How Aut(G) acting transitively on a finite group G^* can lead G to be elementary abelian group? G^*=G{1}.
closed as too localized by Martin Brandenburg, Simon Thomas, Pete L. Clark, Ben Webster♦ Mar 30 '11 at 16:27This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question. 


A way to see this is: a. An automorphism maps the center of $G$ to the center of $G$. Thus, under your condition the center of $G$ is the full group and it is thus abelian, or the center is trivial. b. Note that an automorphism preserves the order of an element. Thus the transitivity implies that all elements of $G$ except $1$ have the same order $n$. And this $n$ (thus) has to be prime. In particular, the center is nontrivial. Thus, $G$ is a finite abelian $p$group in which all elements (except $1$) have order $n$ and $n$ is prime. That is, it is an elementary abelian pgroup. Note: The other answer appeared while this was basically typed; I still post it as perhaps the additional details are useful. 


G must have exponent p where p is prime; and G must be characteristically simple. These are both easy arguments. Isn't that enough? 

