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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 $n$.

Dominik
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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 08 '13 at 14:25
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    @user1729 no, not every element need to be a generator. – Tobias Kildetoft Feb 08 '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 08 '13 at 14:30
  • @Tobias: Well, okay, every non-trivial element. But that was implicit! – user1729 Feb 08 '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 08 '13 at 15:00
  • @DanielRust: It does generate the group. It generates it along with any other non-trivial element. A generator is an element which can be used to generate the group (as part of a minimal generating set). I do not mean that the group is cyclic. – user1729 Feb 08 '13 at 15:14
  • @user1729 I have never before seen the term generate used that way. I have seen the term non-generator to refer to an element that can always be removed from a generating set, but calling a non-non-generator a generator seems like a bad idea. – Tobias Kildetoft Feb 08 '13 at 15:52
  • @Tobias: You're right. However, I did not spend overly long composing my hint, and I believe I still got the point across, nit-picking aside! Also, the term "generator of a group" is used in presentations to mean an element which is part of a given generating set, which is what I am most familiar with, and what I was meaning (although that can be nit-picked about too). – user1729 Feb 08 '13 at 16:05

1 Answers1

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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.

Mikko Korhonen
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