How can you tell if a group is Galois or not?
139989
Here is a web site that gives a nice example that might be helpful. Unfortunately, I am not an expert here so i can't give a quick answer, even if one exists. I suspect that one might have a group that is isomorphic to a Galois group, and not know it.
324840
Well, the basic problem you've run into is that the conjecture isn't true.  If a = 6, b = 3, and c = 9, then a*b = 18, and 18/9 = 2, but neither 6 nor 3 is divisible by 9.\n\nTo prove that something is true you must prove it is true for all possible values of the variables, in this case a, b, and c.  However, to prove something is false, you need only supply a single counter-example.  Thus the above example disproves the conjecture that for all integers a, b, and c if a*b is divisible by c then a or b is also divisible by c.\n\nNote that a ≠ cm and b ≠ cn does not imply that ab ≠ c(cmn).  For example, if a = 6, b = 8, c = 2, m = 4, and n = 3, then cm = 2*3 = 8 ≠6 so cm ≠ a, and cn = 2 * 3 = 6 ≠ 8 so cn ≠ b.  However, (cm)(cn) = 2 * 4 * 2 * 3 = 48, and a * b = 6 * 8 = 48, so ab = (cm)(cn).  This proves that you cannot multiply not-equalities (i.e. statements with "doesn't equal).\n\nAlso, concluding that if c does not divide a or b implies that c does not divide ab is not the same as concluding that if c does divide ab it must also divide a or b.  This logical fallacy is equivalent to the statement that for all plants, not a tree implies not a rock, so being a rock implies being a tree.  The first statement is true: all plants that are not trees are also not rocks.  However, the second statement is clearly false, as no rocks are also trees.
443950
Is the = really a minus?  If so, the GCF of each term is\n4a^2b, because the GCF of 4 and 8 is 4, and the GCF of variables has the lower of the exponents of each variable.\n\nDivide each term by this GCF (when you divide, subtract the exponents) and write the GCF outside () with the division answer inside (distributive property in reverse)\n\n4a^2b(b^2 - 2ac)
877374
G/H = {xH: x is in G}.\n\nAction: g.(xH) = (gx)H for every g and x in G.\n\nThis is indeed a left action because: (i) the group operation is associative (so (g''g').(xH) = g''.(g'.(xH)) for every g' and g'' in G and (ii) ex= x (so e.(xH) = xH).\n\nIt is transitive because, to satisfy g.(xH) = yH, it is enough to take g = y(x^-1)).\n\nNote: need H to be normal if the notation G/H is to be legit because it means the quotient group. H need not be normal for G to left-act on the *set* {xH: x is in G}, but this set won't be a group w/o normality of H.
168776
The construction of negative numbers and rational numbers is much more complex than you think. So 3 over -1 is not an arithmetic operation and using the rules of abstract mathematics it can easily be seen it is true.
940614
i think your question is how to find the GCF of monomials\nfor the GCF you will take the lowest power of ech variable for the GCF\nexample\nif you have to find the GCF of 23x^3y^2z,2xy^4z^4,8x^3y^3z^3\nthen first of all find out the GCF of the coefficients \nGCF is 1 as there is no common factor for 23,2 and 8\nthe lowestpower inwhich x appears is x\nthe lowest power of y is y^2and\nthe lowest power of z is z\nso now theGCF is xy^2z\nhope that it helps
1094077
I'd solve this one by actually determining every single homomorphism from Z/mZ to Z/nZ completely.  Here's a couple of hints to get you going.\n\n(1) Since the group Z/mZ is cyclic, every homomorphism out of Z/mZ is completely determined by where the element 1 is sent (more precisely, the element 1+mZ).\n\n(2) Because m times 1 is 0 in Z/mZ, any homomorphism f : Z/mZ to another group G must send 1 to a number f(1) in G such that m times f(1) is 0 in G.\n\nThese two comments will give you a small set of possible homomorphisms from Z/mZ to Z/nZ (hint: the number of such homomorphisms depends upon gcd(m,n) ).  It's then not hard to show that everything in that set really works as a homomorphism and that they form a cyclic group.
460866
Not true. It is true for continuous homomorphisms. In that case a homomorphism is of the form f(x)=ax for some constant a. THen a^n=1, so a=+-1.
231905
I'm assuming this is a group isomorphism. I will use "e_G" for the identity element of G, and "e_H" for the identity element of H.\n\nLet G and H be groups and f:G->H be an isomorphism. Let a be an element of G of infinite order. Suppose on the contrary that f(a) does not have infinite order. Then there exists some positive integer n such that e_H=f(a)^n=f(a^n), so a^n is in the kernel of f. Since f is an isomorphism, the kernel of f is {e_G}, thus it must be that a^n=e_G, contradicting that a has infinite order.
71377
Say X is a simple set with two elements, X={x,y}.  Then the free group on X is just all words using x,y,x^, and y^ with the rule that xx^=x^x=yy^=y^y=empty word.  The empty word is the identity.  Basically, it is the most general group generate by two elements.\n\nThis can be stated more formally by saying that Free(X), the free group on X, has a map i:X->Free(X) and the following property:\n\n   For every f:X->G, where G is a group, there exists a\n   unique group homomorphism F: Free(X)->G so that\n   f = F composed with i\n\nIn other words, Free(X) is somehow a universal case of a group containing X.\n\nThis can be generalized in lots of other areas. In Category Theory, Free is an 'adjoint functor' of the forgetful functor from Groups to Sets.  'Adjoint functors' arise in lots of interesting places, and this is one of the first ones you'll learn.
904826
f(g(x)) = g(x) + 1 = (x^2 - 1) + 1 = x^2  - 1 + 1 = x^2\n\ng(f(x)) = (f(x))^2 - 1 = ( x + 1 )^2 - 1 = (x^2 + 2x + 1) - 1 \n                                                         = x^2 + 2x + 1 - 1 \n                                                          = x^2 + 2x
252085
Yes it is , Just ask a part of it in english  so it wont be reported the main point here is to share our answers so you´ll figure it out ....
759151
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.
457089
G=uys\nF=***\nE=verthing
1150969
a = 3^(1/3)\nb = 4^(1/3)\n c = a+b\n If you use the primitive element theorem  you will see that c is a primitive element for Q(a)(b)\nso Q(a)(b)=Q(c), which implies that the degree of the minimal polynomial is 3*3 = 9\n By primitive element theorem a is written as a rational fraction in c, a = f/g(c),\nso a^3 = f/g^3(c), or g^3(c)*a^3 = f^3(c), g and  f polynomials in Q.\n so the minimal polynomial will be 3g^3-f^3\n( hoping that g and f have at most degree 3 in c)\ng and f can be computed as in the proof of primitive element theorem\ni.e compute gcd (x^3-3, (c-x)^3-4))\n Maybe there is a better way...
553656
No, a monomial is a type of polynomial.\nA monomial is a polynomial with only one term.\nPolynomials can only have non-negative integers as exponents.\n1/a = a^(-1)\nThe exponent is negative.\nThus, this is not a polynomial, which implies that it is not a monomial.
108416
none if they are false