Group theory

Can a group be closed under more than one operation?

Wrong terminology my man. When talking of a group, a single operation is implied. And it is already implied that the set in the group structure is closed by it. The correct way of asking your question, if I understood you correctly, is "Can a set be closed under more than one operation?"

Welcome to Infinity. (Translation delay = \Pi - \Phi + 1 = 1/2)

I'm not sure why, but you really bast my ass. I'm usually immune to obvious trolling, but this is really making me butt bludgeoned. Good job.

A field is a group under both addition and multiplication.

他們是誰?

我说中文. Well, not really. Just a little. And I'm pretty sure 他們 is plural. It should just be 他是誰. You've just outted yourself as using google translate and not being a real chinese speaker.

Incorrect

u r thinking of a ring, friend

yes, a semiring

>A field is a group under both addition and multiplication.
Wrong.

oh because 0 isn't invertible under multiplication
fuck I forgot that little point

If there's no constraint on how the two operations have to interact, depending on the order of the group you can define many operations. If it needs to satisfy some distributive property between the two it depends.

季節性數學 Because English is superior.

Of course you can define different group structures on the same set. Simplest example would be [math]C_4[/math] and [math]V_4[/math]. But when you're defining the operation directly as a function on the set, it being closed under the operation is a given, and I think it is a mistake to list it among properties of groups. In general subgroups of a given group can be considered closed under an operation, and the group itself is trivially closed under its operation.

you need a 会 in there..
我会说汉语

>u r thinking of a ring, friend
Incorrect
You aren't thinking about the fact that 0 doesn't have a multiplicative inverse, friend.

This thing oscilates back and froth on the screen, make it symmetric please :D

Well for appropriately chosen n, R^n can have a lot of different non-isomorphic group structures. Eg: R^(2n^2) has the usual closedness under vector addition, but then you can isomorph it into the complex C^(n^2) which has a different addition, and further into M(n,R), the set of nxn matrices

>doesn't have a multiplicative inverse
... do (You) therefore conclude the set of integers a is not a ring, friendo?

您的中国问候语:字沙拉。

Are you actually this dumb? They're not saying you can't have a ring without multiplicative inverses, they're saying even a field isn't a group under its product because 0 doesn't have an inverse.