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If $ G $ has no non-trivial automorphism, then $ G $ is abelian and $ g^2 = e $ for all $ g \in G $ .

With the assumption, I dont know how to start the proof.

If there is no non-trivial automorphism, then there is only trivial automorpism, the identity morphism. But how can I show $ g^2 = e $ for all $ g \in G $ with it?

user73309
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    Begin with considering inner automorphisms. – Boris Novikov Apr 25 '13 at 15:38
  • Boris said it well...With your assumption, G has no non-trivial automorphism...what does that mean? – Eleven-Eleven Apr 25 '13 at 15:45
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    The first step is to show that the group is Abelian, using the fact that the inner automorphisms are trivial. Then, you have won something: if (and only if) a group is Abelian, the inversion $x \mapsto x^{-1}$ is an automorphism. Using the hypothesis, you then have $\forall x \in G, x^{-1} = x$, which is equivalent to all elements of the group being involutions. – PseudoNeo Apr 25 '13 at 16:15

3 Answers3

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For every elements $a\in G$, consider $$\phi_a:G\rightarrow G$$ sending an element $b$ to its conjugate $a^{-1}ba$. You can easily check that all $\phi_a$ are automorphisms of $G$, called inner automorphisms. By assumption, $\phi_a=id$, which means $$\phi_a(b)=b$$ for every $a,b\in G$, so that $ba=ab$.

Alternatively, use $$\frac{G}{Z(G)}\cong Inn(G)$$ where $Z(G)$ is the center of $G$ and $Inn(G)$ the group of inner automorphisms, which is trivial in your hypothesis.

As for $g^2=e$, you can use abelianity we have just proved. Since $(ab)^{-1}=b^{-1}a^{-1}=a^{-1}b^{-1}$, the inversion $g\rightarrow g^{-1}$ is an automorphism, which by assumption is trivial. Multiplying $g=g^{-1}$ by $g$ gives $g^2=e$

Federica Maggioni
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Hint: If $Inn(G)$ is the group of inner isomorphisms, which we are taking to be trivial, apply the following theorem:

$G/Z(G)\cong Inn(G)$.

Recall: If $G$ is Abelian, what is the relationship between $G$ and $Z(G)$?

luke
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After the observations above, note that in case $G$ is finite (for infinite groups a similar reasoning applies), it must be an elementary abelian 2-group of rank say $n$. Then $Aut(G) \cong GL(n,2)$. Hence if $|Aut(G)| = 1$ then this can only be the case if $G$ is trivial, or $n = 1$ that is $G \cong C_2$, the cyclic group of order 2.

Nicky Hekster
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