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I am wondering if the nonzero rationals with operation $x\circ y = xy/7$ a group. I think it may be and I found the identity to be 7 and found inverses for each element. Mark

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You probably mean to ask "is $(\mathbb R\setminus\{0\},\circ)$ a group, where $a\circ b = ab/7$"? – Myself Mar 17 '11 at 13:10
A group is a set together with an operation. You provided an expression, which is not a group. What is the set, and what is the operation on the set? – Arturo Magidin Mar 17 '11 at 13:17
Yes sorry I meant to specify that 0 is not in the set and that the operation is in Q* . I hope this makes the question a little clearer Thnx Mark – user8379 Mar 20 '11 at 1:18

The question is: Is $(\mathbb{Q}^\times, \circ)$ a group if $a \circ b = \dfrac{ab}{7}$?

Yes, it is a group. Firstly, note that this is a commutative operation, so that if it is a group, then it is abelian.

  1. $\circ$ is easily seen to be associative and closed: $(\frac{a}{b}\frac{c}{d})\frac{e}{f} = \frac{ac}{7bd}\frac{e}{f} = \frac{ace}{49bdf} = \frac{a}{b}\frac{ce}{7df} = \frac{a}{b}(\frac{c}{d}\frac{e}{f})$
  2. $7$ is an identity: $\frac{p}{q} \circ 7 = \frac{7p}{7q} = \frac{p}{q}$
  3. $\left( \frac{p}{q} \right)^{-1} = \frac{49q}{p}$: $\frac{p}{q} \circ \frac{49q}{p} = \frac{49pq}{7qp} = 7$

So it's a group.

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The inverse of $\frac{p}{q}$ is $\frac{49q}{p}$ surely? – fretty Aug 15 '12 at 17:22
@fretty: whoops! You're exactly right! – mixedmath Aug 15 '12 at 17:24

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