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Possible Duplicate:
$|G|>2$ implies $G$ has non trivial automorphism

I am doing this exercise:

Find all groups $G$, with $\text{Aut}(G)=\{1\}$.

What has been clear to me is the group $G$ should be abelian group. Because we will have $G=Z(G)$ and I see that at least all $\phi_g(x)=g^{-1}xg$ are just identity. Any help is appreciated!

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  • $\begingroup$ Both of the answers to the earlier question give complete solutions. $\endgroup$
    – Brian M. Scott
    Dec 1, 2012 at 9:47
  • $\begingroup$ @BrianM.Scott: I didn't aware of this link. Indeed, sometimes it is hard to find out if your question was asked here or not. Anyway Thanks. $\endgroup$
    – Mikasa
    Dec 1, 2012 at 9:55
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    $\begingroup$ No problem: I know how hard it can be. There have been times when I couldn’t find a question that I’d answered! I actually found this by a Google search on group with trivial automorphism group. $\endgroup$
    – Brian M. Scott
    Dec 1, 2012 at 9:58
  • $\begingroup$ @BrianM.Scott: Sorry Brian, but closing the question means a bad action done by me here? I've not done such this. Sorry for asking. $\endgroup$
    – Mikasa
    Dec 1, 2012 at 10:21
  • $\begingroup$ No, you didn’t do anything wrong; as you said, it can be very hard to find out whether a question has already been asked, and it was a perfectly good question in itself. I only found the earlier one by accident. $\endgroup$
    – Brian M. Scott
    Dec 1, 2012 at 10:25

1 Answer 1

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All the inner automorphisms of $G$ are trivial, so $G$ is abelian.

But for an abelian group $G$, the map $g\mapsto g^{-1}$ is an automorphism. So this must be the identity map, i.e. every nonidentity element has order $2$.

In other words, $G$ is a vector space over $\mathbb{Z}/2\mathbb{Z}$. Standard linear algebra techniques show that any space of dimension at least $2$ has a non-trivial automorphism. Thus $G$ must have dimension $0$ or $1$, i.e. $G$ is the trivial group or cyclic of order $2$.

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  • $\begingroup$ In fact, the map $g\to g^{-1}$ would be a nonidentity auto. if the order of $g$ is grater than 2? $\endgroup$
    – Mikasa
    Dec 1, 2012 at 9:51
  • $\begingroup$ @BabakSorouh: Not necessarily. Consider the group $G = \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z}$. In general, if every non-identity element of $G$ has order $2$, then $g \mapsto g^{-1}$ is the identity map. $\endgroup$
    – Michael Joyce
    Dec 1, 2012 at 13:32

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