[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 6 hours ago | parent | next [-]

> 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 6 hours ago | parent [-]

> 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 6 hours ago | parent [-]

> 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 5 hours ago \| parent [-]
		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 6 hours ago | parent | prev [-]

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 6 hours ago \| parent [-]
		Sorry, I had a mental skip. I was thinking of solutions to a+a+a=0, not a+a+a=a.