Cyclic Multiplicative Group : Field of Order 2^7

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• Fields and generators

  This implies that the order of x+2 is not 2, 4, 8, 16, 3. Moreover: The divisors of 48 are 2, 3, 4, 6, 8, 12, 16, 24. . Therefore x+2 is a generator. (c) . The order of the cyclic group is 7, a prime.

• Find four elements of the field that form a subfield of order 4

  The 15 elements construct the multiplicative group of . Now we find the subfield with order 4 of . We know that is a multiplicative group of . Then is a subgroup of and . Since is a cyclic group, then is also a cyclic group.

• Multiple Table for Four Elements

  group F* has order 8, and is therefore cyclic; so it remains to compute the orders of the elements of F* that are not constant.

• Prove that there is no field of order 6.

  Then we consider its multiplicative group F*=F-{0}. We know that |F*|=6-1=5 is a prime number and thus F* must be a cyclic group. We can suppose that F*=={1,a,a^2,a^3,a^4}. Now we consider the additive inverse of a, which is -a.

• Problems in Group Theory and Quaternions

  To celebrate the discovery, he immediately carved this equation into the stone of the Brougham Bridge: i^2 = j^2 = k^2 = ijk = -1.] 2. Recall that Z/nZ has a unique (cyclic) subgroup of order d for each d|n.

• Prove or disprove the existence of non-commutative division ring of order 4.

  If there exists a non-commutative division ring R of order 4, then all nonzero elements of R construct a multiplicative group, denoted as R*.

• Finding a Primitive Element of GF(49)

  The multiplicative order of the roots +/- 3i of x^2 + 2 have multiplicative order 12 since (+/- 3i)^2 = -2, which has multiplicative order 6 mod 7. Thus, neither of these roots are primitive either.

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  <5> = {1; 5; 7; 8; 4; 2}; <7> = {1; 7; 4;}; <8> = {1; 8} Since ord(G) = 6 and we have elements whose order is 6, G is a cyclic group.

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  Thus, the order of the group is p - 1, which means for any element c in this group, c^(p - 1) = 1 (the multiplicative identity).