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Math Help - Prove that the order of Z(G) is either <e> or G

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    Prove that the order of Z(G) is either <e> or G

    Let  G be a group of order  pq , with  p and  q (not necessarily distinct) primes. Prove that the center  Z(G) = <e> or  Z(G) = G .

    I started off by doing a proof by contradiction.

    Suppose that  Z(G) \neq <e> ,  Z(G) \neq G

    Then  |Z(G)| divides either  p or  q , which implies that  |Z(G)| = p or  |Z(G)| = q

    We know that  Z(G) is a normal subgroup.

    Consider the quotient group  G/Z(G) .

    Using Larange's Theorem, we know that  |G|/|Z(G)| (*).

    I broke it into two cases: 1:  |Z(G)| = p , 2:  |Z(G)| = q .

    If under case 1, (*) evaluates to  p . Otherwise, in case 2, (*) evaluates to  q .

    This is as far as I've gotten and I'm pretty stumped as how else to follow up after.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by crushingyen View Post
    Let  G be a group of order  pq , with  p and  q (not necessarily distinct) primes. Prove that the center  Z(G) = <e> or  Z(G) = G .

    I started off by doing a proof by contradiction.

    Suppose that  Z(G) \neq <e> ,  Z(G) \neq G

    Then  |Z(G)| divides either  p or  q , which implies that  |Z(G)| = p or  |Z(G)| = q

    We know that  Z(G) is a normal subgroup.

    Consider the quotient group  G/Z(G) .

    Using Larange's Theorem, we know that  |G|/|Z(G)| (*).

    I broke it into two cases: 1:  |Z(G)| = p , 2:  |Z(G)| = q .

    If under case 1, (*) evaluates to  p . Otherwise, in case 2, (*) evaluates to  q .

    This is as far as I've gotten and I'm pretty stumped as how else to follow up after.
    Lemma: If G/\mathcal{Z}(G) is cyclic then, we have that G is abelian.

    Proof: Let x,y\in G, then clearly x\mathcal{Z}(G),y\mathcal{Z}(G)\in G/\mathcal{Z}(G). But, since G/\mathcal{Z}(G) is cyclic we have that

    x\mathcal{Z}(G)=g^m\mathcal{Z}(G) and y\mathcal{Z}(G)=g^n\mathcal{Z}(G) for some g\in G. Thus, we have that x=g^m\text{ } z_1,y=g^n\text{ }z_2 for some z_1,z_2\in\mathcal{Z}(G) (this is because cosets form a partition). And so

    xy=g^m\text{ }z_1\text{ }g^n\text{ }z_2=z_2 \text{ }g^{m+n}\text{ }z_1=z_2\text{ }g^{n+m}\text{ }z_1=z_2\text{ }g^n\text{ }g^m\text{ }z_1 =g^n\text{ }z_2\text{ }g^m\text{ }z_1=yx (since z_1,z_2\in\mathcal{Z}(G)). The conclusion follows. \blacksquare


    Now since \mathcal{Z}(G)\lhd G we have, by Lagrange's theorem that \left|\mathcal{Z}\left(G\right)\right|\mid\left|G\  right|=pq\implies \left|\mathcal{Z}\left(G\right)\right|=1,p,q,pq. If it's the first or the last we're done. So, assume that it could be either p or q. Then

    G/\mathcal{Z}(G)=p,q and since all prime ordered groups are cyclic it follows from the lemma that G is abelian and so \mathcal{Z}(G)=G which contradicts the assumption that \left|\mathcal{Z}(G)\right|=p,q.

    The conclusion follows.
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