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Proof. Let e be another identity element such that e e = e . As e is also an identity, then e e = e . Thus e = e . 3 Proposition. If xy = xz in a group G , then y = z . Analogously, yx = zx , then y = z . Proof. If xy = xz , then x−1 (xy) = x−1 (xz) . By associativity, (x−1 x)y = (x−1 x)z . Hence, ey = ez and, finally, y = z . If yx = zx , then y = z , which can be proved in the same way. 4 Proposition. In an arbitrary group G , the inverse of any element is unique. Proof. Let x be another inverse element of the element x .

Both G and {e} are subgroups of (G, ·) , called improper subgroups . All other subgroups are called proper. The subgroup {e} is called the trivial subgroup and it is usual to denote it, by abuse of notation, only by e where e can also be denoted as 0 or 1 or any other notation that represents the identity element of the group under consideration. 18 Proposition. The intersection of subgroups of G is a subgroup of G . Proof. Let {Hi }i∈I be a collection of subgroups of G , indexed by a set I of indices.

Xmq = x12 = 0 Study at Linköping University and get the competitive edge! Interested in Computer Science and connected fields? Kick-start your career with a master’s degree from Linköping University, Sweden – one of top 50 universities under 50 years old. com 51 Click on the ad to read more An Introduction to Group Theory Cyclic Groups form a cyclic subgroup of (x) of order m. , 12 · 1 = 0} Analogously, for q = 2 , m = 6 , we obtain in notation additive is 1·2 and we get, exactly (1) = Z12 . , 12 · 1 = 0} = {2, 4, 6, 8, 10, 0} = (2) .

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