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Let $G$ be a finite group such that for each $a,b \in G \setminus \{e\}$ there is an automorphism $\phi:G \rightarrow G$ with $\phi(a)=b$. Prove that $G$ is isomorphic to $\Bbb Z_p^n$ for some prime $p$ and natural number $n$.

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First hint: Every element has to be a generator for the group. Also, the automorphisms are transitive. What does this tell you about the order of the elements? – user1729 Feb 8 '13 at 14:25
@user1729 no, not every element need to be a generator. – Tobias Kildetoft Feb 8 '13 at 14:27
A good first step would be to show that the groups $\mathbb{Z}_p^n$ have the property in the statement of the question. – Dan Rust Feb 8 '13 at 14:30
@Tobias: Well, okay, every non-trivial element. But that was implicit! – user1729 Feb 8 '13 at 14:54
The non-trivial element (1,0) doesn't generate the group $\mathbb{Z}_2^2$. Infact, $\mathbb{Z}_p^n$ is only generated by a single element if $n=1$. – Dan Rust Feb 8 '13 at 15:00
up vote 11 down vote accepted

Hint 1: If $a, b \in G \setminus \{e\}$, then $a$ and $b$ have the same order.

Hint 2: Using the previous hint, show that $G$ has order $p^n$ for some prime $p$ and that every nonidentity element has order $p$.

Hint 3: In a $p$-group, the center is a nontrivial characteristic subgroup.

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