• Support PF! Buy your school textbooks, materials and every day products Here!

|G|=pq then |G| is abelian or Z(G)=1

  • Thread starter ArcanaNoir
  • Start date
  • #1
768
4

Homework Statement


Show that if [itex]|G|=pq[/itex] for some primes p and q, then G is abelian or Z(G)=1.


Homework Equations



|G| = pq =⇒ |Z(G)| = 1, p, q, or pq. Prove: |Z(G)| = p and |Z(G)| = q are impossible. If
|Z(G)| = p then |G/Z(G)| =|G|/|Z(G)| =pq/p = q. But then, since G/Z(G) is cyclic of prime order,
|G/Z(G)| = 1. Thus q = 1, a contradiction since q > 1. Similarly, |Z(G)| = q =⇒ p = 1, a
contradiction to the definition of p


The Attempt at a Solution



I am trying to understand the part of the proof that says "G/Z(G) is cyclic". Why is this so?
 

Answers and Replies

  • #2
Zondrina
Homework Helper
2,065
136

Homework Statement


Show that if [itex]|G|=pq[/itex] for some primes p and q, then G is abelian or Z(G)=1.


Homework Equations



|G| = pq =⇒ |Z(G)| = 1, p, q, or pq. Prove: |Z(G)| = p and |Z(G)| = q are impossible. If
|Z(G)| = p then |G/Z(G)| =|G|/|Z(G)| =pq/p = q. But then, since G/Z(G) is cyclic of prime order,
|G/Z(G)| = 1. Thus q = 1, a contradiction since q > 1. Similarly, |Z(G)| = q =⇒ p = 1, a
contradiction to the definition of p


The Attempt at a Solution



I am trying to understand the part of the proof that says "G/Z(G) is cyclic". Why is this so?
I believe it's because groups of prime order are cyclic. This stems from Lagrange's theorem.
 
  • #3
768
4
Omg yes. Thank you, I had all those pieces just wasn't putting them together for some reason. :)
 

Related Threads on |G|=pq then |G| is abelian or Z(G)=1

Replies
8
Views
5K
  • Last Post
Replies
4
Views
3K
Replies
5
Views
726
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
1
Views
930
  • Last Post
Replies
14
Views
2K
Replies
1
Views
588
  • Last Post
Replies
2
Views
1K
Replies
1
Views
1K
Top