4 ms·
A simple example where 1 + 1 + 1 = 1 is ℤ₂, the group of integers modulo 2 under addition. In fact, in any group with binary operation, say +, the identity ele
by susam 11mo ago
A simple example where 1 + 1 + 1 = 1 is ℤ₂, the group of integers modulo 2 under addition.
In fact, in any group with binary operation, say +, the 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).
There are plenty of groups with elements a satisfying a + a = 0. ℤ₂ as mentioned above has its unique non-zero element of order 2. The Klein group V₄ has three non-identity elements, each of order 2. Dihedral groups D₂ₙ (the symmetry groups of regular n-gons) contain reflections, all of which have order 2. Symmetric groups Sₙ (n ≥ 2) contain transpositions, each of which has order 2.
For example, in the dihedral group D₈, if we let a be a reflection of the square, then a + a = 0 and a + a + a = a. But this is conventionally written in multiplicative notation as a² = the identity element, so a³ = a.
Similarly, in the symmetric group S₃ under the binary operation of composition, if a denotes the transposition (12), then (12)(12) is the identity element and (12)(12)(12) = (12). In other words, applying a transposition three times is the same as applying it once.
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. It is conventional to use + in some subjects (coding theory, additive groups of integers modulo n, etc.) and · in others (permutation groups, dihedral groups, etc.). Often + is used for the binary operation in abelian groups and · in non-abelian ones. I'm sure none of this is particularly insightful to someone who has studied group theory, but still I wanted to share a few concrete examples here.
- 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 agoYou 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 agoI'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 agoSorry, I had a mental skip. I was thinking of solutions to a+a+a=0, not a+a+a=a.
- thaumasiotes 11mo ago> The Klein group V₄ has three non-identity elements, each of order 2. Unrelated, but this calls out for a link to the classic song Finite Simple Group (of Order Two) by the Klein Four: https://www.youtube.com/watch?v=BipvGD-LCjU https://www.youtube.com/watch?v=BipvGD-LCjU
- vbsd 11mo ago> A simple example where 1 + 1 + 1 = 1 is ℤ₂, the group of integers modulo 2 under addition. That’s a good example of an algebra where 1 + 1 + 1 = 1, but the article is specifically about systems where in addition to that condition, this second condition is also true: 1 + 1 != 0 (not equal!). ℤ₂ is not an example of that.
- deleted 11mo ago[deleted]