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Proof. It is immediate to see that the group axioms hold in G/H with eH = H as the identity element and x−1 H as the inverse of xH . As p(xy) = (xy)H = (xH)(yH) = p(x)p(y) and p is surjective, then it is an epimorphism. Finally, ker(p) = {x ∈ G|p(x) = eH = H} = = {x ∈ G|xH = H} = {x ∈ G|x ∈ H} = H. 7 Corollary. e. H = ker(g : G −→ G ) for a group G . Proof. As H is normal, then it is the kernel of an epimorphism, as in the previous theorem. 8 Proposition. If H = ker(g : G −→ G ) for a group G then H G.

When we consider non abelian groups we will usually use the multiplicative notation and when the groups are abelian we will usually use additive notation. However, we will use multiplicative notation for cyclic groups (which are abelian). We have the following properties (known as the laws of exponents) in multiplicative notation xn xm = xn+m , (xn )m = xnm , x−n = (xn )−1 and, in additive notation nx + mx = (n + m)x, m(nx) = (mn)x, (−n)x = −(nx). If a group G is abelian, the following holds n(x + y) = nx + ny.

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) . For q = 3 , m = 4 , we get {11·3 , 12·3 , 13·3 , 14·3 = 112 = 0} which, in additive notation is q = 4, For we obtain m = 3, {3 · 1, 6 · 1, 9 · 1, 12 · 1 = 0} = {3, 6, 9, 0} = (3) . 1·4 2·4 3·4 12 which, in additive notation is {1 , 1 , 1 = 1 = 0} {4 · 1, 8 · 1, 12 · 1 = 0} = {4, 8, 0} = (4) .