# Doesn't look like an automorphism to me…

I am self-studying algebra using Dummit and Foote. One early exercise asks me to prove that for each fixed nonzero $k \in \mathbb{Q}$, the map $\varphi:\mathbb{Q} \to \mathbb{Q}$ defined as $\varphi(q) = kq$ is an automorphism. It's certainly a bijection, but I am having difficulty persuading myself it is a homomorphism. For let $p,q \in \mathbb{Q}$. Then \begin{align} \varphi(pq) & = kpq \end{align} while \begin{align} \varphi(p)\varphi(q) & = k^2pq, \end{align} and unless $k = 1$, these are not equal. Evidently I'm missing something obvious, but I can't figure out what.

• The group you're asked to do prove should be $(\mathbb{Q},+)$ and not $(\mathbb{Q}\setminus\{0\},\times)$? – Frenzy Li May 1 '18 at 0:46
• You are right, this is not a $\mathbb{Q}$-automorphism if we're talking about field automorphisms. Maybe they were talking about group automorphisms if you see $\mathbb{Q}$ as a group under addition. – Antoine Giard May 1 '18 at 0:46
• Well, that's clearly the obvious thing I was missing. Thanks, all. – solitaireartist May 1 '18 at 0:48
• As a group, under addition, this is an automorphism, since scalar multiplication distributes over addition, etc. However, you're correct in that it is not a homomorphism when lookin gat $\mathbb{Q}\setminus\{0\}$ under multiplication, since for any $k\ne 1,$ we have $k1\ne 1,$ and so it does not preserve the identity of $\mathbb{Q}\setminus\{0\}.$ – Chickenmancer May 1 '18 at 0:49

## 1 Answer

I presume the question implicitly means for the semigroup operation to be addition, not multiplication.