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4400algebra2point5

4400algebra2point5 - ALGEBRA HANDOUT 2.5 MORE ON...

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ALGEBRA HANDOUT 2.5: MORE ON COMMUTATIVE GROUPS PETE L. CLARK 1. Reminder on quotient groups Let G be a group and H a subgroup of G . We have seen that the left cosets xH of H in G give a partition of G . Motivated by the case of quotients of rings by ideals, it is natural to consider the product operation on cosets. Recall that for any subsets S, T of G , by ST we mean { st | s S, t T } . If G is commutative, the product of two left cosets is another left coset: ( xH )( yH ) = xyHH = xyH. In fact, what we really used was that for all y G , yH = Hy . For an arbitrary group G , this is a property of the subgroup H , called normality . But it is clear – and will be good enough for us – that if G is commutative, all subgroups are normal. If G is a group and H is a normal subgroup, then the set of left cosets, denoted G/H , itself forms a group under the above product operation, called the quotient group of G by H . The map which assigns x G to its coset xH G/H is in fact a surjective group homomorphism q : G G/H , called the quotient map (or in common jargon, the “natural map”), and its kernel is precisely the subgroup H . Theorem 1. (Isomorphism theorem) Let f : G G be a surjective homomor- phism of groups, with kernel K . Then G/K is isomorphic to G . Proof. We define the isomorphism q ( f ) : G/K G in terms of f : map the coset xK to f ( x ) G . This is well-defined, because if xK = x K , then x = xk for some k K , and then f ( x ) = f ( x ) f ( k ) = f ( x ) · e = f ( x ) , since k is in the kernel of f . It is immediate to check that q ( f ) is a homomorphism of groups. Because f is surjective, for y G there exists x G such that f ( x ) = y and then q ( f )( xK ) = y , so q ( f ) is surjective. Finally, if q ( f )( xK ) = e , then f ( x ) = e and x K , so xK = K is the identity element of G/K . In other words, a group G is (isomorphic to) a quotient of a group G iff there exists a surjective group homomorphism from G to G . Corollary 2. If G and G are finite groups such that there exists a surjective group homomorphism f : G G , then # G | # G . Proof. G = G/ ker f , so # G · #(ker f ) = # G . 1

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2 PETE L. CLARK Remark: Suitably interepreted, this remains true for infinite groups. Corollary 3. (“transitivity of quotients”) If G is isomorphic to a quotient group of G and G ′′ is isomorphic to a quotient group of G , then G ′′ is isomorphic to a quotient group of G . Proof. We have surjective group homomorphisms q 1 : G G and q 2 : G G ′′ , so the composition q 2 q 1 is a surjective group homomorphism from G to G ′′ . 2. Cyclic groups Recall that a group G is cyclic if there exists some element g in G such that every x in g is of the form g n for some integer n . (Here we are using the conventions that g 0 = e is the identity element of G and that g n = ( g 1 ) n .) Such an element g is called a generator. In general, a cyclic group will have more than one generator, and it is a number-theoretic problem to determine how many generators there are.
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4400algebra2point5 - ALGEBRA HANDOUT 2.5 MORE ON...

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