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Can someone give me an example of an group endomorphism that is injective, but not surjective?

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Clearly $G$ has to be infinite. What's the first infinite group you think of? – Chris Eagle Feb 11 '13 at 21:29
The integers under addition? – user61882 Feb 11 '13 at 21:30
$\lim\limits_{\text{learning effect} \to 0} = \text{posting a complete solution to this question}$. – Martin Brandenburg Feb 11 '13 at 22:49

$f : (\mathbb{Z},+) \to (\mathbb{Z}, +), \quad x\mapsto 2x$

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What a simple but good answer. – dinoboy Feb 11 '13 at 21:44

If $f: G \rightarrow G$ is injective but not surjective, then $f(G)$ is a proper subgroup of $G$ and $f(G) \cong G$.

Furthermore, if $H$ is a proper subgroup of $G$ and $H \cong G$, then there exists an isomorphism $\phi: G \rightarrow H$. Since $H$ is a proper subgroup, $\phi$ is a homomorphism $G \rightarrow G$ that is injective but not surjective.

Thus finding an example of a homomorphism $f: G \rightarrow G$ that is injective but not surjective is equivalent to finding a proper subgroup $H$ such that $H \cong G$. In azimut's answer, you have the example $\mathbb{Z} \cong 2\mathbb{Z}$.

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Another example: $$f:\mathbb Q_+\to\mathbb Q_+, f(q) = q^2$$ where $\mathbb Q_+$ is the group of positive rationals under multiplication.

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