Automorphism of the Additive Group of Rationals

Join Trial or Access Free Resources

Any automorphism of the group Q under addition is of the form x → qx for some q ∈ Q.

True

Discussion: Suppose f is an automorphism of the group Q. Let f(1) = m (of course 'm' will be different for different automorphisms). Now $f(x+y) = f(x) + f(y)$ implies $f(x) = mx$ where m is a constant and x belongs to set of integers (Cauchy's functional equation).

Now suppose x is rational. Then x = p/q where p and q are integers. Hence $f(p) = mp$. But $p = qx$ hence $f(p) = f(qx) = f(x+x+ ... + x) = qf(x)$

Therefore, $mp = qf(x)$  implies $ m \times {\frac{p}{q} }= f(x) \implies f(x) = mx $ where $m = f(1)$

More Posts

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

One comment on “Automorphism of the Additive Group of Rationals”

© 2010 - 2025, Cheenta Academy. All rights reserved.
linkedin facebook pinterest youtube rss twitter instagram facebook-blank rss-blank linkedin-blank pinterest youtube twitter instagram