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

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    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
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    Quote Originally Posted by apple2009 View Post
    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).
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    Quote Originally Posted by apple2009 View Post
    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 View Post
    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 $\displaystyle G=\langle g\rangle\quad\blacksquare$.
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