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I want to show that a group of order $9$ is abelian without using "Sylow Theorem". I thought of using the theorem"If $$\frac{G}{Z(G)}$$ is cyclic then $G$ is abelian". Since $Z(G)$ is normal in $G$ , the factor group is well defined. Moreover since $|G|=9$, $|Z(G)|$ can be $1$,$3$ or $9$. If $ |Z(G)|=9$, then there is nothing to show. If $|Z(G)|=3$, then $$|\frac{g}{Z(G)}|=3$$ and since any group of prime order is cyclic, $G$ is abelian.

I am stuck with the case where $|Z(G)|=1$. All i need to show is that $|Z(G)| \gt 1$ and I will be through.

This is where I am stuck at the moment.

Any help will be appreciated.

tattwamasi amrutam
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1 Answers1

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That is correct. This is what you need.

It is true that a group of order $p^n$ is has a non trivial center: By the class equation:

$$p^n=|Z(G)|+\sum [G:C_G(x_i)]$$

by Lagrange's we have that $[G:C_G(x_i)]$ is a number multiple of $p$, and looking at the above modulo $p$, we have that $|Z(G)|\equiv 0\pmod{p}$, in particular $|Z(G)|>1$

Daniel Montealegre
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