is not a group, for example, because you can't subtract $6$ from $3$: some elements are ``larger'' than others. By requiring an inverse element to exist, you get rid of this issue. (You also need identity for this; it's hard to define inverses without it.) Even more abstruse comment: \Cref{thm:cayley_theorem} shows that groups are actually shadows of symmetric groups. This makes rigorous the notion that ``groups are very symmetric''. \section{\problemhead} \begin{problem} What is the joke in the following figure? (Source: \cite{img:snsd}.) \begin{center} \includegraphics[height=8cm]{media/love-proper-isomorphic-subgroup.jpg} %\caption{$\heartsuit$ is a group, $G \subsetneq \heartsuit$ a subgroup and $G \cong \heartsuit$.} \end{center} \begin{hint} Orders. \end{hint} \begin{sol} The point is that $\heartsuit$ is a group, $G \subsetneq \heartsuit$ a subgroup and $G \cong \heartsuit$. This can only occur if $\left\lvert \heartsuit \right\rvert = \infty$; otherwise, a proper subgroup would have strictly smaller size than the original. \end{sol} \end{problem} \begin{problem} Prove Lagrange's theorem for orders in the special case that $G$ is a finite abelian group. \begin{hint} Copy the proof of Fermat's little theorem, using \Cref{lem:group_mult_biject}. \end{hint} \begin{sol} Let $\{g_1, g_2, \dots, g_n\}$ denote the elements of $G$. For any $g \in G$, this is the same as the set $\{gg_1, \dots, gg_n\}$. Taking the entire product and exploiting commutativity gives $g^n \cdot g_1g_2 \dots g_n = g_1g_2 \dots g_n$, hence $g^n=1$. \end{sol} \end{problem} \begin{problem} Show that $D_6 \cong S_3$ but $D_{24} \not\cong S_4$. \begin{hint} For the former, decide where the isomorphism should send $r$ and $s$, and the rest will follow through. For the latter, look at orders. \end{hint} \begin{sol} One can check manually that $D_6 \cong S_3$, using the map $r \mapsto (1 \; 2 \; 3)$ and $s \mapsto (1 \; 2)$. (The right-hand sides are in ``cycle notation'', as mentioned in \Cref{subsec:cycle_notation}.) On the other hand $D_{24}$ contains an element of order $12$ while $S_4$ does not. \end{sol} \end{problem} \begin{sproblem} Let $p$ be a prime. Show that if $G$ is a group of order $p$ then $G \cong \Zc p$. \begin{hint} Generated groups. \end{hint} \begin{sol} Let $G$ be a group of order $p$, and $1 \neq g \in G$. Look at the group $H$ generated by $g$ and use Lagrange's theorem. \end{sol} \end{sproblem} \begin{problem} [A hint for Cayley's theorem] Find a subgroup $H$ of $S_8$ which is isomorphic to $D_8$, and write the isomorphism explicitly. \end{problem} \begin{dproblem} \onechili Let $G$ be a finite group.\footnote{In other words, permutation groups can be arbitrarily weird. I remember being highly unsettled by this theorem when I first heard of it, but in hindsight it is not so surprising.} Show that there exists a positive integer $n$ such that \begin{enumerate}[(a)] \ii (Cayley's theorem) $G$ is isomorphic to some subgroup of the symmetric group $S_n$. \ii (Representation Theory) $G$ is isomorphic to some subgroup of the general linear group $\GL_n(\RR)$. (This is the group of invertible $n \times n$ matrices.) \end{enumerate} \label{thm:cayley_theorem} \begin{hint} Use $n = \left\lvert G \right\rvert$. \end{hint} \begin{sol} The idea is that each element $g \in G$ can be thought of as a permutation $G \to G$ by $x \mapsto gx$. \end{sol} \end{dproblem} \begin{problem} \onechili Find the smallest integer $n$ such that the symmetric group $S_n$ has a subgroup isomorphic to the dihedral group $D_{2018}$ of order $2018$. \begin{hint} For the lower bound, consider orders (note that $1009$ is prime). For the upper bound, consider a $1009$-gon. \end{hint} \begin{sol} The answer is $n = 1009$. This solution uses the fact that $1009$ is prime. To show that no smaller $m$ is possible, note that $D_{2018}$ has elements of order $1009$, a prime. Since $S_n$ has no elements of this order for $n < 1009$, we need $n \ge 1009$. To give a construction from $n = 1009$, note that $D_{2018}$ can be thought of the symmetries of a $1009$-gon. If one labels the vertices of the $1009$-gon by $S \coloneqq \{1,2,\dots,1009\}$, then elements of $D_{2018}$ induces permutations on $S$, and the set of permutations achieved is the desired subgroup. \end{sol} \end{problem} \begin{problem} [IMO SL 2005 C5] \twochili There are $n$ markers, each with one side white and the other side black. In the beginning, these $n$ markers are aligned in a row so that their white sides are all up. In each step, if possible, we choose a marker whose white side is up (but not one of the outermost markers), remove it, and reverse the closest marker to the left of it and also reverse the closest marker to the right of it. Prove that if $n \equiv 1 \pmod 3$ it's impossible to reach a state with only two markers remaining. (In fact the converse is true as well.) % https://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=90046&p=3573800 \begin{hint} Draw inspiration from $D_6$. \end{hint} \begin{sol} We have $www = bb$, $bww = wb$, $wwb = bw$, $bwb = ww$. Interpret these as elements of $D_6$. \end{sol} \end{problem} \begin{problem} \onechili Let $p$ be a prime and $F_1 = F_2 = 1$, $F_{n+2} = F_{n+1} + F_n$ be the Fibonacci sequence. Show that $F_{2p(p^2-1)}$ is divisible by $p$. \begin{hint} Look at the group of $2 \times 2$ matrices mod $p$ with determinant $\pm 1$. \end{hint} \begin{sol} Look at the group $G$ of $2 \times 2$ matrices mod $p$ with determinant $\pm 1$ (whose entries are the integers mod $p$). Let $g = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ and then use $g^{\left\lvert G \right\rvert} = 1_G$. \end{sol} \end{problem} %\begin{problem}[Hard] % Exhibit two groups $G$ and $H$ which are not isomorphic with the property that % for every positive integer $n$, % the number of elements $g \in G$ with $\ord g = n$ % equals the number of elements $h \in H$ with $\ord h = n$. % ```