Let K be a normal subgroup of G and |K|=k. If k and [G:K] are relatively prime, prove that K is the unique subgroup of G of order k.
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You can use the Sylow theorem to show this fact.\n\nLet p be a prime dividing k.\nFirst let us show that any p-Sylow subgroup of G lies in K.\nBy your hypothesis, any p-Sylow subgroup H of K is also a p-Sylow subgroup of G.\nYou know, by Sylow's theorem, that any p-Sylow subgroup H' of G is a conjugate of H. So that, K being a normal subgroup of G, H' must also lie in K.\nSo that the claim is proved : any p-Sylow subgroup of G lies in K.\n\nBut any subgroup K' of G of order k is generated by its p-Sylow subgroup (p varying among the divisors of k).\nSo, by the above claim, you can see that K' must be a subgroup of K.\nSince these groups K' and K have the same order, it means that they're equal.\n\nHINT FOR ADDITINAL QUESTION : the special case when k is a power of a prime number p says that if a p-Sylow subgroup is normal then it is the only p-Sylow subgroup.\nSo you can surely mimic the part of the proof of Sylow's theorem that deals with the fact that all p-Sylow subgroups are conjugated...
173885
I apologize if this is too simple, but it seems this is the obvious answer:\n\nSU(2) is a collection of 2x2 matrices, so the sequence { s_j } is just a subset.  The diagonal matrices in the special unitary group consist of [[a,0],[0,conj a]] where |a| = 1.  A "diagonal subsequence" should be a subset of this group.  You should be able to identify such a subsequence with complex numbers which lie on the unit circle.
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Let n,m in N and h,g in H\nwhat you need to show is  that\nnh(mg)^(-1) is in NH.\nbut nh(mg)^(-1) = nhg^(-1)m^(-1)\n But  hg^(-1)m^(-1) is in hg^(-1)N.\nBecause N is normal, this last set is included in N(hg^(-1))^(-1)) = Ngh^(-1) so it is further included in NH.\nThen hg^(-1)m^(-1) is in NH so \nnhg^(-1)m^(-1) is even more in NH.\nProof done.
184858
To begin with, an answer.\nLocker N will be open only if N is a square number.\nhas an odd number of divisors, including 1 and N. (1,4,9,16,,...)\n\nNow why it is true:\n\n1) Easy to see, that locker will be open if it was touched by students (for opening or closing) an odd number of times.\n\n2) First student touched every locker. Second - every second , for which N must be divisible by 2. Third - every third, for which N must be divisible by 3.... m-th touched every m-th, for which N must be divisible by m. \nSo, N-th locker was touched by m-th student only if N is divisible by m.\nSo, to be touched an odd number of times, N must have an odd number of divisors.\nNow let's prove that it is true only for square numbers.\n\n3) Let's build a prime factorization of N:\nN = p_1^a_1 * p_2^a_2 * ... * p_n^a_n, where p_1=2, p_2=3, p_3=5... - sequence of prime numbers, a_1...a_n - powers of them. There is only one possible such factorization.\nEasy to see, then m is divisor of N only when\nm = p_1^b_1 * p_2^b_2 * ... * p_n^b_n and\nb_1 <= a_1, b_2 <= a_2, b_3 <= a_3, ... b_n <= a_n.\nThere are a_1 + 1 possible values of b_1.\nThere are a_2 + 1 possible values of b_2.\n...\nThere are a_n + 1 possible values of b_n.\nAnd all of (b_1, ... b_n) are independent.\nSo, the number of divisors is:\nS = (a_1 + 1)*(a_1 + 1)*...*(a_n + 1).\n\nIf any of a_1...a_n is odd, the product S is even.\nSo, S can be odd only if ALL a_1...a_n are even.\nBut it means that N = p_1^a_1 * p_2^a_2 * ... * p_n^a_n is a square number.\n\nQED
394112
Without knowing the relationship between successive elements, I think the best you can do is this:\n[k=0, n-1]∑(x_(k+1) - x_k)²\n[k=0, n-1]∑(x_(k+1)² - 2x_k*x_(k+1) + x_k²)\n[k=0, n-1]∑x_(k+1)² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n[k=1, n]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n\nx_0 + x_n + 2[k=1, n-1]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1))
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If n is contains an odd prime factor, this is part of an unsolved problem called Beal's conjecture. This says that if any\nsuch solution exists, then a and b cannot be coprime.\nIf n is a power of 2, it is a multiple of 4, since n> 2.\nA glance at Wikipedia also shows that this\ncase is also part of Beal's conjecture.\nAt the present time no counterexamples to\nBeal's conjecture are known.\nIncidentally, there is a $100000 prize for the\nfirst proof or counterexample to Beal's conjecture.
418375
If h(x) is I-able, then h(x) is defined and exists for any x in between [a,b]. So Int{x=a until b} of h(x)*dx=H(b)-H(a). And back: If h(x) is defined in [a,b], then h(x) is I-able, being |H(b)-H(a)| a definite number.\nNow if both f(x) & g(x) are defined, then h(x)=f(x)*g(x) is also defined, thus h(x) is I-able.\nIf g(x)=f(x), then f^2(x) is also I-able.\nmay it help you Kevin!
212445
I think so, yes.\n\nBy contradiction, if g is not injective then exists x0,x1 member of B such that g(x0)=g(x1)\nAs f is bijective, there exists y0,y1 member of A st f(y0)=x0, f(y1)=x1)\n\nSo the composition f0g maps y0,y1 onto the same point in C\nThis contradicts the assumption of f0g being bijective\n\nI'm pretty sure you can do a similar thing for surjectivity
386126
a. This is true. \nThe example is mod n group over integers.\nm mod n is the remainder whe m is devided by n.\nThis mod n group will have exactly n number of elements.\n\neg. mod 3 = {0,1,2} and  mod 5 = {0,1,2,3,4,5}\n\nb. This is false .\n\nThe example are mod 3 and the set {i, -1, 1}\nThey dont have same group structure, also both have different operations for group structure. One had remainders while other has complex number multiplication, addtion etc.\n\nc. This is true.\nThis is by definition of subgroups.
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Looks like three questions to me. Each one might be worth 10 pts, but not all three together. Break it up.\n\nHint:\n3. X is separable. Pick a point from a countable dense set in each open set.\n1+2. Show that any uncountable set has a condensation point.
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Every group has a compact normal subgroup, namely the identity.  \nAs noted, the scalar matrices form the center of your group G.  Therefore the two-element subgroup {plus/minus I} is also normal in G.\n\nAs noted, the group SL_n is normal in G.  Therefore if H is also normal in G, its intersection H' with SL_n is normal in SL_n.  It is known that if H' is a normal subgroup of SL_n, then H' is contained in the scalar matrices, which are trivial if n is odd and {plus/minus I} if n is even.\n\nIf H is not equal to H' it must contain a matrix of determinant not equal to 1.  If this determinant is not plus/minus 1, then H is not compact.  Let us note that the set of matrices with determinant plus/minus 1 is a normal subgroup of G containing SL_n.\nFor that matter, also note that the set of matrices with positive determinant is normal in G.\n\nThis discussion should enable you to classify the normal subgroups of GL_n.  If you mean *Lie* subgroups we are done.  \nIn particular, there are exactly two compact ones, already noted.
201812
Isomorphisms between groups are not the same thing as isomorphisms between, for example, rings.  There is a fully general definition of isomorphism that stems from universal algebra and mathematical logic, but I risk boring you if I develop it here.  So let me tackle the specific case of normed vector spaces, as I suspect it is in this setting that you came across the term "isometric isomorphism".\n\nTwo normed vector spaces are said to be isomorphic if there is a bijective map f between them that is homogeneous and linear, and for which there is a constant C>0 such that (1/C) ||x|| < ||f(x)|| < C ||x|| for all x.  An isometric isomorphism is an isomorphism for which that constant C can be taken to be 1, provided you reinterpret the symbol < as less than or equal--it's not easy to type the appropriate symbol.  In such a case, distances are not only roughly preserved (as in the case in which one has an isomorphism with constant C>1): they are indeed actually exactly preserved.  Thus the name isometry.
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Necessary and sufficient condition is a is not the neutral element  and m is a prime superior to 2
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Yes. What's more, K is a normal subgroup of S_4 (the group of all permutations). \nOne has to show that p^(-1)kp is in K for every k in K and every permutation p. Now each k in K is a pair of disjoint transpositions, and K consists of ALL such pairs of swaps (and identity). What happens when p^(-1)kp is applied is that the elements of {1,2,3,4} get "renamed", then arranged in two pairs and swapped within each pair as per k, and finally renamed back. This permutation is just another pair of swaps, so it is in K. \nMore formally, if k = (ij)(mn) then p^(-1)kp = (p(i)p(j)) (p(m)p(n)). The argument is placed on the left, as usual with permutations, so the 1st to apply in p^(-1)kp is p^(-1), then k then p. So p(i) and p(j) are first mapped to i and j, then swapped, and then mapped back to p(j) and p(i). The result is that p(i) and p(j) get swapped. Same with m and n.
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sigma(2^k) = (2*2^k - 1)\nsigma(2*2^k-1) = 2*2^k when 2*2^k - 1 is prime.\nSo whenever 2*2^k - 1 (or 2^k+1 -1) is prime and k is even this is true.\nSo this works for A = 2, 4, 8, and possibly other powers of 2.
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By the Sylow Theorems all of the three element subgroups are conjugates of each other, so for one of them to be normal there must only be one such subgroup.  Likewise for the subgroup of order five.\n\nIf we denote these two subgroups as { e, a, a^2 } and { e, b, b^2, b^3, b^4 } then we see that\n\n1 element { e } has order 1\n2 elements { a, a^2 } have order 3\n4 elements { b, b^2, b^3, b^4 } have order 5\n\nWhat is the order of the remaining 15 - (1 + 2 + 4) = 8 elements.  It is not 1, 3, nor 5 but it divides 15.  So every 15 element group has eight elements of order 15 and so is cyclic.
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A subgroup H is generated by a set S if H is the smallest subgroup that contains S. This will mean that every element of H can be written as a finite product of elements of S and their inverses.\n\nIn your case, H will consist of {1, (12), (34), (12)(34)}. No other products are possible with (12) and (34) along with their inverses. \n\nTo show that H is normal in G, you need to show that xHx^-1=H for all x in G. Consider \n\nK={x in G:xHx^-1=H}.\n\nShow this *is* a subgroup of G (easy) and then show that K contains (12), (34), and (13)(24). If you do this, K will have to contain G and hence will equal G. \n\nTo show [G:H]=2, just notice that (12) and (34) are already in H and that the square of (13)(24) is in H. This will show there are two cosets of H in G. The rest follows.