binary operation

• Mar 14th 2010, 06:04 AM
alexandrabel90
binary operation
What does it mean that to check if S( sub script A) is a group, we need to check that * is a binary operation?

i thought we only need to check that S( sub script A) satisfy the group axioms of associativity, inverse and identity?

how do i check that * is a binary operation then?

thanks!
• Mar 14th 2010, 06:31 AM
tonio
Quote:

Originally Posted by alexandrabel90
What does it mean that to check if S( sub script A) is a group, we need to check that * is a binary operation?

i thought we only need to check that S( sub script A) satisfy the group axioms of associativity, inverse and identity?

how do i check that * is a binary operation then?

thanks!

Most authors now define: binary operation on a set $\displaystyle A$ is a map $\displaystyle A\times A\rightarrow A$...as simple as that. This means that a bin. op. is a CLOSED operation of two variables from A to itself.
It can be associative, with inverse and etc., but that's another matter.

Tonio
• Mar 14th 2010, 06:58 AM
alexandrabel90
does it mean that i need to show that is h, h' are in A then hh' is in A hence it is closed?
• Mar 14th 2010, 09:09 AM
tonio
Quote:

Originally Posted by alexandrabel90
does it mean that i need to show that is h, h' are in A then hh' is in A hence it is closed?

No, you don't need to do that IF you're given a BINARY OPERATION AND the definition is what I told you.

Now, if you're trying to prove something is a bin. Op. then yes: you need to show closedness.

Tonio