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Suppose we have four elements $a,b,c,d$ of $S_n$ as in your case. Then we can "combine" them in various ways: $$((ab)c)d$$ $$(a(bc))d$$ $$a((bc)d)$$ $$a(b(cd))$$ $$(ab)(cd)$$ The problem is: are all those combinations equal? Suppose they are (indeed they are), then we can simply define $abcd$ to mean any one of the preceding coupling, because ...

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The gist of your question was how to "break out" an element of a product of three terms. Associativity gives you $$\begin{eqnarray} (a(bc))d=((ab)c)d=(ab)(cd)=a(b(cd))=a((bc)d) \end{eqnarray}$$ directly. The second equality uses associativity applied to the product of $ab$ (a single element of the group), $c$, and $d$. The third uses it applied to the ...

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I won't calculate the exact result but let me give you some ingredients: You should know how to transform a matrix to another via a change of basis. This way we can work with a preferred one. Mine is the standard basis. Since you used $e_i$ already lets call them $v_1,v_2,v_3$. You hopefully know how the permutation matrices that you called ...

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Your question (and the answer you gave) show that you've missed an essential point. In group theory at a very early point one remarks that the associativity axiom can be iterated to show that any two pure product expressions (no inverses) of the same string of variables, but grouped differently by parentheses to show which sequence of multiplications is ...

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Totally correct, associativity means that the order of the operations doesn't matter (as long as the elements operated on stay in the same order), so you can drop parenteses (no need to tell in which order operations are done, if the result is the same regardless). But you can go further, as $f f^{-1}$ is the identity: $$(f^{-1}gf)(f^{-1}hf) = ... 3 Indeed, @Oliver hit the answer. A fact you may find useful is: If x and y are two permutations of a set \Omega, such that x=(\xi_1,\xi_2,...,\xi_k) then$$y^{-1}xy=(\xi_1^y,\xi_2^y,...,\xi_k^y)=x^y

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Just a sketch of a proof; sorry, this is all I can do with the time I have right now. If $m\in\mathbb N$ is arbitrary, then an $m$-reciprocal polynomial means a polynomial $p \in \mathbb Z\left[q\right]$ whose coefficient before $q^i$ equals its coefficient before $q^{m-i}$ for every $i\in\mathbb Z$ (this implies that the degree of $p$ is $\leq m$). Let ...

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Think about a circular conveyor belt where $n$ boxes are running in circles constantly. Some guy is standing at a fixed position at the belt and can only swap the two boxes right in front of him. By waiting for different boxes to come by he can swap any two neighboring boxes. Can he arrange the $n$ boxes in any given order? How does this relate to ...

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As you note, every permutation can be expressed as a product of disjoint cycles. The order of such a product is given by the least common multiple of the cycle lengths. The smallest $n$ such that $S_n$ contains a permutation of that form is the sum of the cycle lengths. Since 101 is prime, the only disjoint cycle expression of a permutation of order 101 ...

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I fill in Ted's exquisite answer. This is tricky - the two cases look the same, but they're not. The first one is a right action $v \cdot (p_1 \cdot p_2) = (v \cdot p_1) \cdot p_2$, while the second one is a left action $p_1 \cdot (p_2 \cdot f) = (p_1 \cdot p_2) \cdot f$. To see why, consider these 2 permutations: $\color{tomato}{p_1(1) = 1, p_1(2) = 3, ... 2 You ask about the element$(2,3,1)$. What this element means is that the second position is sent to the third position, the third to the first, and the first to the second. But notice this is the same as$(1,2,3)$and$(3,1,2)$. So$(2,3,1)$is in$S_3$, and you'd already mentioned it - but you called it$(1,2,3)$above. 0 The intuition behind say$S_3$is just how many bijective map you can have from a set$\{1,2,3\}\to \{1,2,3\}$I leave it to you that there can be$6$such function, and you must write them explicitly to get a clear picture.$f_1=Id,f_2,f_3,f_4,f_5,f_6$say and then we use usual composition of function as group operation.$f_2=(123)$say so$f_2$sends ... 1 The symmetry group is$A_5$, which does not contain an element of order$6$. 1 It seems to be easier just to observe that the six sides of one of the hexagons are not all equal -- three of them border on pentagons and three of them border on hexagons. So a 60° rotation like you describe would take pentagons to hexagons and vice versa -- it wouldn't be a symmetry at all. 0 Here is a more nuts-and-bolts answer to the question of a path to understanding frieze groups: a quick outline of the proof of the classification theorem(s). The structure of the proof is basically the same for frieze groups, and for 2d and 3d crystallographic groups, although of course there are many more cases to consider as one ups the dimensions. To keep ... 1 In a word: symmetry. For example, you may have already encountered the set of all symmetries of the square: all the ways of placing a square on top of itself. There are 8: four rotations (including the 360 degree rotation), plus four reflections (flips). These form a group under composition (first do one symmetry, follow it with another). The frieze groups ... 2 To find Sylow$p$-subgroups in a group in which you can generally find$p$-elements and normalizers the following algorithm is sometimes used: Assume$P_i$is a$p$-subgroup of$G$(we can take$P_0 = \{1_G\}$to be the trivial subgroup). If$P_i$is a Sylow$p$-subgroup, then yay, return$P=P_i$. Otherwise we know by theory* that$P_i$is not a Sylow ... 2 Let$f:C_8\rightarrow S_3$be non-trivial homomorphism. Then one can see that$\frac{C_8}{Kerf}$is isomorphic to a subgroup of$S_3$. Note that$|\frac{C_8}{Kerf}|$is power of$2$. Thus,$\frac{C_8}{Kerf}\cong \langle \sigma \rangle$where$o(\sigma)=2$. Now, Assume that$C_8=\langle x \rangle$. We can define homomorphisms$f_1(x)=(1,2)$,$f_2(x)=(1,3)$... 1 Let$\sigma_1,\sigma_2,...,\sigma_p$be a collection of pairwise commuting$p$cycles in$S_{p^2}$, where$\sigma_i =(a_{i,1},a_{i,2},..,a_{1,p})$. Choose a$p^2$cycle$\tau =(a_{1,1},a_{2,1},..,a_{p,1},a_{1,2},a_{2,2},..,a_{p,2},...,a_{1,p},a_{2,p},..,a_{p,p})$. Clearly this$\tau^p=\sigma_1.\sigma_2..\sigma_p \in <\sigma_1,\sigma_2,...,\sigma_p>$... 4 The additional generator you could take to generate a Sylow$p$-subgroup of$S_{p^2}$is$(1,p+1,2p+1,\ldots,(p-1)p+1)(2,p+2,2p+2,\ldots,(p-1)p+2) \cdots (p,2p,3p,\ldots p^2)$For example, for$p=3$, this would be$(1,4,7)(2,5,8)(3,6,9)$. This normalizes the elementary abelian subgroup of order$p^p$that you have already found: check that it conjugates ... 6 In general, if the prime$p$divides$n$, then the permutation representation of$S_n$over the field${\mathbb F}_p$has two irreducible constituents of degree$1$and one of degree$n-2$and, for$n \ge 5$, this results in an embedding$S_n \to {\rm GL}(n-2,p)$. Also,$S_4$embeds into$GL(2,{\mathbb Z}_4)$: you can take the images of$(1,2)$and ... 3 As the following shows, GAP can work out generators of the kernel. Perhaps you were put off by the fact that it said "no generators known". That just means that it has not calculated them yet. The other possibility is that the example you gave it was too large. If you tried mapping$F_{20}$to$S_{20}$for example, then$K$would have far too many ... 4 By the standard theorems on covering spaces, conjugacy classes of maps$\varphi : F_n \to S_k$are naturally in bijection with$k$-fold covers$Y$(not necessarily connected) of a wedge$X = \bigvee_{i=1}^n S^1$of$n$circles. The cycle type of the image of a generator of$F_n$tells you what the preimage of the corresponding copy of$S^1$should be in this ... 4 I think the$S_5$work is correct (sorry, haven't looked at the$A_5$work). I would do it a somewhat different way: Since 5, but not 25, divides 120 (the size of$S_5$), the Sylow-5 subgroups of$S_5$must be cyclic of order 5. There are 24 5-cycles in$S_5\$, 4 of them in each of these subgroups, so, 6 Sylow-5 subgroups. Similarly, the Sylow-3 subgroups ...

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Comparing two infinite groups is a very hard problem. If you know that one of them is crystallographic, then GAP has packages Cryst - Computing with crystallographic groups and CrystCat - The crystallographic groups catalog. It might happen that these may help here, but I suggest to ask this in the GAP Forum or GAP Support to reach package authors (Just ...

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