abelian-groupsabstract-algebragroup-theoryp-groups
I know there is a proof using these theorems:
The center of a finite pāgroup is non-trivial
For any group G ,
$G/Z(G)$ is cyclic iff $G$ is abelian, or in otherwords: the quotient
$G/Z(G)$ can never be non-trivial cyclic.
But is there a proof not using these theorems?
Rotman's Introduction to the Theory of Groups (theorem 6.9 on page 130) contains such a proof, attributed to Schenkman, but this proof is not exactly as I have outlined in the question.
The book The Theory of Finite Groups: An Introduction, by Kurzweil and Stellmacher, does what I had in mind in Chapter 2.
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 $GL(n,\Bbb{Z}_3$)
Best Answer
If there is an element of order $p^2$, it's cyclic and thus abelian. Suppose there is no element element of order $p^2$. Then, the order of of the elements of $G$ are either $1$ or $p$. Let $h_1,h_2\in G$ two elements of order $p$ s.t. $h_2\notin\left<h_1\right>$. Then, $\left<h_1,h_2\right>$ is of order $p^2$ and is s.t. $|\left<h_1,h_2\right>|\geq p+1$. Therefore, $|\left<h_1,h_2\right>|=p^2$, and thus $G=\left<h_1,h_2\right>$. Therefore, $G$ is abelian.
We can show that $\left<h_i\right>$ are normal in $G$. Then, $[G:H_i]=p$ and thus $G/H_i$ are cyclic, and thus abelian. Let consider $$\pi: G\longrightarrow G/H_i,$$ defined by $\pi(g)=gH_i$. Take an element of $[G,G]=\left<ghg^{-1}h^{-1}\mid g,h\in G\right>$. You have that $$\pi(ghg^{-1}h^{-1})=\pi(g)\pi(h)\pi(g^{-1})\pi(h^{-1})\underset{G/H_i\ cyclic}{=}H_i$$ and thus and thus $[G,G]\leq H_i$, and since $H_1\cap H_2=\{1\}$, we get $[G,G]=\{1\}$. Therefore $G=G/[G:G]$ is abelian.