Is the following subset of $\mathbb R \times \mathbb R$ with the indicated operation a group? Is it an abelian group?

Question

$(a, b) * (c, d) = (ad + bc, bd),$ on the set $\{(x, y) \in \mathbb R \times \mathbb R: y \neq 0\}$.

$(a, b) * (c, d) = (ad + bc, bd) = (cb + da, db) = (c, d) * (a, b).$ Commutativity holds.

$((a, b) * (c, d)) * (e, f)) = ((a, b), f) + ((c, d), e) , ((c, d), f) \neq (a, (e, f)) + (b,(c, d)), (b, (e, f)) = (a, b) * ((c, d)) * (e, f)).$ Associativity fails.

$(a, b) * (e_1, e_2) = (ae_2 + be_1, be_2) = (a, b),$ so $e_2 = 1, e_1 = 0.$ Then, $(e_1, e_2) * (a, b) = (a, b).$ So, there's an identity in this set.

$(a, b) * (a', b') = (ab' + ba', bb') = (e_1, e_2),$ so $b' = \frac{1}{b}, a' = - \frac {a}{b^2}$ Thus $(a', b') = (-\frac {a}{b^2}, \frac 1b).$ Now, $ (a', b') * (a, b) = (ab' + ba', bb') = (\frac ab - \frac ab, 1) = (e_1, e_2).$ So, there's an inverse.

This set is not a group.

I don't really understand what you did in the associativity. — Miguelgondu, Jan 31 '15 at 20:07
You seem to have messed up something in you notation. $((a,b),f)$ does not make sense in this context. — Lukas Juhrich, Jan 31 '15 at 20:11

score 0 · Accepted Answer · answered Jan 31 '15 at 20:13

0

You have made a mistake -- associativity holds: $$ [(ab)*(d,d)]*(e,f) = (ad+bc,bd)*(e,f) = (adf+bcf+bde,bdf)\\ (a,b)*[(c,d)*(e,f)] = (a,b)*(cf+de,df) = (adf + bcf + bde,bdf) $$ Also your demonstration of the identify $(0,1) and the inverse are both right so the operation does form an abelian group.

In the words of Peter vanNieuwenhuizen, "Your work is riddled with an error."

answered Jan 31 '15 at 20:13

Mark Fischler

41,743

Associativity seems to be sooo difficult. I am determined to conquer it :) – AbstraktAlgebra Jan 31 '15 at 20:19

Is the following subset of $\mathbb R \times \mathbb R$ with the indicated operation a group? Is it an abelian group?

1 Answers1