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

$(a, b) * (c, d) = (ad + bc, bd),$ 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

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.