Let Gpq for some primes p q show that either the center is t

Let |G|=pq for some primes p &q. show that either the center is trivial or the group is abelian.

Solution

Question: Let |G|=pq for some primes p &q. show that either the center is trivial or the group is abelian.

Answer:

Let us assume G is abelian. Then we know that Z(G)=G, thus the centre must have order pq.

Now, let us assume that G is non-abelian. By Cauchy’s theorem we know there must exist elements x and y of respectively order p and q. Without loss of generality we can assume p q, because [G:y]=p, and p is the smallest prime divisor of G, y must be a normal subgroup of G.

            Let us claim that x is a complete set of representatives for y in G. We know, because p is prime, that x is cyclic and therefore x={xk:k{0,…,p1}}. We can assume i< j. If xiy= xjy, then x ixj= xj iy. But xj ix, thus of order p, hence xj i can\'t be in y. As so, x gives a full set of representatives.

          Now, we can write every element in G as xmyn, with m{0,…,p1} and n{0,…,q1}. Furthermore, every element other than the identity element has order p or q, or else G would be cyclic, and hence, the group is abelian.

           Furthermore, we have to prove that there exists an element in G which doesn\'t commute with anything in G, except for the identity element. Here, the order of Z(G) must divide the order of the group G. So either it must be 1,p,q or pq. However it can\'t be pq, or else the group would be cyclic and thus abelian. Nor can it be p, since then [G:Z(G)]=q, and thus it\'s cyclic, so by a specific theorem G must be abelian. The same argument is for q. Therefore, when the order of Z(G) must divide the order of the group G, then, it be 1,p,q and therefore, centre is trivial or the group is abelian.

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