G is cyclic and |G| = p^n, p is prime <=> H,K Subgroups, H⊆K or K⊆H

In summary, it is shown that the conditions 1 and 2 are equivalent for a finite group G, where G is cyclic and has an order of p^n where p is prime and n≥0, if and only if any subgroups H and K of G satisfy either H⊆K or K⊆H. This is proven by considering the negation of 1 and showing that it leads to subgroups that do not follow the given condition. Additionally, it is shown that G must be of the form p^k for a prime p and an integer k, and it is implied that G must also be cyclic.
  • #1
tonit
55
1

Homework Statement



Show that the following conditions are equivalent for a finite group G:

1.[itex]G[/itex] is cyclic and [itex]|G| = p^n[/itex] where [itex]p[/itex] is prime and [itex]n\geq 0[/itex]
2.If [itex]H[/itex] and [itex]K[/itex] are subgroups of [itex]G[/itex], either [itex]H⊆K[/itex] or [itex]K⊆H[/itex].


The Attempt at a Solution



1 => 2.

Let [itex]H,K[/itex] be subgroups of [itex]G = <g>[/itex] where [itex]o(g) = p^n[/itex]. We have [itex]H = <g^a>[/itex] and [itex]K = <g^b>[/itex] where [itex]a[/itex] and [itex]b[/itex] divide [itex]p^n[/itex]. Since [itex]p[/itex] is prime, [itex]a = p^s[/itex] and [itex]b = p^t[/itex]. If [itex]s \leq t[/itex], this means [itex]a|b[/itex] whence [itex]H⊆K[/itex]. Similarly, if [itex]b \leq a[/itex] we have [itex]K⊆H[/itex].

Now I'm stuck at 2 => 1. Any help is appreciated :rolleyes:
 
Physics news on Phys.org
  • #2
Maybe if you can use a fact like, for any integer that divides the group G, there is a subgroup of that order? Do you have a fact like that that you are permitted to use? Maybe only for prime divisors. That result I think comes sooner.

In other words, a strategy that might have lead you to this is to think about the negation of 1, so what if the order of G is not a power of a prime. And how this might lead to subgroups which are not "one a subset of another".
 
  • #3
If [itex]H[/itex] and [itex]K[/itex] are subgroups of [itex]G[/itex], suppose [itex]H⊆K[/itex]. So we have that [itex]|H|[/itex] divides [itex]|K|[/itex], and they both divide [itex]|G|[/itex]. If it would happen that [itex]|G| = p^ka^r[/itex] where [itex]p, a[/itex] are different primes, then [itex]G[/itex] would have two subgroups [itex]M[/itex] and [itex]N[/itex] such that [itex]|M|[/itex] divides [itex]p[/itex], [itex]|N|[/itex] divides [itex]a[/itex], [itex]M ⊄ N[/itex] and [itex]N ⊄ M[/itex], thus contradicting the condition.
So [itex]|G| = p^k[/itex] for a prime [itex]p[/itex] and an integer [itex]k[/itex].

Is it correct?
Now how do I show that [itex]G[/itex] is cyclic?
 

Related to G is cyclic and |G| = p^n, p is prime <=> H,K Subgroups, H⊆K or K⊆H

1. What does it mean for a group G to be cyclic?

A cyclic group is a group where all of its elements can be generated by a single element, called a generator. In other words, every element in the group can be written as a power of the generator.

2. How is the order of a cyclic group related to its generator?

The order of a cyclic group is equal to the order of its generator. This means that the number of elements in the group is the same as the number of powers of the generator.

3. What does it mean for the order of a group to be a prime number?

A prime number is a positive integer that is only divisible by 1 and itself. When the order of a group is a prime number, it means that the group has no nontrivial subgroups.

4. How can we determine if a group G is cyclic and has order p^n, where p is prime?

If the order of a group is p^n, where p is prime, and there are no nontrivial subgroups, then the group is cyclic. This is because the group must have a generator that generates all of its elements.

5. Can a group G have multiple subgroups H and K where H is a subgroup of K or K is a subgroup of H?

Yes, it is possible for a group G to have multiple subgroups H and K where H is a subgroup of K or K is a subgroup of H. This is because a subgroup is a subset of a group that also forms a group under the same operation.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
924
  • Calculus and Beyond Homework Help
Replies
4
Views
842
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Math POTW for University Students
Replies
1
Views
190
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
409
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
Back
Top