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Mathematics
Linear and Abstract Algebra
Finding the injective hull
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[QUOTE="mathwonk, post: 6035165, member: 13785"] i also read that for an abelian group divisible is equivalent to injective. but we should try to understand this. i.e. try proving that injective implies divisible say. recall that injectivity means that all maps into it extend. so take e.g. the map of the subgroup nZ into an abelian group. If we want to extend this to all of Z, we are going to need to send 1 to 1/n times whatever n went to, so we have to be able to divide by n in the target group, or in multiplicative notation, to take nth roots. so that direction looks easy. the other direction is probably harder and may even use zorns lemma or some such nonsense. try it for finite abelian groups, i.e. direct sums of cyclic groups. or first just for cyclic groups like Z, as I just used. just try to get a feel for it. have you had the basic courses on abstract algebra? e.g. have you done the classification of finite abelian groups? [/QUOTE]
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Mathematics
Linear and Abstract Algebra
Finding the injective hull
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