How do I define a multiplication operator here so that this mapping is isomoprhic?

Question

Part a) of this is fine, but I'm really stuck on part b) and I have a test on this in an hours time, does anyone have any hints?

Answer 1

I think the best place to start is to first give things names, e.g.

$$(a,b,c) \ast (\lambda, \mu, \nu) =: (x,y,z) $$

And then simplify the RHS of the expression

$$ F(x,y,z) = F(a,b,c) \cdot F(\lambda, \mu, \nu)$$

to get an expression for $x$, $y$ and $z$. You'll use part a) for this.

EDIT: $(\alpha, \beta, \gamma)$ were not the best choice of letters!

Good luck!

Answer 2

$(a\alpha^2+b\alpha+c)(d\alpha^2+e\alpha+f)=ad(\alpha^4)+(bd+ae)\alpha^3+(af+be+cd)\alpha^2+(bf+ce)\alpha+cf$

Then use the relation $\alpha^3=-3\alpha+24/5$ to reduce this to something like

$ad(-3\alpha^2+24/5\alpha)+(bd+ae)(-3\alpha+24/5)+(af+be+cd)\alpha^2+(bf+ce)\alpha+cf =(-3ad+af+be+cd)\alpha^2+(24ad/5-3bd-3ae+bf+ce)\alpha+(24bd/5+24ae/5+cf)$

which will give us a way to compose $(a,b,c)*(d,e,f)=(-3ad+af+be+cd,24ad/5-3bd-3ae+bf+ce,24bd/5+24ae/5+cf)$