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homomorphism. Pick out the true statements

Let $f : (\Bbb Q,+) \to (\Bbb Q,+)$ be a non-zero homomorphism. Pick out the true statements:

a. $f$ is always one-one.

b. $f$ is always onto.

c. $f$ is always a bijection.

d. $f$ need be neither one-one nor onto.

no idea at all how to sove this problem. can anyone help.

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marked as duplicate by DonAntonio, William, Michael Greinecker, Thomas, tomasz Sep 25 '12 at 2:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

hint: think about how $f(1)$ affects what the homomorphism could be. – David Wheeler Sep 17 '12 at 7:12
To elaborate a bit on David Wheeler's hint: Assume that you have decided what $f(1)$ would be. What choices do you have for $f(2)$, $f(1/2)$, $f(3)$, $f(1/3)$, $f(2/3)$, $f(3/2),\ldots$? – Jyrki Lahtonen Sep 17 '12 at 7:24

Let $p=f(1)$.

  1. Prove that $f(2)=2p$; you’ll use the fact that $f$ is a homomorphism.

  2. Prove by induction that $f(n)=np$ for every non-negative integer $n$.

  3. Use the result of (2) and the fact that $f$ is a homomorphism to prove that $f(n)=np$ for every negative integer $n$ as well.

  4. Use (2) to prove that if $n$ is a positive integer, then $f\left(\frac1n\right)=\frac{p}n$.

  5. Prove that for any $q\in\Bbb Q$, $f(q)=qp$; you’ll find it useful to write $q$ in the form $\frac{m}n$ for integers $m$ and $n$.

  6. Conclude that if $f$ is non-zero, then $f$ is a bijection.

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In 4. do you mean $f(\frac{1}{n}) = \frac{p}{n}$? – Arthur Sep 17 '12 at 7:30
@Arthur: I sure do. (I changed an earlier version and typoed it in the process.) – Brian M. Scott Sep 17 '12 at 7:32
Typos aside, what all these steps do is 1. show that any homomorphism is multiplication by some number, and 2. show that multiplication (by a non-zero number) is a bijection in $\Bbb Q$ – Arthur Sep 17 '12 at 7:34
@Arthur: Absolutely; in my notation, by $p$. – Brian M. Scott Sep 17 '12 at 7:40
thank you.............. – mintu Sep 17 '12 at 7:41

Let $f(1) = a$. Then $f(n) = na$ for every integer $n$. Let $n, m \in \mathbb{Z}$, $m > 0$. Then $mf(n/m) = f(n) = na$. Hence $f(n/m) = (n/m)a$. Hence $f$ is bijective, since $a \neq 0$(otherwise $f$ must be $0$).

So the answer is $a, b, c$.

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