Decide whether each of the given sets is a group with respect to the indicated operation.

1- For a fixed positive integer n, the set of all complex numbers x such that x^n=1(that is, the set of all nth roots of 1),with operation multiplication.

2-The set of all complex numbers x that have absolute value 1, with operation multiplication. Recall that the absolute value of a complex number x written in the form x= a + bi, with a and b real, is given by lxl=l a+bi l=Sqrta^2+b^2

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(1) this is a group

if x^n = 1 and y^n = 1, then certainly (xy)^n = 1, using the laws of exponents, so the set if closed under the group
operation

the group operation is also associative, since [(xy)z]^n= [x(yz)]^n = 1 where x^n = 1, y^n = 1, and z^n = 1

also, since x^n = 1, then take the inverse of both sides to get x^{-n} = 1, which ...

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Whether given sets are groups is determined. The solution is detailed and well presented. The response was given a rating of "5/5" by the student who originally posted the question.

Modern Algebra
Group Theory (XIV)
Symmetric Set of Permutations
Symmetric Set of Permutations : Find order of all elements in S_3, where S_3 is the symme

3 (a)
(i) Let G=Z12(sub12 don't know how to put it), the group of integers modulo 12. Prove that H= {0, 6} AND K= {0, 4, 8} are subgroups of G. Calculate the subset H+K formed by adding together all possible pairs of elements from H and K, i.e.
H+K= {h+kh is a subgroup of H, k is a subgroup of K}
Prove that this is also a

Question: Let G be a nonempty set closed under an associative product, which in addition satisfies:
(a) There exists an e in G such that e.a = a for all a in G.
(b) Given a in G, there exists an element y(a) in G such that y(a).a = e.
Then G is a group under this product.

Show that all automorphisms of a group G form a group under function composition.
Then show that the inner automorphisms of G, defined by f : G--->G so that
f(x) = (a^(-1))(x)(a), form a normal subgroup of the group of all automorphisms.
For the first part, I can see that we need to show that f(g(x)) = g(f(x)) for x in