Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
3  
Clearly $G$ has to be infinite. What's the first infinite group you think of? –  Chris Eagle Feb 11 '13 at 21:29
2  
The integers under addition? –  user61882 Feb 11 '13 at 21:30
5  
$\lim\limits_{\text{learning effect} \to 0} = \text{posting a complete solution to this question}$. –  Martin Brandenburg Feb 11 '13 at 22:49
add comment

3 Answers

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

share|improve this answer
    
What a simple but good answer. –  dinoboy Feb 11 '13 at 21:44
add comment

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 $\mathbb{Z} \cong 2\mathbb{Z}$.

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.