Date: 03/10/2003 at 23:16:44 From: Karen Subject: Cyclic Groups Hi Dr. Math, I hope you can help me out with this question. Prove that an infinite group must have an infinite number of subgroups. I believe that the order of this group is infinite. Then H (subgroup) must have the form a^(n/k) where n is the order of the infinite group and k is a divisor. Please tell me if I'm on the right track. Karen
Date: 03/11/2003 at 08:51:24 From: Doctor Jacques Subject: Re: Cyclic Groups Hi Karen, I'm afraid you cannot say "n is the order of the infinite group" : if the group is infinite, its order is infinite by definition, and you cannot use it as a number since infinity is not a number. Let G be an infinite group, and assume, by contradiction, that it contains only a finite number N of subgroups. Consider the cyclic subgroups of G. There is also a finite number M (<= N) of them. If any of these is the infinite cyclic group, it is isomorphic to Z under addition, and Z contains an infinite number of subgroups (all nZ for integer n). In this case, G also contains an infinite number of subgroups - a contradiction. This means that all the cyclic subgroups of G are finite. As there is only a finite number of them (M), there is a maximum order, i.e. there is a number L such that each of these M cyclic subgroups contains at most L elements. On the other hand, every element of G belongs to at least one cyclic subgroup (the cyclic subgroup generated by the element itself). Can you continue from here? Please feel free to write back if you are still stuck. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/
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