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Forget about the external binaries (or the ACE package, or in fact any packages at all) at the moment -- they can give runtime improvements, but that is not what you need here. Also the web pages you link are not the best references. Go to http://www.gap-system.org. CosetTable or "coset enumeration" refers to a technical concept (a particular way to write ...

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Since you are working with $\mathbb{Z}_{7}^{*}$ you only need to consider $x = 1,2,3,4,5,6$. For each of these $x$ you need to find another $x$ in the same set which when multiplied together yields $1$ mod($7$). For instance, $2\cdot 4 = 8 \equiv 1$ mod($7$). Once you have found one inverse, you have also found another one. Consider the case $2 \cdot 4 = 4 ... 4 This is not even possible on the level of objects, let alone morphisms; not every group is a commutator subgroup. 2 The point is you're working modulo 7, so you need, for all$x<7$, to find a$y$such that$xy\equiv1\mod7$.$\mathbb{Z}_7^\ast=\{1,2,3,4,5,6\}$. So here goes: \begin{gather*} 1\cdot1=1; \\ 2\cdot4=8\equiv1\mod7; \\ 3\cdot5=15\equiv1\mod7; \\ 6\cdot6=36\equiv1\mod7 \end{gather*} Summarizing: $$\begin{matrix} \text{Element} & \text{inverse} \\ 1 ... 1 First of all, as Matt Samuel pointed out, A_{3}=<(123)> is the only subgroup of order 3 of S_{3}. The other subgroups of S_{3} are <(12)>,<(13)> and <(23)>, the trivial 1 and S_{3} itself. So, it is reasonable to get H=A_{3} as Hagen von Eitzen noticed. Now, let's show this by using group action by left ... 2 Note that for all g\in G, gHg^{-1} is a subgroup of order d. Thus gHg^{-1} = H for all g\in G and so H is normal. 2 If g\in G then gHg^{-1} is a subgroup of same order as H, hence by assumption gHg^{-1}=H for all g\in G. 0 then observe f(a^n)=\alpha^n hence range is generated by \alpha hence range will be same as G if \alpha is a generator which is possible only if \alpha=a or \alpha={a}^{-1} 1 We know that \alpha=a^n for some n\in\mathbb Z. As f has an inverse and \alpha must be a generator of G, we also conclude that a=\alpha^m for some m\in\mathbb Z. Hence a^{nm}=a. This is only possible if nm=1 as otherwise we'd have a finite cyclic group. The only solutions to nm=1 with n,m\in\mathbb Z are n=m=\pm1. 2 Let G = \mathbb Z/4\mathbb Z and let X = \{0,1\}. Let G act on X via the action$$g\cdot x = g+x\pmod 2$$Then G acts transitively on X. However, the only subgroup of G of order 2 is \{0,2 \pmod 4\}, which does not act transitively on X. 1 What is problem in visualization? How many faces it has? Can you pair up them in some nice way? For each pair, consider rotations which take the face of that pair to itself. How many rotations will there be? What it gives? 1 If P is one of the pentagons, then the rotations which preserve P (including the identity) are five, and they form a cyclic group of order 5. Since a non-identity rotation can maps two faces to themselves (and the two faces are opposite on the solid), we see that G has at least as many Sylow 5-subgroups as there are pairs of opposite faces in the ... 3 "Natural" can be given a precise meaning using the concept of natural transformations. For example, loosely speaking, for any group G there is a "natural" homomorphism [G, G] \to G, where [G, G] is the commutator subgroup of G, and similarly there is a "natural" homomorphism G \to G/[G, G]. In fact there is a "natural" short exact sequence$$1 \to ... 0 If$G$is finite group, then there is no problem in proof: If$K/M_1$is nilpotent, and$K/M_2$is nilpotent then$K/(M_1\cap M_2)$is also nilpotent. Since$K/M_1$is nilpotent, there exists$n\geq 2$such that$\gamma_n(K/M_1)=1$i.e.$\gamma_n(K)\subseteq M_1$. (Then, this subset-relation also holds for$n+1,n+2,\cdots$.) Similarly there exists ... 4 Since spherical harmonics was mentioned then this is most likely the Wigner$3j$symbol. This symbol is useful if you want to descibe the product of two spherical harmonics expanded in a series of spherical harmonics $$Y^{m_1}_{l_1}Y^{m_2}_{l_2} = \sum_{l,m}\sqrt{\frac{(2l_1+1)(2l_2+1)(2l+1)}{4\pi}}\begin{pmatrix} l_1 & l_2 & l \\[8pt] 0 & ... 0 This usually represents the function f with domain \{l_1,l_2,l_3\}, such that f(l_1)=m_1, f(l_2)=m_2, and f(l_3)=m_3. (l is a horrible name for a variable. It looks like a 1 when you draw it. \ell is better.) 2 k^{\times} is an affine variety, but it's not an affine subvariety of k. Instead it's an affine subvariety of k^2, under the embedding$$k^{\times} \ni x \mapsto (x, x^{-1}) \in k^2.$$It's the fact that this second coordinate is x^{-1} that allows you to take m \in \mathbb{Z}. More abstractly, the coordinate ring of k^{\times} is k[x, ... 2 This notation is often used to write a permutation \sigma so that l_i \mapsto m_i. 0 That 1-1 correspondende preserve indexes, that is (G:H) = (G/<a>:H'), where H' is the subgroup of G/<a> corresponding to H. This group H' can be identified with H/<a> and therefore preservation of indexes follows from Lagrange's theorem for instance. 3 From the comments, you're interested in dealing with the first part, which I'll show. For x,y\in G, x(yx)^2y = (xy)^3 = x^3 y^3, hence (yx)^2 = x^2y^2 by cancellation. So if a, b\in H, say a = x^6 and b = y^6, then$$ab = x^6y^6 = (x^3)^2(y^3)^2 = (y^3x^3)^2 = [(yx)^3]^2 = (yx)^6\in H.$$1 It is proven by Swenson in E. Swenson, A cut point theorem for CAT(0) groups, J. Differential Geom. 53 (1999), no. 2, 327–358. that each infinite CAT(0) group contains an infinite order element. 2 I think that is the "hard way" to show isomorphism. An easier way is to use the Fundamental Isomorphism Theorem, and display a homomorphism: \psi: D_{2n} \to D_{2k} with \text{ker }\psi = \langle r^k\rangle. I suggest taking \psi(r^j) = r^{j\text{ mod }k}, and \psi(s) = s. To prove \psi is a homomorphism, it will suffice to show \psi(r^n) = 1 ... 1 This is related to the first isomorphism theorem (for groups). However to prove directly what you ask for: Let f(g) = g'. Then you want to show that f^{-1} (g') = g \ker(f) (i.e. the pre-image under f of g' should coincide with a left coset of \ker(f)). Let h \in f^{-1}(g') \subseteq G, thus f(h) = g' = f(g). Hence f(g^{-1}h) = 1_{G'} thus ... 1 I will explain the first two equalities, and leave the rest to you. Let \Phi = \Phi(G). From the definitions, we have A \le A^* \le Z(\Phi), and they are all abelian. So d(A_i) = d(A \cap Z_i) \le d(A^* \cap Z_i) \le d( Z(\Phi) \cap Z_i), and we want to prove that these are all equalities. For any abelian p-group Q, we have d(Q) = ... 3 The group is isomorphic to S_5, the permutation representation is that on the cosets of the intransitive subgroup S_2\times S_3 with orbits of length 2 and 3. This was done using GAP in the following way: Construct the group, verify that it is transitive on the 10 points moved, and identify it in the list (using a prior classification) of transitive ... 0 Are homomorphisms elementary enough? The map x\mapsto gxg^{-1} is a bijective homomorphism, so it maps subgroups onto subgroups and preserves the order of all subsets, in particular of subgroups. The verification it is a homomorphism is easy; bijectivity follows from considering that the map x\mapsto g^{-1}xg is its inverse. 1 Hint Prove that the map$$\phi: h\mapsto ghg^{-1}$$is bijective. 3 This is not true. For example, S_{10} can be generated by the three involutions (2,3), (3,5)(4,7)(6,9)(8,10), (1,2)(3,4)(5,6)(7,8). Of course for transpositions the answer is easy, as you need n-1 transpositions to be transitive on n points. 1 It is not true. S_3 can be generated by two involution for example. <\{(1,2),(2,3)\}> =S_3. 1 K can partition H and H can partition G. Thus the number of K-cosets is a multiple of the number of H-cosets. 0 This follows from the observations below: Every subgroup of \mathbb Z/n\mathbb Z corresponds to a subgroup of \mathbb Z that contains n \mathbb Z. The subgroups of \mathbb Z that contain n \mathbb Z are precisely d\mathbb Z, with d a divisor of n. Every subgroup of \mathbb Z is an ideal. The canonical group homomorphism \mathbb Z \to ... 1 An ideal has to be a subgroup (of the additive group of the ring) to begin with. So if you prove that all subgroups are already ideals, you are done. PS Just correct the statement every sub-group is mZ_n for 0<m<n to every subgroup is of the form m Z_{n} for m \mid n. 2 I don't think so being that g\{x|x\in H and x\in K\}=\{gx|x\in H and x\in K\}\neq \{gx| gx\in H and gx\in K\}. Since a set is usually written as \{elements|conditional\} changing the elements in the set (in this case gH\cap K) doesn't change the conditional (in this case x\in H\cap K which isn't the same as gx\in H\cap K). Here is how ... 2 The definition of a subgroup K of a group G is that K is a group and that K\subseteq G. Since K is a subgroup of H it must be a group and H\subset G, can you see how K is a subgroup of G. Your proof looks okay but the ordering is a bit weird. Start with the fact that K\subset G and then prove that a,b\in K\implies ab^{-1}\in K. Which ... 1 Okay, that's just bad wording. The elements of H are h_1,h_2,\dots,h_m, they are all distinct so if i\ne j, then h_i \ne h_j. The elements of the coset Ha are h_1a,h_2a,\dots,h_ma and we want to show again that they are all distinct, that is, if i\ne j, then h_ia\ne h_ja. For contradiction, suppose there is some i,j such that i\ne j and ... 1 Let x\in G with x\neq e. By Lagrange theorem, the order of x is 3, 9 or 27. If x has order 3, take H = <x>. If x has order 9, then x^3 has order 9/3. Take H = <x^3>. If x has order 27, then x^9 has order 27/9. Take H = <x^9>. 0 To see the error in your proof consider the case where the x you chose has order 9. You should be able to correct your proof though, as you are very close to a solution. Simply say: if x does not have order 9, then ... and if x does have order 9, then ..... 3 you have a^{-1}ba=b^i. So multiply a^{-1} and a on left and right respectively to get a^{-2}ba^2=a^{-1}b^{i}a=(a^{-1}ba)^i=b^{i^2}. Now proceed by induction. 2 For a solvable group being Noetherian is equivalent with being polycyclic. Similarly, a finitely generated nilpotent group is Noetherian. You also have the "standard subgroup-quotient-extension-closeness", i.e. subgroups of Noetherian groups are Noetherian, quotients of Noetherian groups are Noetherian and if a normal subgroup and the corresponding quotient ... 0 I'm not sure how you could prove this without characterizing what the subgroups of a cyclic group look like. Your result follows directly from the fact that a (finite) cyclic group (of order n) has a unique subgroup of order k for each divisor k of n. [so the number of divisors = the number of distinct subgroups.] To prove this fact about ... 0 You need to show that the composition of f_a and f_b is again of the same form, so is in H as well. This shouldn't be too hard. Then you know H is closed under the operation of composition (in S_G). The identity of H is also of this form, namely it's f_a for a = ? And the inverse of f_a \in H is also of this form, namely.... Then H is ... 1 G has an element of order n, say g. Also, since |G|=2n, G has an element of order 2 (by Sylow Theorem), say h. Since G is commutative,$$\langle g\rangle\langle h\rangle =\langle g\rangle \times \langle h\rangle.$$Furthermore, this group has exactly$2n$elements, so it is equal to$G$. Therefore$G$is cyclic. 1 This is not true if$n = 2, G = \Bbb Z_2\times \Bbb Z_2$, and$a = t$is any element apart form the identity element, or if$G = \Bbb Z_4$and$a = t = 2$. Also, if$G = \Bbb Z_2$and$a = t = 1$, it is false that$G = N \cup tN$. Otherwise it's true. Assume that$a \neq e$and that$t \in \langle a\rangle$. That means that there is a$k$such that$t = ...

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Since $N$ is cyclic, it is Abelian, so if $t\in N$, then $a^{-1}=tat=att=a$, which would imply that the order of $a$ divides $2$. Since $t\in\langle a\rangle$, we must have $|a|=2$. In that case, the statement is false for $t=a=2$ in $\Bbb Z_4$, or for $t=a$ in $V$, the Klein four group.

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Yes, you will need to use the fact that $G$ has an element of order $2$. Let $x \in G$ have order $2$, and $y \in G$ have order $n$. Here is a hint. Using the fact that $G$ is abelian and the fact that $\gcd(2, n) = 1$, show that the order of $xy$ is $2n$, whence $G$ must be cyclic. Here is a proof that any group of even order must have an element of order ...

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Let $r_1 = \dfrac{a}{b}$ and $r_2 = \dfrac{c}{d}$. Then $G=\{n_1r_1+n_2r_2 \mid n_1,n_2\in\mathbb{Z}\} = \left\{\dfrac{n_1ad+n_2cb}{bd} \;\middle|\; n_1,n_2\in\mathbb{Z}\right\}$ By Euclid's Algorithm, for any integers $A$, $B$ there exist integers $m_1$, $m_2$ such that $m_1A + m_2B =\gcd(A, B)$ (where “gcd” means greatest common divisor). So $G= ... 3 An element$(n, m) \in \Bbb Z_{27}\times M$has order$\operatorname{lcm}(|n|, |m|)$, which is a power of$3$iff$|m|$is a power of$3$(remember that$|n|$is either$1, 3, 9$or$27$). 1 You're proof is fine, but more can be said. The set$P_n$of$n\times n$permutation matrices is a group under matrix multiplication, and as such$S_n\cong P_n$. Indeed, let$\{v_1,\ldots,v_n\}$be a basis for$\mathbb{R}^n$(you can take any field here, but nevermind). For each$\sigma\in S_n$, we define a linear transformation$P_\sigma$by ... 3 The proof starts with: Let$|x|=n$and... This defines$n$to be the order of$|x|$. It is not an assumption. 1 As Cameron Williams noted, but one can generalize his argument : using cardinality arguments, any group whose cardinality is not divisible by$24$will work. This is because a quotient of a subgroup of$G$has cardinality dividing$|G|$, thus if$24$does not divide$G\$, this cannot happen. I guess you can have fun finding lots of finite groups with such ...

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