Let $G$ be a non abelian group with 27 elements.

(a) Prove that $|Z(G)|=3$.

(b) Prove that for all $x\in G$ we have $x^3\in Z(G)$.

For (a), note that $Z(G)$ is a subgroup of $G$, therefore it's order must divide 27. The order of $G$ is a prime power, therefore $|Z(G)|\neq 1$, also $|Z(G)|\neq 27$ since $G$ is non abelian, and $|Z(G)|=9$ would give rise to a cyclic quotient $G/Z(G)$ which again contradicts that $G$ is non abelian. Therefore $|Z(G)|=3$.

For (b), I noted that the elements of $x$ are again of order $1,3,9,27$. For order $1,3$ this is clear. $G$ does not contain an element of order 27, since it would be abelian. We are left with the case that the order of $x$ is $9$. I can't see why this should be true, but I bet there is an easy argument. Could someone point me in the right direction?


1 Answer 1


Since the order of $Z(G)$ is $3$, the order of $H= G/Z(G)$ is $9$. Hence the order of the image of an element $x \in G$ in the quotient group $H$ is either $1$, $3$, or $9$. But we know there is no element of order $9$ in $H$, as then $G$ would be abelian. Hence the only possible orders are $1$ and $3$. This implies that any element $x \in G$ is either already in $Z(G)$ or $x^3 \in Z(G)$.

  • 1
    $\begingroup$ Exactly what I was looking for. Shokran. $\endgroup$ May 23, 2018 at 17:31

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