Find the number automorphism of the  symmetric  group  of  order 3?
336113
S3 is isomorphic to the dihedral group of order 6, which consists of maps over the equilateral triangle with unique sides. There are only two basis elements, the flip over a central axis, and a single rotation. All of the other 4 maps can be written as products of these two actions. Since a homomorphism is completely defined by its action on the basis, we have 2! automorphisms (ways of mapping the basis into itself).
73405
Since 2 is relatively prime to the order of the group it will generate the group, so it has an order of p.\n\nIf this is hard to recognize in the abstract, take an example like U(5).  \n\nIn U(5), 2 = 2\n         2^2 = 4\n         2^3 = 8 which is 3.\n         2^4 = 3x2 = 6 which is 1\n         and 2^5 brings you back to 2 again.\n\nI'd forgotten all my "Advanced Abstract Modern Algebra"... thanks for the chance to refresh my memory!
676633
54\n*45\n____\n\n 270\n\n2160\n_____\n\n2430
269936
4 to the 3rd is 4 multiplied by itself 3 times\n\n4 to the 3rd = 4 x 4 x 4\n\n2 to the 6th = 2 x 2 x 2 x 2 x 2 x 2
757625
1) \n\n1+2+3+4+5+....n +(n-1)+(n-2)+...+3+2+1...regrouping\n\n=2(1+2+3+4+5+....(n-1)) +n...regrouping the term in parenthesis\n\n=2( (1+n-1)+(2+n-2)+(3+n-3)...)+n\n=2(n+n+n+...(n-1) terms)+n\n=2(n)(n-1)/2 +n\n=n^2-n+n\n=n^2\n\n\n2)\n 1=10^0 and 11=(10^1+10^0) and 111=(10^2+10^1+10^0)...\n\n(10^n+10^(n-1)+....10^2+10^1+10^0) times itself\n=(10^0+10^2+10^3+....10^n)^2\n=10^2n+2*10^(2n-1)+.....\n\n\n3)\n\nn^2-n=n(n-1)...one of them must be even and is therefore divisible by two.\n\nn^3-n=n(n^2-1)=n(n+1)(n-1)=(n-1)(n)(n+1) or three consecutive integers.\n\nIf the product is three consecutive integers, then one of them must be divisible by three and therefore the product is divisible by three.
864795
8(t-2sqrt(a))(t+2sqrt(a))
49923
Try:  http://www.internet4classrooms.com/eoc_algebra2.htm
410291
http://en.wikipedia.org/wiki/Permutation
483173
((6p - 18)/(9p)) / ((3p - 9)/(p^2 + 2p))\n((6(p - 3))/3(3p)) / ((3(p - 3))/(p(p + 2)))\n((2(p - 3))/(3p)) / ((3(p - 3))/(p(p + 2)))\n((2(p - 3))/(3p)) * ((p(p + 2))/(3(p - 3))\n(2p(p - 3)(p + 2))/((9p)(p - 3))\n(2p(p + 2))/(9p)\n(2(p + 2))/9\n(2p + 4)/9
1144759
I don't know the answer to your question, but here is a good website that might help:\n\nhttp://www.purplemath.com/modules/asymtote3.htm
297626
The roots are of the form root(8,3)*omega^j where omega is a primitive 8th root of unity. Now, we can independently send root(8,3) to any root and omega to any primitive root of unity. There are four such primitive roots, so the group will have 8*4=32 elements. The subgroup that leaves root(8,3) fixed will be a Klein 4-group, while the subgroup leaving omega fixed will be cyclic of order 8. You go from there.
771836
Simple substitution.\n\n5(3) - 2(2) + 4(4)\n15 - 4 + 16\n27
25946
=3fg*(f^2g)+3fg*5f-3fg*4\n=3(f^(1+2g))*g+15(f^2)g-12fg
187387
(n)(3n-1)(3n-2).......(n+1)
638209
1, 2, 3, 6, 7, 14, 21, 42
575490
I- + 2e- -------> I2\n\n4P + 3e-  -------------> P4\n\nthere ya go :)
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.