- #1
bjnartowt
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Homework Statement
prove that an infinite group must have an infinite number of subgroups.
Homework Equations
The Attempt at a Solution
there are two cases to address regarding the order of the subgroups whose quantity is in question,
Case 1: those subgroups, themselves, are either finite or infinite
Case 2: those subgroups, themselves, are strictly finite.
suppose there was a member of G that had infinite order, which we will call x,
[tex]\exists x \in G{\rm{ }}\left| x \right| = \infty [/tex]
Make iterations of that member x iterated,
[tex]\left\langle {{x^n}} \right\rangle [/tex]
That constitutes an infinite subgroup. But there are more such infinite subgroups! There should be one such infinite subgroup every time we consider,
[tex]\exists \left\langle {{x^{{p_1}}}} \right\rangle [/tex]
in which p1 is a prime number. Since there is an infinite number of primes, there is, correspondingly, an infinite number of [tex] \left\langle {{x^{{p_1}}}} \right\rangle [/tex] that exist, and thus an infinite number of distinct subgroups of G, each of which are infinite in order!
This proof would be easy if only I could assume that an infinite group always has some element of infinite order. Then, you could chop up that infinite order into just-as-infinite subgroups (able to be iterated infinity times), and there would be one DISTINCT infinite subgroup per prime integer...of which there are infinite. Distinctness would fall from primeness.