Proving: Infinite Group Has Infinite Subgroups

In summary, an infinite group must have an infinite number of subgroups because for every prime number, there exists an infinite subgroup of the group generated by an element with that prime as its exponent. Additionally, subtracting an infinite number of finite subgroups from an infinite group still leaves an infinite group. This can be proven by considering the order of the subgroups and the concept of infinity.
  • #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.
 
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In fact, one author seems to imply that subtracting an infinite number of finite groups from an infinite group gives you a still-infinite group leftover after the aforementioned subtraction from it. I find that dubious!
 

Related to Proving: Infinite Group Has Infinite Subgroups

1. What is an infinite group?

An infinite group is a mathematical concept that refers to a set of elements with a binary operation (such as addition or multiplication) that follows certain rules, such as closure, associativity, and identity. The set must also have an infinite number of elements.

2. What does it mean for a group to have infinite subgroups?

A subgroup is a subset of a group that also follows the group's binary operation and rules. For a group to have infinite subgroups, it means that there are an infinite number of possible subsets that can also function as groups within the larger group.

3. How do you prove that an infinite group has infinite subgroups?

To prove that an infinite group has infinite subgroups, you must first show that there are an infinite number of elements in the group. Then, you must demonstrate that there are an infinite number of possible subsets of these elements that follow the group's binary operation and rules. This can be done through mathematical proofs or examples.

4. Can an infinite group have a finite number of subgroups?

Yes, an infinite group can also have a finite number of subgroups. This can occur if the group has a limited number of elements or if the subsets formed by these elements do not follow the group's binary operation and rules.

5. What are some real-world examples of infinite groups with infinite subgroups?

One example of an infinite group with infinite subgroups is the set of all positive integers with the binary operation of addition. Another example is the set of all real numbers greater than 1 with the binary operation of multiplication. Both of these groups have an infinite number of elements and an infinite number of possible subsets that follow the group's rules.

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