(1) Let G be a non-abelian group of order 50. Prove that G has at least one element of order 5 and at least one subgroup of order 5.\n\n(2) Prove that a non-cyclic group of order 55 has at least one element of order 5 and at least one element of orfer 11. (I think this has something to do with the number of elements in G of a prime order being of the form m(p-1) where m>=0)\n\n\n\nHi guys,  am hoping someone can help me out with these questions for my honours algebra course (exam in 3 weeks!!!) I do have a vague answer sheet for them but I was hoping someone could explain them to me a bit better. Thanks very much.
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Both are results of the CAUCHY THEOREM which says that for any prime p and finite group G, if p||G|, then G contains an element of order p.  Now note that for any element a in any group G, {a^j| j is an integer} is ALWAYS a subgroup of G because for any element a^j, its inverse a^(-j) is in this subgroup, it contains the identity element (a^0), it's closed because a^ia^j = a^(i+j) and the integers are closed, and it inherits its associativity from G.  In this case, we have an element a of order p so that a^p = 1 and hence for all integers j, a^j = a^t for some nonnegative t < p.  This is because any integer j can be expressed as mp+r for some integer m and 0 <= r < p so that a^j = a^(mp+r) = a^(mp)a^r = (a^p)^m*a^r = 1^m*a^r = a^r.  Therefore this subgroup will contain the elements, 1,a,a^2,...,a^(p-1) and hence p elements as desired.
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1 4 2 5 3\n\nBut it's cyclical, so the sequence is 1 4 2 5 3 1 4 2 5 3 1 4 2 5 3 ... and so on.\nYou can start anywhere in the sequence. The first answer you received is correct, but that person chose to start with 3 instead of 1.
185784
How many ways to pick 5 from 20 if order isn't important?\nAnd, how many ways to pick 2 from 7 if order isn't important?\n\nFor the second question, you have... 1,2 1,3...1,7 or 6 that include 1. Then 2,3 2,4... 2,7 which is 5 more. Then 3,4 3,5...3,7 which is 4 more. Keep doing this and you will see the answer is 6+5+4+3+2+1 = 21\n\nFor the first part 5 out of 20, note that the total number of arrangements of 20 things is 20! which is 20 x 19 x 18 x17 ..... x 2 x 1. But for every arrangement that you take the first 5 of, there will be 15! ways to arrange the rest.\nAnd for those first 5, you can get them in any order which means that 1 2 4 3 5 is the same for you as 2 3 1 4 5. So, you need to divide the 20! by both 15! and 5! which looks like this-  20!/(15! x 5!) and equals 15,504.\n\nNow for each pick from the cable houses (15,504 possible) you have 21 different ways to pick a no cable set. Therefore, multiply 15,504 by 21 to get your final answer. 325,584
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Wow this is a long question.... hehe.\n\n1) Firstly consider any integer divided by 6, we must get one of the residues ( remainders ), r = 0,1,2,3,4,5. If the remainder of y divided by 6 is r, thus we can write \n y = 6 x "quotient" + "remainder" = 6k + r, where let k denote the value of the quotient. Since r = 0,1,2,3,4,5, we have just shown that any integer y ( we may assume positive ), can be written in one of the "forms"\n y = 6k , 6k + 1, 6k + 2, 6k + 3, 6k + 4, 6k + 5.\n\nNow among these, the ones that are composite ( NOT prime ) \nfor any integer k, are\n      6k , Even\n    6k + 2, Even\n    6k + 3 = 3(2k + 1) , a multiple of 3. \n    6k + 4, Even\n\nRemembering that the only even prime number is 2.\nThus, the only possibilities for forms for primes in here is either 6k+1 or 6k + 5 . All other forms denotes a composite integer. Remember that a composite integer is an integer with at least two distinct prime factors.\n\nIf p is form 6k+1, Let p = 6k+1,\nThus, p^2 + 2 = ( 6k + 1 )^2 + 2\n              = 36k^2 + 12k + 3\n              = 3 ( 12k^2 + 4k + 1 )\n              = A multiple of 3 > 3 \n              = A Composite number.  \n\nIf p is form 6k+5, Let p = 6k+5,\nThus, p^2 + 2 = ( 6k + 5 )^2 + 2\n              = 36k^2 + 60k + 27\n              = 3 ( 12k^2 + 20k + 9 )\n              = A multiple of 3 > 3 \n              = A Composite number.\n\nThus, in both cases, if p is prime then p^2 + 2 is always composite.\n\n2)\nAssume that the number of primes are finite.\nNow the integer N is NOT divisible by any of the primes in the list p_1, p_2, ..., p_n.\nWhy? because say p_k divides N. Then, p_k must divide the following integer : p_1*...*p_n where it does NOT contain p_k. Which is a contradiction .\nThus, primes  must be infinite for if not, they produce the contradiction shown above.\n\n3)\n\nI will not use bertrand's conjecture because its NOT proven yet... So, i'll skip this one... its redundant. I will not  use something that has not been proven yet... i.e, a conjecture.
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1) Let's first consider not the combination but the permutations without repetition (considering the order).\n\nSuppose to have 39 numbered cards in your hands and shuffle them well. \nIf you get one card and you drop it on the table you have 39 different possibilities, obviously.\n\nNow, from the remaining 38 cards in your hands, If you drop another card on the right side of the previous card, you'll have 38 different possibilities for just that second choose, but If you consider the order of both cards you'll have exactly 39*38 = 1482 different possibilities that the 2 card will be ordered and choosen in that way.\n\nIf you get a 3rd card from the remaining 37 in your hand you'll have on the table in the right side of previous exraction, a group of 3 ordered cards choosen by 39*38*37 = 54834 different ways will be on the table.\n\nRecursively go to choose until 6 ordered cards will be the table and you'll have 39*38*37*36*35*36 = 2349088560 possibile ways if you consider the order. \n\n2) Now let's see what happend when the order of the selection matters as you asked in your question. \nSo let's make this consideration:\n\na) If you have one card (A) you don't mind about ordering problems:  {A} = (A)\n\nb) if you have two cards (A, B) you have to consider the couple (A, B) = (B, A) so we write {A, B} are commutative;\n\nc) if you have three cards "A", "B" and "C" you can group (A, B) as one element {A, B} because order doesn't matter (point b).\n\nSo we can consider the 3 cards (A, B, C) as:\n ( {A, B}, C ) = ( C, {A, B} ) as we made in the point b.\nYou could also group (A, {B, C}) = ( {B, C}, A)  or {A, C} and then swapping with B.\n\nSo with 3 elements we have 3 different ways to create couple to swap so we can create 3*2 = 6 different combination.\n\nIf we have 4 cards we know it's possible to choose 4 different ways to create list of 3 cards. Let's stop one each time with your finger and you can see that you can group easily the remaining other 3's.\nSo the possible combination with 4 cards is 4 times the possible combination I have with 3 cards. So it's  4 * (3 * 2).\n\nWith 5 cards you'll have 5 times the number of combinations we found with 4 cards, so it's  5 * (4 * 3 * 2).\n\nFinally, with 6 cards in a table arrange in 6 * 5 * 4 * 3 * 2 = 720 different ordered ways.\n\nNow, your question was how to do if the order of selection matters and in the previous example we find that the number of possibilities we have when we select 6 cards ordered is this number 39*38*37*36*35*34 = 2349088560. And we know now that this number is 720 times greater than the possibilities wether the order of selection doesn't matter. (720 = 6 * 5 * 4 * 3 * 2).\n\nIn other words, 39 * 38 * 37 * 36 * 35 * 34 is the number of permutations without repetition of 39 elements grouped by a list of 6 elements considering the order.\n\nAnd now we know with 6 elements you can have 6*5*4*3*2 possible different positions of those elements. \n\nSo Just divide 39*38*37*36*35*34  by 6*5*4*3*2 and you'll have the result of the combinations.\n\nSo to see how many combination of 6 cards bye a set of 39 just divide 39*38*37*36*35*34 / 6*5*4*3*2 = 3262623.\n\nthe number 6*5*4*3*2 could be wrote "6!" using the the factorial function.\n\nThe factorial function of positive integer n, is defined as the product of first n numbers:\n\nn! = 1 * 2 * 3 * ... * n\n\nand is a convention that the product of no numbers at all is 1.\n\n0! = 1\n\n\nThe factorial is also defined recursively by the follow definition:\n\n(n+1)! = (n+1) * n!\n\n\nSo if you have n object and you want to know how many permutation (the order matters) without repetition (each object can be chosen only once) the objects makes with a number k of selection the formula is:\n \n  n! / (n-k)!\n\nso it's the product of k terms starting from n and descending until reached (n-k+1).\n\nn*(n-1)*(n-2)*.
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:D i think the person asking this question is not stupid, as most people here make him out to be, but rather is having fun, and seeing how far ppl will go to be picked :D ... anyways\nHere's my version of the proof, i'm going to do it in a step by step manner\n\nGiven: 5*5\nProof: the value is equvalent to 25\n\nStatement                |         Reason\n1)5*5                    |  1)Given\n2)=25                    |  2)Multiplicative Property of                      \n                         |     Arithmetic
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!!, !!!!!!!! are "multifactorials". They are described in the first attached article.\n\nBriefly,\nn!! = 1 for n=0 or 1,\nn!! = n * (n-2)!! for n>=2.\n\nn!!...! (k "!" signes) = 1, for n<k\nn!!...! (k "!" signes) = n * (n-k)!!...! (k "!" signes)\n\n(2n)!! = n! * 2^n\n(kn)!!...! (k "!" signes) = n! * k^n\n\nSo, we have\n1) 6! = 1*2*3*4*5*6 = 720\n2) 6!! = 2*4*6 = 48\n3) 0! = 1\n4) 5!! = 1*3*5 = 15\n(5!!)!! = 15!! = 1*3*5*7*9*11*13*15 = 2027025\n\nOther factorials are too huge to compute precisely, but we can evaluate them using Stirling formula (second article)\nln n! ~= n*(ln n) - n + (ln(2*pi*n))/2 + 1/(12*n)\n\n5) 5! = 120\n(5!)! = 120! =~ e^(120*ln(120) - 120 + (ln(2*pi*120))/2 + 1/(12*120)) =~ e^457,81 =~ 6.69*10^198\n6) 1234!!!!!!!! = 1234 * 1233 * 1232!!!!!!!! =\n= 1521522 * (154*8)!!!!!!!! =\n= 1521522 * 154! * 8^154 = e^972.43 =~ 2.1*10^422\n\nThus, 6!! - 6! + (5!)! - 1234!!!!!!!! - 0! + (5!!)!! =\n=~ 48 - 720 + 6.69*10^198 - 2.1*10^422 - 1 + 2027025 =~ -2.1*10^422\n\nAnswer: about -2.1*10^422
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1, 2, 3, 4, 5,  and so on are all just labels for items.  We have defined the group of integers by 1+1=1'=2, 2+1=2'=3, and so on where for any n≠0 n•1=1+1+1+ . . . +1+1≠0 (with n 1s).  Then we have the negative integers so that for every number a, there is a solution to the equation x+a=0.\n\nOne thing that we can do instead (as is learned in slightly more advanced algebra) is basically set all of the integers equal to 0.  Then 0=1=2=3=4=5= . . .  If we do this, and keep the standard addition that we use everyday, we still have a mathematical object called a group.  It is just like the integers, except that the integers are said to be a abelian group of infinite order, and the group where all the integers are set equal is an abelian group of order 1.  In particular, 1=2.  It is perfectly valid, but since the group of order 1 isn't really that interesting.  But to answer your question, yes 1 can be equal to 2, just not in a field of characteristic 0.
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1. Set A is subset of set B if and only if any element in A is in B. For example, {2,3,5} is a subset of {1,2,3,4,5}.\n An integer cannot be subset of any set, since itself is not a set.\n\n 2. Composite number is a positive integer, which can be written in product of two positive integers that are bigger than 2. Any positive number other than prime numbers and 1(1 is neither a prime number nor composite number) is composite number.\nFirst few composite numbers are 4,6,8,9,10,12,14,15,16,18,20,...\n\nThe last two questions are not clear. Can you clarify it?
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1) [1/x] <= 1/x, for all x>0\n2) [1/x] = 1/x, iff 1/x is an integer, so x=1/n for some integer n.\n3) [1/x] < 1/x, for all x != 1/n for any n (x>0).\n4) So, f(x) = [x*[1/x]] = [x*1/x] = [1] = 1 for all x=1/n for some integer n.\n5) Also, x*[1/x] < x*1/x = 1 for all x != 1/n for any n (x>0),\nthen f(x) = [x*[1/x]] = [<1] = 0 for all x != 1/n for any n (x>0).\n\nlim f(x)(x->a) = b, iff for ALL sequences {a_n->a} f(a_n)->b.\n\nLet's take 2 sequences:\n{a_n = 1/n} f(a_n)=1 for all n (see 4), so f(a_n)->1.\n{b_n = 2/(2*n+1)} f(b_n)=0 for all n (see 5), so f(b_n)->0.\nEasy to see that {a_n}->0, {b_n}->0.\n\nThus, there is NO limit of f(x) when x->0.\n\nPS: "iff" means "if and only if"
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You can prove it by induction.\n\nIt is obviously true when n = 1.\n\nInduction pypothesis:  Suppose it is true for n\n\nWe need to show  is true for (n+1).\n\nDoes 24 divide (5^(2*(n+1))-1))?\n\nWe can rewrrite this as:\n\n25*(5^2n)-1  Subtract 24 from this to get 25*(5^2n)-25 \n\nThis is just equal to 25*(5^2n)-1).\n\nThis has to be divisible by 24 -- since one of the factors is (5^2n)-1) and our induction hypothesis says that it is.  \n\nAdd 24 to it and we get something that is also divisible by 24 -- but it is (5^(2*(n+1))-1)) -- so our induction is done.
25664
This question is not very well-posed and I believe that is why you have no answers yet. \nfirst, what's with the n(B) = 2.....  ? the dots are leaving out something important.\n\nby AnB i assume you mean the intersection of A and B, not the union AuB. Also, what is the expression A'nB' mean?\n\n-too much time on my hands....
54139
That's an extremely good question. \n\nThe answer is simple. Before that, i will explain what your question means to those who don't get. Well, it means that we must get 24 using 1,3,4 and 6 any number of times however involving operations +,-,* and / only once. \n\n  Then, the answer is very simple :\n\n  {[(6*6)/(3-1)]+[6]}  =  24\n\n---------------------------------------------------------\nWOKRING :\n\n {[(6*6)/(3-1)]+[6]}   =  {[(36)/(2)]+[6]}\n                       =  18 + 6\n                       =  24\n\n----------------------------------------------------------\n\nI hope you post more questions like these. They evoke a lot of thought. I hope the solution serves th purpose.
656952
INTRO:\n\nC-\nC5-E5 (4x)\n\n   \ne:-------------|----------------------------|-------------|-5~-4~-2~-----8--7--5~--|-------|\nB:-------------|----------------------------|-------------|-6~-5~-3~-OR-10~-9~-7~--|-------|\nG:-------------|----------------------------|-------------|-5~-4~-2~----10~-9~-7~--|-------|  (2X)\nD:-------------|---0-2-3~~-2~~-0-2~-0-------|-------------|-------------10~-9~-7~--|-------|\nA:-3~~-0-------|-0---------------------xx---|-3~~-0-------|--------------8--7--5---|-----0-|\nE:-------3~~-0-|-----------------------xx---|-------3~~-0-|------------------------|-3~~---|\n\nFirst Stanza:\n\nC             em\nSige, pag kasama ka naman,\n          Am                    F\nKitang-kita ko ang ating kasiyahan\nC                  em\nSige, wag na nating pigilan\n   Am               F\nAt di magtatagal, tayo ay liligaya\n\nRefrain:\nem         Am            em   Am  \nOkey lang naman ang ating usapan\n              em   Am           F\nHindi na lang babalikan ang nakaraan\n         G\nAng nakaraan\n\nChorus:\nC                em\nAyos lang, basta't kasama \nAm(break)\nKonting alak lang,\n  F\nKahit walang pulutan\nC                em\nAng minsan, naaalala\nAm                         F\nDi magtatagal, tayo ay liligaya\n\nSecond Stanza: (do first stanza chords)\n\nSige, pagpatuloy niyo lang\nUnti-unting lunurin sa kasiyahan\nSige, pagpasensiyahan na lang\nMga pumipigil sa ating ligaya\n\n(Repeat Refrain)\n(Repeat Chorus)\n\n\nADLIB: Do Chorus chords\n\ne:--------------------------------------------------------------------------------------------------|\nB:--------------------------------------------------------------------------------------------------|\nG:----------------------------------------------------------------------------12--------------------|\nD:-10-------12-------14-------15-------14----12----12-------14-------15----------------15----14-----|\nA:----10-10----10-10----10-10----10-10----10----10----10-10----10-10----10-10----10-10----10--------|\nE:--------------------------------------------------------------------------------------------------|\n\ne:--------------------------------------------------------------------|\nB:----------------------------------------------------13-13s17-15~----|\nG:-------12-------12-------12-------12---12b~~~-12s14-----------------|\nD:-14h15----14-15----14-15----14-15-----------------------------------|\nA:--------------------------------------------------------------------|\nE:--------------------------------------------------------------------|\n\ne:----------12-15-12----------------------------12-15-12-------------------------------------------|\nB:----13-15----------13-15--(4x)-THEN-----13-15----------15p13h15br15-15---13----------------------|\nG:-12----------------------------------12-------------------------------------14-12-----12fb~~~~---|\nD:----------------------------------------------------------------------------------13-------------|\nA:-------------------------------------------------------------------------------------------------|\nE:-------------------------------------------------------------------------------------------------|\n\n(Repeat Refrain)\n(Repeat Chorus 2x)\n(intro)
315772
Well, 351=3^3*13.\n\na) Many answers here. The main idea is to make sure the factors of 3 are distributed among the factors of the group. So, for example, Z_9xZ_39.\n\nb) Since there are non-abelian groups of order 27, just take one of these, H, and construct HxZ_13.\n\nc) By Sylow, if a Z_13 is not normal, there will be 27 subgroups of order 13.  Past that, I'll have to think. ;)
384646
First, start listing the prime #'s starting with the number 2.\nAnd let's only go up to 50, because 2 x 50 = 100 and we need 3 #'s.\n\nPRIME #'S LESS THAN 50:\n\n2\n3\n5\n7\n11\n13\n17\n19\n23\n29\n31\n37\n41\n43\n47\n53 ........oops!  STOP, too far.\n\nOkay.  Now we want the product of the 3 #'s needs to be 100.\n\n100 is an even #, so one of the numbers must be even.  \nThat eliminates, quite a lot.\n\nActually, that eliminates all of the #'s but the number 2.\nSo, 2 is the 1st prime number less than 100.\n\nYa, know. . . .\n \nI gave it a try.\n\nAnd I can't seem to find the other two numbers.\n\nYou got me stumped.\n\nPerhaps I read the question wrong.\n\nDid you mean:\nWhat three different prime numbers, multiplied together, have a product of 100?\n\nIf so, you stumped the professor!  :o)
373799
Problem 1:\nWhich one is a RELATION but not a function\n\nChoices:\n1) {(-10,1),(1,8),(-4,-2)}\nfunction\n2) {-4,1),(1,1),(-2,-2)}\nfunction\n3) {(-5,1),(1,-7),(21,-2)}\nfunction\n4) None\n\n\nProblem 2:\nWhich of the following is a function\n\nChoices:\n1) {(-4,1),(1,8),(-4,-2)}\nrelation\n2) {(-4,1),(1,-8),(-2,-2)}\nfunction\n3) {(-4,1),(1,-8),(1,-2)}\nrelation\n4) None\n`