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Math Help - 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 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.
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