Modern algebra question on isomorphisms?
960669
the symmetry group is not commutative(one rotation followes by  a reflection is not equal with a reflection followed by a rotation) but Z_4 X Z_2 is commutative.
553317
To show the order of the automorphism of even, it suffices to find an involution i.e., an automorphism of order 2.  Consider the map g -> g^{-1}.  If the group is commutative this map is a homomorphism, and it obviously has order 2.
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.
858799
This depends on the specific Lie group. The exponential map for GL^+(n,R) is not surjective (for n=2, [[-1 0][0 -2]] is not an exponential), but is is surjective for GL(n,C). Notice that it cannot be surjective unless the group is connected [GL(n,R) is not]. The map tends not to be injective. For example exp[[0 0][0 0]] and exp[[2pi*i 0][0 0]] are both the identity. Compactness considerations give negative answers for bijectivity for all compact Lie groups.\n\nThe exponential map is bijective on a neighborhood of the identity for every Lie group. The reason is that the differential map is essentially the identity and so the Inverse Function Theorem applies. Cleartly it is both injective and surjective for R^n.
905291
The previous result is that the number of elements in the preimage of each element in f(K) (or more generally in f(G)) is n. All those sets are disjoint and their union is K, so the total number of elements in K is n * (Number of elements in f(K)).\n --> |K| is a multiple of |f(K)|\n\nAnother way of looking at is that if N is the kernel of f, f(K) is isomorphic to the quotient group K/N which always divides the order of K.
64159
It's debatable.  Here's an interesting column that addresses the question:\nhttp://mathforum.org/library/drmath/view/57188.html
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.
757825
There has been no development in this field, but you can check out the following result pertaining to set theory and group theory.\n\nThe power set P(x) of a set X, is a not a group under (U) union operation.\n\nProof : \nTo prove this we try to prove three properties \n1. Associativity \nFor A, B, C are in P(x) then \n(AUB)UC = A U (BUC)  [This is more or less obvious or simple to prove]\n\n2. Identity element\nThere is an an element(e) in P(x) such that for all A in P(x)\ne U A = A U e = A\nAnd, this element for power set is 'empty set' . [Sorry, but I dont have 'phi' symbol here.]\n\n3. Inverse element\nFor each A  in P(x) there exist another element B in P(x) such that\n(AUB) = (BUA) = e(empty set)\n\nThis property is not possible because you can not produce an empty set with union operation with any other set(irrespective wether its empty or not)\n\nSo, its not a group under Union.\n\n\nPlease noe that : The power set P(x) of a set X, is a not a group under intersection operation.
909313
if u meant make x the subjct then tis is how u wd go abt it.\n\na^2 - ax = b^2 - bx\nax-bx=a^2 - b^2\nx(a-b)= a^2-b^2\nx        =(a^2-b^2)/(a-b)\nx        =[(a-b)(a+b)]/(a-b)    (using difference of two squares)\nx        =a+b
818651
(a) Suppose f is NOT injective, that is, f( x') = f(x'') for some x' and x'' in X. Then also g(f( x')) = g(f( x'')), which means that g o f is NOT injective. (This is proof by transposition: instead of showing directly that A implies B, we show that Not B implies Not A.)\n\n(b) Take any t in T. Then t = g(f(x)) for some x (since g o f is surjective). This means that t = g(y) for some y, namely, for y = f(x). So g is surjective.\n\n(c) By Part (a), f is injective. By Part (b), applied to f o g in place of g o f, f is surjective. Beig both injective and surjective, f is bijective.
761029
So you just need to show that for any x,y in A, f(x)=f(y) implies that x=y.  So let x,y be elements of A so that f(x)=f(y).  Since f(x)=f(y), and g is a function, then g(f(x))=g(f(y)).  Since gof is injective by assumption, this implies that x=y.  So we start with f(x)=f(y), and this implies x=y.  Therefore f is injective.  Cheers!
1068245
One problem you are having is that you have decided there should only be two terms for the symmetry here. Also, it is good to write z=f(x,y) so we get all three variables into the picture.\n\nThe symmetry f(-x,-y)=f(x,y) is a symmetry through the line x=y=0. This is a vertical line through the origin.\n\nThe symmetry f(-x,-y)=-f(x,y) is a symmetry through the origin (0,0,0).\n\nThe symmetry f(-x,y)=f(x,y) is a symmetry through the yz-plane.\n\nThe symmetry f(-x,y)=-f(x,y) is a symmetry through the line x=z=0.\n\nTo do a more extensive investigation would require some group theory. Groups are ways of describing the symmetries of a situation and are generally studied in an abstract algebra course.
348293
There are two seminar papers by Magnus Hestenes that should give you a good start:\n\n"A  Ternary Algebra with Applications to Matrices and Linear Transformation"\n\n"On a Ternary Algebra"
246622
1 st isomer of a compound = n-(name of compound)\n2 nd isomer = iso-(")\n3 rd isomer = neo-(")\nFor example \nBUTANE C5H12\nIsomers of BUTANE\nn-BUTANE\nCH3-CH2-CH2-CH2-CH3\n\nISO-BUTANE\n   CH3\n     |\nCH3-CH-CH2-CH3\n\nNEO-BUTANE\n   CH3\n     |\nCH3-C-CH3\n     |\n   CH3
464936
Commutative operation.
406468
D = {x | x < -3 U x > 4 }
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.