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[My post below is wrong] > In fact, in any group with binary operation +, identity element 0, and a non-identity element a, we have a + a + a = a if and only i

by HWR_14 11mo ago

[My post below is wrong] > In fact, in any group with binary operation +, identity element 0, and a non-identity element a, we have a + a + a = a if and only if a + a = 0 (i.e. a has order 2). The "if" is correct. The "only if" is not. (I assume that '+' and '0' are used as shorthand for "any binary operation" and "the identity of that binary operation", as I don't recall cases where "+" and "*" are used for specific types of binary operations).

susam 11mo ago
> The "if" is correct. The "only if" is not. Both "if" and "only if" are correct. Let a + a + a = a. Adding the inverse of a to both sides, we get a + a = 0. Let a + a = 0. Adding a to both sides, we get a + a + a = a. > I assume that '+' and '0' are used as shorthand for "any binary operation" and "the identity of that binary operation" Yes. As I mentioned in my previous comment, "In the last two examples, it is conventional to use product notation instead of +, although whether we use + or · for the binary operation does not matter mathematically." In multiplicative notation, the statement becomes: a·a·a = a holds if and only if a·a = e, where e denotes the identity element.
- HWR_14 11mo ago
  > mentioned this in my previous comment You did. I'm sorry I glossed over the ending to your comment. I was focused on a counterexample I was working on and went only on my memory of group theory. > Adding the additive inverse of a, i.e., -a from both sides, we get a + a = 0. That assumes associativity, but that's a nitpick, not a real objection. In reality, I got a bit tired and mentally shifted the question to a + a + a = 0, not a + a + a = a. That of course has numerous examples. But is irrelevant. Thanks for taking the time for the thoughtful, and non-snarky, response. Sorry if I was abrupt before.
  - susam 11mo ago
    > That assumes associativity, but that's a nitpick, not a real objection. I don't think that is a valid nitpick. My earlier comments assume associativity because a group operation is associative by definition. If we do not allow associativity, then the algebraic structure we are working with is no longer a group at all. It would just be a loop (which is a quasigroup which in turn is magma). > Thanks for taking the time for the thoughtful, and non-snarky, response. Sorry if I was abrupt before. No worries at all. I'm glad to have a place on the Internet where I can talk about these things now and then. Thank you for engaging in the discussion.
    - HWR_14 11mo ago
      You are again right. I misrecalled a group as a loop. Thank you again. It's been too long since I've had to use this knowledge and am happy to have the opportunity to (try to) use it.
patrickthebold 11mo ago
I'd be good to give an example of where the 'only if' doesn't apply. If only to clear up the confusion.
- HWR_14 11mo ago
  Sorry, I had a mental skip. I was thinking of solutions to a+a+a=0, not a+a+a=a.

by HWR_14 11mo ago

susam 11mo ago

> The "if" is correct. The "only if" is not. Both "if" and "only if" are correct. Let a + a + a = a. Adding the inverse of a to both sides, we get a + a = 0. Let a + a = 0. Adding a to both sides, we get a + a + a = a. > I assume that '+' and '0' are used as shorthand for "any binary operation" and "the identity of that binary operation" Yes. As I mentioned in my previous comment, "In the last two examples, it is conventional to use product notation instead of +, although whether we use + or · for the binary operation does not matter mathematically." In multiplicative notation, the statement becomes: a·a·a = a holds if and only if a·a = e, where e denotes the identity element.

HWR_14 11mo ago
> mentioned this in my previous comment You did. I'm sorry I glossed over the ending to your comment. I was focused on a counterexample I was working on and went only on my memory of group theory. > Adding the additive inverse of a, i.e., -a from both sides, we get a + a = 0. That assumes associativity, but that's a nitpick, not a real objection. In reality, I got a bit tired and mentally shifted the question to a + a + a = 0, not a + a + a = a. That of course has numerous examples. But is irrelevant. Thanks for taking the time for the thoughtful, and non-snarky, response. Sorry if I was abrupt before.
- susam 11mo ago
  > That assumes associativity, but that's a nitpick, not a real objection. I don't think that is a valid nitpick. My earlier comments assume associativity because a group operation is associative by definition. If we do not allow associativity, then the algebraic structure we are working with is no longer a group at all. It would just be a loop (which is a quasigroup which in turn is magma). > Thanks for taking the time for the thoughtful, and non-snarky, response. Sorry if I was abrupt before. No worries at all. I'm glad to have a place on the Internet where I can talk about these things now and then. Thank you for engaging in the discussion.
  - HWR_14 11mo ago
    You are again right. I misrecalled a group as a loop. Thank you again. It's been too long since I've had to use this knowledge and am happy to have the opportunity to (try to) use it.

patrickthebold 11mo ago

I'd be good to give an example of where the 'only if' doesn't apply. If only to clear up the confusion.

HWR_14 11mo ago
Sorry, I had a mental skip. I was thinking of solutions to a+a+a=0, not a+a+a=a.