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How Aut(G) acting transitively on a finite group G^* can lead G to be elementary abelian group? G^*=G-{1}.

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closed as too localized by Martin Brandenburg, Simon Thomas, Pete L. Clark, Ben Webster Mar 30 '11 at 16:27

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This is an exercise in elementary algebra; I vote for "too localized". –  Martin Brandenburg Mar 30 '11 at 13:44

2 Answers 2

up vote 2 down vote accepted

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 non-trivial.

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 p-group.

Note: The other answer appeared while this was basically typed; I still post it as perhaps the additional details are useful.

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Thanks and sorry for my so simple question. –  Babak Sorouh Mar 30 '11 at 14:04

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

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Thanks Charles for the answer. I should have imagined the group G as a vector space on Z_p first. So, that would certaily lead me to what you mentioned. –  Babak Sorouh Mar 30 '11 at 13:56

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