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A $p$-group is a group where the order of every element is a power of $p$. A direct consequence of this is that if the order of every element is a power of $p$, then the order of the group must be a power of $p$. This follows from Lagrange's theorem: the order of a subgroup of a group divides the order of the group. So, one can say that a $p$-group is a ...

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When $n=2$, the braid group $B_n$ is just the free group on one generator $\sigma$, and the representation is determined by the matrix via which $\sigma$ acts on $\mathbf{C}^2$; for $z=-1$ this is $$\sigma \mapsto \left( \begin{matrix} 2 & -1 \\ 1 & 0 \end{matrix} \right).$$ This matrix is conjugate to $$\left( \begin{matrix} 1 & 1 \\ 0 & 1 ... 1 Hint: Take the characteristic homomorphism \rho: \mathbb Z \to F, where \rho (n) = n 1. Notice that F is a domain and its characteristic must be a prime p and \rho (\mathbb Z) \simeq \mathbb Z_p, because its kernel is I(p), that is, the ideal generated by p is the smallest subfield of F. 0 Yes, the set of values 0,1,1+1,1+1+1,\dots, is closed under multiplication and addition (why?), and it must be a subgroup under addition. Since it is necessarily cyclic and finite, it is of the form \mathbb Z/n\mathbb Z. Use field properties to show that n must be prime. 0 Yes it's true that every finite field contains \mathbb F_p for one value of p. In fact every field contains either \mathbb F_p for some p, or \mathbb Q. 1 HINT: Given f\in F, consider the following functions: for each m\in\Bbb N let g_m\in F be defined by$$g_m(k)=\begin{cases} f(k),&\text{if }k\ne m\\ f(k)+1,&\text{if }k=m\;. \end{cases}$$1 Let n=|\sigma|, then n|p so n=1 or p. If \sigma\ne e, then |\sigma|=p. Now since 1\le r<p, r and p are coprime and so |\sigma^r|=p. Finally let i\in\{1,\ldots, p\}. The orbit of i consists of i,\sigma(i),\sigma^2(i),\ldots,\sigma^{p-1}(i), therefore \sigma is a p-cycle. 2 For the first question, it is possible for G to be finitely presentable. The free group F on two generators a,b is the semidirect product of the normal closure of a in F with the infinite cyclic subgroup generated by b. (This is the same example that you gave. In your example, F is freely generated by t and x_0.) The answer to the second ... 3 Since there is an answer up, I would do$$ax^2a=x^3x^2=a^2x^2a^2=ax^3a=x^3axaaxa=x^{-1}ax^2a=x^{-2}=x^3$$Leaving you with some bits to fill in. 0 Exercise: let G be a simple group and fix a field K. Then every nontrivial linear representation of minimal dimension is faithful and irreducible unless G is cyclic of order p and p=0 in K. In particular, if the simple group G admits a nontrivial linear representation over K (e.g., G is finite), then it admits a faithful irreducible ... 5 a^{-1}x^2a=x^3 \implies a^{-1}x^4a=x^6 \implies a^{-1}x^6a=x^9 but x^6=x^3x^3=(a^{-1}x^2a)(a^{-1}x^2a)=a^{-1}x^4a \implies a^{-1}(a^{-1}x^4a)a=x^9 \implies a^{-1}a^{-1}x^4aa=x^9 now using a^2=a^{-2}=e we have x^4=x^9 so x^5=e and because 5 is prime x=e or x have order 5. 1 (We only consider the complex representation) Suppose that G is non-Abelian. Then there is an irred repr. \rho s.t. it is d-dim'l, d>1. By dimension theorem, we get d=p,q, or pq, but we know that d must be p; otherwise by the property that \sum_{\chi'\in\text{Irr}(G)}d_\chi^2=|G|, we will get d^2>pq because of the condition ... 2 Every finite group is isomorphic to a subgroup of S_n. Now take G a subgroup of odd order included in S_n. Let g\in G\subseteq S_n, write its unique decomposition into cycles :$$g=c_1...c_r\text{ then } \epsilon(g)=\epsilon(c_1)...\epsilon(c_r) $$Now you know how to compute the signature of a cycle, namely if c is a cycle of length l then ... 2 Hint Say, left multiplication by any element g of a finite group G permutes the elements of G (and by uniqueness of the identity element this permutation is distinct for each element g), so we can always regard G as a subgroup of S_G \cong S_{\# G}. For any m there is an injective homomorphism S_m \hookrightarrow A_{m + 2}. Remark Putting ... 1 As noted, the quotient is an abelian group of order 48. The element x:=(1,01) is of order 24 in the quotient since$$ 24(1,0,1)=(24,0,24)\equiv(24,16,32)=8(3,2,4) but if the order of x was less then 24 then by Lagrange it would have an order dividing 24. Since 24=2^{3}\cdot3 |x|\mid24 and x\neq24 imply |x|\mid12 or |x|\mid8 but ... 2 Yes. It follows easily from the fact that maximal subgroups of finite supersolvable groups have prime index. 1 An answer in the negative for this question would immediately prove the Smith conjecture: Does the finite part of the automorphism tower of a group G eventually repeat? (Smith 1964) I have spent some time wrestling with this conjecture, and the pattern I observed was that this did occur and typically very rapidly (only a few steps in, I was using GAP to do ... 1 (a) If H is a proper subgroup of G, then the order of H is a divisor of 2n. If n is prime, the only possible divisors of 2n are 1, 2, n or 2n. Since the last one is not possible,.... (b) Let x \in G be x \neq e. Then the order of x is a divisor of n^2, thus 1,n or n^2. The order cannot be 1. If the order of x is n then ... 1 First you need to check whether it is a group. Just checking closure under addition is not enough. First closed under addition (done), second associativity ((ab)c=a(bc)), then existence of identity ((1,1)), then existence of inverse ((a^{-1},b^{-1})). Communitivity is OK, but it seems not to be the natural way people write. (a,b) \oplus (c,d) = (c, ... 0 You forgot to find a inverses for (a,b) and you forgot to find an identity, you also forgot to prove the operation is associative. Also not this method for creating a new group from two old ones can be generalized and is called the direct product (sometimes called direct sum when working with abelian groups). 1 To show that f is a homomorphism, we must show that f((a,b) \oplus (c,d))=f((a,b))+f((c,d)). So, \begin{align} f((a,b) \oplus (c,d)) &= f((a+c,b+d))\\ &=a+c-(b+d)\\ &=a+c-b-d\\ &=a-b+c-d\\ &=f((a,b))+f((c,d)) \end{align} So, f is a homomorphism. To show f is onto (surjective), we must show that for every y \in \mathbb Z, there ... 1 |ab| does not in general divide lcm(|a|,|b|). See the earlier post, with a link to a nice proof. Order of product of two elements in a group 0 This is the direct product of four copies of C_2 (the cyclic group with 2 elements). So you can think of this group as \mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z. The elements are (a,b,c,d) with a,b,c,d \in \{0,1\} 2 Unfortunately, no. In fact, it's even worse! The quaternion group Q_8 has subgroups of all possible orders, and every subgroup is normal, but the group isn't abelian! More generally p-groups, groups whose orders are a power of a prime, will have subgroups of all orders, yet need not be abelian. There is a partial converse to Lagrange's theorem, about ... 10 The answer to your question is no, that is not possible. This is a consequence of the following slightly more general result. Let d(n) is the number of divisors of n. Theorem Let G be a finite group of order n. Then for any divisor m of n, there exist at least d(m) subgroups of G of order dividing m. Proof We prove the statement by ... 1 In short, you are correct. (15)(36) is the right representation of a as a product of disjoint cycles, and a^{24} is the identity while a^{25} is just a again, because a has order 2. Note that it doesn't matter if we write (51)(63) or (15)(36) - in general cyclic permutations represent the same cycle. 2 I like the construction Tobias has posted. Here is a more hands-on method. The Sylow theorems tell us that a group of order 21 has a normal subgroup of order 7. Let this be generated by a, and let b be an element of order 3. Then, because a generates a normal subgroup, we have bab^{-1}=a^r for some 1\le r \le 6. If r=1 the group is ... 3 Here is a way to construct the nonabelian group of order 21 without referring to semidirect products. Let F be the field with 7 elements, which we will identify with the set \{0,1,2,3,4,5,6\}. Note that the elements H = \{1,2,4\} form a subgroup of the units of F of order 3. Now define G = \{f: F\to F\mid f(x) = ax + b,a\in H, b\in F\}. ... 0 If I understand the question you are being asked, your scenario is the following: (1) The set S is your domain; in particularly, S = \mathbb{R^+}. (2) The binary-operation you are working with is a \circ b = 4ab^2 (3) You are being asked to find the identity for this operation. The difficulty is here is that you will not find an identity ... 0 For finite groups, your guess is correct; not sure about the infinite case. For a proof, let G = HK, suppose that H \unlhd G, K \unlhd G, and that G/H \cong K. To show that G\cong H \times K, we need only show that H \cap K =1. We appeal to the diamond isomorphism theorem, which says in this case that G/H=HK/H \cong K/H \cap K. But since G/H ... 0 Notice that D_4/Z(D_4) is a group of order 4 (in particular, its the Klein 4-group), and thus abelian. The map in question is simply conjugation in an abelian group, so it is the identity. 0 In general, it is not valid. a^2=e does not imply a=e. Even in D_4 it is not true I'm afraid. The condition and conclusion is right here though. \overline{\phi}=\mathrm{Id} just means that the quotient group is commutative. Also note thatD_{2n}=\langle r,s |r^n=s^2=e, s^{-1}rs=r^{-1}\rangle$$or in your notation it means ... 2 We need these conditions: 1.) G=HK 2.) H and K normal in G. 3.) H \cap K= \{1\} Then G=HK \cong H \times K 6$$1=1-r^n=(1-r)(1+r+\ldots+r^{n-1})$$1 Your second guess is a correct description of \Bbb Q(\pi) (as long as you specify that the denominator is not the zero polynomial). The first guess describes the ring \Bbb Q[\pi], which does not have inverses. In particular, \pi has no inverse since \pi f(\pi) \ne 0 for any polynomial f\in \Bbb Q[x]. It's not hard to see that these rational ... 0 It is your second option. The first is called \Bbb Q[\pi]. In the example they make F=\Bbb Q and \alpha=\pi. 9 If \pi\in S_5 is a permutaion of \{1,2,3\}, then either \pi or \pi\circ (4\,5) is an even permutation, i.e., \in A_5. This allows us to embed S_3\to A_5. Or have fun with geometry: A_5 is the symmetry group of the dodekahedron. It is possible to select 4 of its 20 vertices that make up a regular tetrahedron T_1. A rotation of the ... 2 Every element in a group has an inverse. In your case, there must be some element x \in \{a,b\} for which b * x = a. To see this another way, note that no two elements in a row are equal. This is because if b * x = b * y, then by multiplying on the left by b^{-1}, we have x = y. 1 Although you said that you didn't want a yes/no answer, my hint to you is that the answer is yes, so you should just try to construct a subgroup, H, of order 6. Look for elements \sigma and \rho of order 2 or 3. A priori there are two possibilities for H - namely \mathbb Z_6=\langle\rho, \sigma | \sigma \rho \sigma ^{-1}=\rho, \quad ... 2 This is kind of a silly example: \mathbb{Z} \subseteq \mathbb{Q}, but the Krull dimension of \mathbb{Z} is one while the Krull dimension of \mathbb{Q} is 0. 1 Look at the answer to this StackExchange question: http://math.stackexchange.com/questions/132729/a-free-submodule-of-a-free-module-having-greater-rank-the-submodule (and also the MO discussion referenced therein). 0 If G is finite, then it is uniquely p-divisible if and only if p does not divide |G|. (This is a generalisation of your example.) See for example: http://groupprops.subwiki.org/wiki/Kth_power_map_is_bijective_iff_k_is_relatively_prime_to_the_order 6 2014 = 2 * 19 * 53 As 2, 19, 53 are all distinct, there is only one Abelian group of order 2014 up to isomorphism, which is \mathbb{Z}_{2014}\simeq\mathbb{Z}_{2}\oplus\mathbb{Z}_{19}\oplus\mathbb{Z}_{53}. Also you can proceed this way without using the classification theorem. By Cauchy's theorem there are elements of 2, 19, 53. Suppose they are ... 0 So let S_i be the words generated beginning with A or A^{-1} and let T_j be the words that beginning with B or B^{-1}, because S_i and T_j can have two options, both m and n are both 2? 1 The group SL_2(\mathbb{R}) has a natural but nonfaithful action on the projective space \mathbb{R}P^1. Letting M \in SL_2(\mathbb{R}) and letting \ell_{\vec v} \in \mathbb{R}P^1 be the line through the origin with a direction vector \vec v, the action is given by$$M \cdot \ell_{\vec v} = \ell_{M \vec v}  The kernel of this action is all ...

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Hint: By Langrange's Theorem if $H < G$ then $|H|$ divides $p$. As $p$ is prime then we must have $|H| = 1$ or $|H| = p$. What can you conclude from that?

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A group of order $p$ has $2$ subgroups: the trivial subgroup and the entire group.

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Let $H$ be elementary abelian of order $p^n$ for an odd prime $p$, and let $G = H \rtimes \langle t \rangle$, with $t^2=1$, where $t$ inverts every element of $H$. (This is soemtimes called a generalized dihedral group.) Then generating sets for $G$ have size at least $n+1$, but $|G/[G,G]|=2$.

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The use of Tietze transformations to prove that $\langle a,b;(ab)^2\rangle\cong \mathbb{Z}*\mathbb{Z}_2$ is correct. There are no mistakes in that. The mistake is to claim that $\langle a,b;(ab)^2\rangle$ is a presentation of the projective plane. In fact, the relation $\langle a,b;(ab)^2\rangle\cong \mathbb{Z}*\mathbb{Z}_2$ together with the Classification ...

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For a concrete example, groups with a non-trivial center acting on their set of subgroups via conjugation work well (in fact, I think Hagen von Eitzen gave you a description of all non-faithful group actions). In the dihedral groups acting on a regular $2n$-gon with $n$ even, the half-rotation is in the center of the group, and fixes every subgroup.

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