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

1. ## 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

2. 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).

3. 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
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 $G$ have order $pq$, where $p,q$ are distinct primes. Prove $G$ is either cyclic or every element $g\in G-\left\{e_G\right\}$ has order $p$ or $q$

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