# Prove G is either cyclic or every element x≠e in G has order p or q

• Nov 4th 2009, 04:02 PM
apple2009
Prove G is either cyclic or every element x≠e in G has order p or q
Let G have order pq, where p and q are distinct primes. Prove G is either cyclic or every element x≠e in G has order p or q
• Nov 4th 2009, 04:06 PM
jmoney90
Quote:

Originally Posted by apple2009
Let G have order pq, where p and q are distinct primes. Prove G is either cyclic or every element x≠e in G has order p or q

One corollary of LaGrange's Theorem is that the order of every element in G is a divisor of the order of G, which means it is either order p or order q, since those are the only divisors of |G|. The only other possibility for order of elements is that an element can have order pq, which by definition means it is cyclic. (not so sure on this part, I think that is correct though).
• Nov 4th 2009, 05:39 PM
Drexel28
Quote:

Originally Posted by apple2009
Let G have order pq, where p and q are distinct primes. Prove G is either cyclic or every element x≠e in G has order p or q

Quote:

Originally Posted by jmoney90
One corollary of LaGrange's Theorem is that the order of every element in G is a divisor of the order of G, which means it is either order p or order q, since those are the only divisors of |G|. The only other possibility for order of elements is that an element can have order pq, which by definition means it is cyclic. (not so sure on this part, I think that is correct though).

Yes. I do beleive that is correct.

Problem: Let $\displaystyle G$ have order $\displaystyle pq$, where $\displaystyle p,q$ are distinct primes. Prove $\displaystyle G$ is either cyclic or every element $\displaystyle g\in G-\left\{e_G\right\}$ has order $\displaystyle p$ or $\displaystyle q$

Proof: As was stated we know that if $\displaystyle a\in G$ that $\displaystyle \text{ord}(a)|\text{ord }G$. Now suppose that $\displaystyle g\in G-\left\{e_G\right\}$, we must have that $\displaystyle \text{ord}(g)|pq$ and since $\displaystyle p,q$ are distinct primes this is only true if $\displaystyle \text{ord}(g)=p,q, \text{or }pq$. If $\displaystyle \text{ord}(g)$ equals the first two we are done. Otherwise we'd have that $\displaystyle \text{ord}(g)=pq$ which means that $\displaystyle \left\{e_G,g,g^2,\cdots,g^{pq}\right\}$ are distinct elements. If there existed a $\displaystyle g_1\in G$ such that $\displaystyle g_1\notin \langle g\rangle$ we'd have that $\displaystyle \text{ord }G\ge pq+1$ which is clearly false. Thus $\displaystyle G=\langle g\rangle\quad\blacksquare$.