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How do you prove a group of order 27 can be non abelian?

From what I tried:

Say $|G|=27$ then it's a p-group therefore the ceneter is non trivial so $|Z(G)|$ could be either $3,9,27$ if $|Z(G)|=27$ then the group is abelian. if $|Z(G)|=9$ then from lagrange $|G/Z(G)|=3$ therefore cyclic therefore $G$ is abelian (which is very strange to me although I'm using the theorems I just found an abelian group $G\neq Z(G)$). and if $Z(G)=3$ then I know that $G/Z(G)$ is abelian because it's of order 9 but I don't know how to continue from here.

Any help will be appreciated.

Xam
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Kim Seel
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  • You may enjoy http://math.stackexchange.com/questions/999247/if-g-zg-is-cyclic-then-g-is-abelian-what-is-the-point – lhf Mar 18 '17 at 15:14
  • Arguing in generalities is the wrong thing to do. It's pointless. It's like saying "There can be a prime between $n$ and $2n$ for $n>1$ because $2n-n>1$." That proves that there's an integer in that range, which could be prime, but it doesn't prove there's a prime, and not every integer in the range is prime. – Matt Samuel Mar 18 '17 at 19:35

1 Answers1

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You should prove by giving an example since not every group of order $27$ are non-abelian.

Take $$G=\{\begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix}\;|\; a,b,c\in \Bbb{Z}_3\}$$ Try to verify that this group is a non-abelian group of order $27$.
In fact it is called the Heisenberg group over $\Bbb{Z}_3$.
It is subgroup of $GL(n,\Bbb{Z}_3$)

Wang Kah Lun
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  • Thank you this will definitely work but I'm looking for a constructive answer. I would use the argument that if $|G| \neq |Z(G)|$ then the group is not abelian but for some reason if $|G|\neq |Z(G)|=9$ this results that G is abelian so I cannot use this argument if $|Z(G)|=3$ do you understand my confusion? – Kim Seel Mar 18 '17 at 12:21
  • @KimSeel In fact for every nonabelian group of order $p^3$, $|Z(G)|=p$, $Z(G)=G'$ and $G/Z(G)\cong \Bbb{Z}_p \times \Bbb{Z}_p$. Do you mean you want to construct a nonabelian group of order $27$ using these informations? – Wang Kah Lun Mar 18 '17 at 12:25
  • I know if the group is non-abelian of order $p^3$ then what you said follows. but if all the information I have is the the group is of order $p^3$,$|Z(G)|=3$, then how do I prove the group is non abelian? – Kim Seel Mar 18 '17 at 13:00
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    @KimSeel A group $G$ is abelian iff $G=Z(G)$. If $|Z(G)|=3$, clearly $G\neq Z(G)$, hence $G$ is nonabelian. – Wang Kah Lun Mar 18 '17 at 13:22
  • @AlanWang 's solution to the question you pose is perfect. – ancient mathematician Mar 18 '17 at 13:24
  • I agree with the theorem that $G$ is abelian iff $G=Z(G)$ but then what's wrong with my proof that when I assumed $|Z(G)|=9$ and I proved that G is abelian? – Kim Seel Mar 18 '17 at 14:22
  • Finally I know what you are confused with. Suppose $|Z(G)|=9$, $G/Z(G)$ is cyclic and hence $G$ is abelian. Hence $G=Z(G)$; a contradiction. This means that "$|Z(G)|=9$" can never be happened. – Wang Kah Lun Mar 18 '17 at 14:26
  • Ok, how can I be sure there's no contradiction with $|Z(G)|=3$ and thus G must be abelian? – Kim Seel Mar 18 '17 at 14:34
  • @KimSeel I can't understand what you mean. – Wang Kah Lun Mar 18 '17 at 14:36