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Possible Duplicate:
homomorphism. Pick out the true statements

[NBHM-2006-PhD Screening Test-Algebra]

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.

How can I solve this problem?

  • 1
    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 $f(2)$, $f(1/2)$, $f(3)$, $f(1/3)$, $f(2/3)$, $f(3/2),\ldots$?2012-09-17

2 Answers 2

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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.

  • 0
    @Arthur: Absolutely; in my notation, by $p$.2012-09-17
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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 $a, b, c$.