Groups do not necessarily have to have only one operation

[Gear300]

To be sure of things, groups do not necessarily have to have only one operation; they may have more, but there is one and only one operation they are identified with. Am I right?

 

[CompuChip]

There may be two operations o and * defined on a set G, such that (G, o) and (G, *) are both groups.
When we say "let G be a group" we're sloppy, you should say what the operation is.

 

[Gear300]

Thus, if I were asked whether the set of all rational numbers Q with addition was a group, would it be valid to assume multiplication as an operation and state that a/b + c/d = (ad + cb)/(bd) ε Q (for a, b, c, d ε Z) in the case of proving that addition is closed with respect to Q.

 

[CRGreathouse]

Thus, if I were asked whether the set of all rational numbers Q with addition was a group, would it be valid to assume multiplication as an operation and state that a/b + c/d = (ad + cb)/(bd) ε Q (for a, b, c, d ε Z) in the case of proving that addition is closed with respect to Q.

Yes, that's valid, but you're not multiplying group elements (which would be bad). You're just expanding the definition of rational addition.

 

[Gear300]

I see. Thanks for the replies.

 

[sutupidmath]

Well, in general there is only one operation that is defined in a group. That is there is only one operation which is applied berween the elements of a group.

In your example, we could say let (Q,@) be a group, where Q is the set of rational numbers, and then we would say that @ is defined in this way:

\frac{a}{b} @ \frac{c}{d}=\frac{a*d+c*b}{b*d}

where + is the natural addition symbol and * multiplication.

But, like it was said above, here you are not multiplying the elements of Q, which we have assumed are of the form

Q={ \frac{a}{b}: a,b \in Z }

 

[HallsofIvy]

But in any case, in
\frac{a}{b}+ \frac{c}{d}= \frac{ad+ bc}{cd}
You are NOT multiplying rational number you are multiplying integers.