In Sets and Groups by Green a question 5 from the chapter 3 reads:
- Write $\theta_{a,b}$ for the map of the preceding exercise [which is $\theta(x)=ax+b=\theta_{a,b}$]. Prove $\theta_{a,b}\theta_{c,d}=\theta_{ac,ad+b}$, for any real numbers a, b, c, d. For which values of a, b, c, d do $\theta_{a,b}$ and $\theta_{c,d}$ commute?
I've done the prove and I'm interested in the last part of question. The answer reads "ad+b=bc+d". I've came up with "when a=c and b=d". Is it wrong?
Note that $$ \theta_{a, b}\theta_{c, d} = \theta_{ac, ad+b}\\ \theta_{c, d}\theta_{a, b} = \theta_{ca, cb + d} $$ We want these to be equal. Well, $ac = ca$ always, so the only thing we need is for the two second parameters, $ad + b$ and $bc + d$, to be equal. This will happen if $a = c$ and $b = d$, but there are other cases too, like $$ \theta_{1, b}\theta_{1, d} = \theta_{1, d}\theta_{1, b} $$ no matter what $b$ and $d$ are (which is to say, $(x+b)+d = (x+d)+b$).
It is, of course, paramount to point out that $$ \theta_{p, q} = \theta_{s, t}\iff (p = s\,\land\,q = t) $$ so the only way that the two composed maps above can be equal is if the parameters are equal.