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The group is abelian

Let \mathcal{G} be a finite group. If forall a, b \in \mathcal{G} \setminus \{e\} there exists f \in {\rm Aut}(f) such that f(a)=b then prove that \mathcal{G} is abelian.

Solution

Since each automorphish preserves the elements’ order , then all elements have the same order. This order ought to be a prime number. It is obvious from Cauchy’s theorem that the order of the group is p^n where n\geq 1. For n=1, 2 it is well known that the group is abelian. For n \geq 3 we work as follows:

Let |\mathcal{G}|=p^n. Then its centre \mathcal{Z} is not trivial. Hence there exists a \in \mathcal{Z} such that a \neq e. It suffices to prove that f(a) \in \mathcal{Z}. Let g \in \mathcal{G}. Hence:

    \begin{align*} f(a) g = g f(a) &\Leftrightarrow f^{-1} \left ( f(a) g \right )= f^{-1} \left ( gf(a) \right ) \\ &\Leftrightarrow a f^{-1}(g) = f^{-1} (g) a \end{align*}

and the last one is true since a \in \mathcal{Z}. Hence \mathcal{G} = \mathcal{Z} and thus the group is abelian and we are done.

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