|Z(G)| for |G|=pq

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Abstract Algebra: Let G be a group of order pq, where p and q are distinct primes. We show that either G is abelian or Z(G) = {e}. We give two proofs: the first uses the class equation, the second uses more elementary methods.
  1. MathDoctorBob
  2. Mathematics
  3. abstract
  4. algebra
  5. group
  6. theory
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Комментарии
Автор

@The1000ankit You're welcome, and thanks for the support! I often have problems with replies, especially when using symbols. I'll check into it.

MathDoctorBob
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Your welcome and thanks for the kind words! That sounds like a perfect argument to me. I'll annotate.

MathDoctorBob
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@cpaniaguam Definitely. I also slipped a third proof in there as an annotation - no class equation or quotient groups, just centralizer subgroups. - Bob

MathDoctorBob
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Hallo Bob greetings from Greece. I just saw something too good to be true but cant find the catch.

If G not abelian and x not in Z(G) then Z(x) not= G but also Z(x) not=Z(G) because Z(x) has x in it. Since Z(G) subgroup of Z(x) subgroup of G, G finite, then their orders have to be three different integers. if order of Z(G) is not 1 then only options for |Z(G)|, |Z(x)| are p and q. So p|q or q|p wich cant happen for two different primes, which means |Z(G)|=1.
Your work is awsome, thank you.

vasilischatzipanagiotou
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Sir is it same answer for this question which is as follows : let g be group or oder pq. p, q are prime H=<a(belong to G, a power p = 1 show that H is normal in G

naturelover