# Math Help - Sanity check about a homomorphism

1. ## Sanity check about a homomorphism

(sighs) I need someone to confirm if I'm going insane again.
Prove that for each fixed nonzero $k \in \mathbb{Q}$ the map from $\mathbb{Q}$ to itself by $q \to kq$ is an automorphism of $\mathbb{Q}$.
If it's an automorphism then it's a homomorphism. But:
$\phi _k : \mathbb{Q} \to \mathbb{Q}: q \mapsto kq$ is not a homomorphism, except for k = 1:

Let $f,g \in \mathbb{Q}$.
$\phi _k (f g ) = kfg$

and
$\phi _k (f) \phi_k(g) = (kf) (kg) = k^2fg$

What am I missing this time? (It's just been that kind of day.)

-Dan

2. ## Re: Sanity check about a homomorphism

Well, it's an automorphism with respect to addition...