# How do I define a multiplication operator here so that this mapping is isomoprhic? [closed] 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?

## closed as off-topic by Toby Mak, Dietrich Burde, user3417, mrtaurho, Yanior WegAug 12 at 19:32

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• This is definitely not the right place to ask if you need the answer in an hour. Can you reach out to anyone you know for help instead? Good luck with your test though. – Toby Mak Aug 12 at 11:53
• @RobbieMeaney You wouldn't say that the mapping is isomorphic (to what?) here; rather you would say that it is an isomorphism. – ಠ_ಠ Aug 12 at 11:56

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!

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

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 $$(a,b,c)*(d,e,f)=(-3ad+af+be+cd,24ad/5-3bd-3ae+bf+ce,24bd/5+24ae/5+cf)$$