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

• ArcanaNoir
In summary, the conversation discusses a proof that states if |G|=pq for primes p and q, then G is either abelian or Z(G)=1. The proof involves showing that |Z(G)| = p or |Z(G)| = q are impossible, and that G/Z(G) must be cyclic of prime order, leading to a contradiction if q>1. This is because groups of prime order are cyclic.

## Homework Statement

Show that if $|G|=pq$ 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?

ArcanaNoir said:

## Homework Statement

Show that if $|G|=pq$ 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.

• 1 person
Omg yes. Thank you, I had all those pieces just wasn't putting them together for some reason. :)