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Does there exist a non-trivial group $G$ without automorphisms, a homomorphism $f:H\to G$ and a homomorphism $g:G\to H$ such that $g\circ f = \mathrm{id}_H$ for some group $H$ with non-trivial automorphisms.

The answer is no:

We have $G\cong C_2$, and $f$ is injective. There are no subgroups of $C_2$ with non-trivial automorphisms.

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Hint; there are only two groups without non-trivial automorphisms! –  Chris Eagle Dec 23 '11 at 21:34

1 Answer 1

up vote 1 down vote accepted

First of all, such groups must be abelian, then $x\mapsto x^{-1}$ is an automorphism of $G$ and in order to be the trivial automorphism, we should have $x^{-1}=x$ for any $x \in G.$ That means that, $G$ is the additive group of a vector space over $\mathbb{Z}/2,$ but it can be seen that if $|G|>2$ the cardinality of the group of automorphism of such vector spaces is either infinite or an even number. Therefore, $|G|$ is $1$ or $2.$

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