39

For instance addition is commutative, but the inverse, subtraction, is not.

$$ 5+2 = 2+5\\ 5-2 \neq 2-5 $$

Same for multiplication/division:

$$ 5\times4 = 4\times5\\ 5/4 \neq 4/5 $$

So is there a group operation $\circ$ with the inverse $\circ^{-1}$ such that

$$ a\circ b = b\circ a\\ a\circ^{-1}b = b\circ^{-1}a $$

Blaisorblade
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Frank Vel
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    Define inverse operation of a given operation, please. – Git Gud Feb 12 '15 at 13:00
  • @GitGud That's also something I was wondering about when I wrote the question. Is the inverse operation always $a\circ^{-1}b := a\circ b^{-1}$ (or $a^{-1}\circ b$ depending on the order), or are there other ways to define it? I figured that should be another question instead of having two questions. But perhaps I should've asked that one first instead... – Frank Vel Feb 12 '15 at 13:35
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    I think the correct definition of an inverse operation here is that $(a \circ b) \circ^{-1} b = a = (a \circ^{-1} b) \circ b$, all $a$ and $b$. If we weren't going to require commutativity anyway, you would probably want to require the same equations with left and right swapped in at least the obvious way. – Erik P. Feb 12 '15 at 20:24
  • @ErikP. can you deduce the standard definition from the one you gave? I think you also need to repeat from $\circ$ the "associativity" law: $(a \circ b) \circ^{-1} c = a \circ (b \circ^{-1} c)$ (and a symmetric version, given again by commutativity in this instance). That way, it's easy to prove that $(a \circ^{-1} a^{-1}) \circ (b \circ^{-1} b^{-1}) = a \circ^{-1} a^{-1} = b \circ^{-1} b^{-1}$ — it seems I need to reassociate twice before applying your law. – Blaisorblade Feb 13 '15 at 19:53
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    @ErikP. Ah, apparently you gave the definition of quasigroups in universal algebra (http://en.wikipedia.org/wiki/Quasigroup); in fact, "an associative quasigroup are either empty or a group". – Blaisorblade Feb 13 '15 at 20:05
  • Kudos @goblin for linking to quasigroups. – Blaisorblade Feb 13 '15 at 20:08
  • fvel, could you please edit your question to specify that you use the standard group definition (which we all assumed)? The more general version of the question is @goblin's. – Blaisorblade Feb 13 '15 at 20:16
  • @Blaisorblade I wouldn't know the difference, and it doesn't seem to have bothered anyone else. But if you think it's necessary you're welcome to do the edit yourself. – Frank Vel Feb 14 '15 at 09:05

6 Answers6

49

On natural numbers, the bitwise XOR operation is commutative, and is its own inverse operation (the neutral element is$~0$).

Marc van Leeuwen
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    Also known as the Nim sum. – GEdgar Feb 12 '15 at 13:12
  • What about normal or? – tox123 Feb 12 '15 at 16:07
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    @tox123 normal or (inclusive or) has no inverse: there's no way to invert $(\ldots \lor \sf true)$. – chi Feb 12 '15 at 16:27
  • Are there any examples on the reals where the operation is not its own inverse? – Ross Presser Feb 12 '15 at 23:38
  • @RossPresser: the other (2-input) bitwise operations are commutative but not their own inverse. (like or, and, xnor) – Roy Tinker Feb 13 '15 at 01:30
  • Clarifying: Are there non-bitwise operators which are commutative, that are not their own inverse, where an inverse exists and that inverse is also commutative? – Ross Presser Feb 13 '15 at 17:54
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    @RossPresser: Other answers (notably the accepted one) show that all elements must be of order (dividing) $2$, which mean the operation is its own inverse. This can be deduced using (in addition to what the question mentions) only associativity. Without associativity there are other examples, see the question that goblin linked to. – Marc van Leeuwen Feb 13 '15 at 19:07
  • @RossPresser If the operation forms a group, then by Fredrik Meyer's answer, all elements of the group have order 2. Therefore the group is isomorphic to a direct sum of copies of $\mathbb{Z}/2\mathbb{Z}$, so up to renaming the elements bitwise XOR on Boolean strings seems to be the only example. – Sasho Nikolov Feb 13 '15 at 19:17
48

Note that $-$ (minus) is really just a short way of writing $+$ something negative.

In fact, what you call $\circ^{-1}$ is just the composition $$G \times G \xrightarrow{id \times inv} G \times G \xrightarrow{\cdot (mult)} G$$

So the condition you are asking for $a \circ^{-1} b= b \circ ^{-1} a$ is equivalent to the condition $a\circ b^{-1}=b \circ a^{-1}$.

But this is equivalent to $a = b \circ a^{-1} \circ b$. If you demand that $G$ is commutative, then this is equivalent to $a^2=b^2$, which for example is true if all elements have order $2$.

EDIT As Klaus Draeger points out below, the implication that all elements have order two does not need commutativity (see his comment). But then again, if all elements have order two, the group must be commutative...

Fredrik Meyer
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22

If we define $\circ^{-1}$ for a group, we have a neutral element $1$. Applying commutativity with b = 1 we get

$$1\circ^{-1} a = a \circ^{-1} 1$$ but this simplifies to $$a^{-1} = a$$

so each element is its own inverse.

Blaisorblade
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ratchet freak
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    Don't know why this got downvoted; this is actually a pretty neat way to see that all elements must have order$~2$; easier than in the accepted answer by Frederik Meyer. It would have been a bit more readable though if it started with the inverse of $a$ on the far left, and had used $1$ as neutral element to match the multiplicative notation, but after all that is just notation. – Marc van Leeuwen Feb 12 '15 at 17:23
  • I wasn't the original downvoter, but this is worded in a needlessly confusing way — the right equation doesn't follow from a∘1=a=1∘a. I see no good reason why reading such a simple proof should require going to the whiteboard: http://math.stackexchange.com/a/1146999/79293 – Blaisorblade Feb 13 '15 at 20:13
9

If $G$ is an abelian group, then $ab^{-1}=ba^{-1}$ for all $a,b\in G$ if and only if each element of $G$ has order at most $2$.

Jonas Meyer
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5

The notion of inverse of an element can still be defined in the setting of inverse semigroups, which generalize groups.

An inverse semigroup is a semigroup $S$ such that, for every $x \in S$, there is a unique $x^*$ such that $x x^* x = x$ and $x^* x x^* = x^*$.

Then it can be seen that the operation $/$ defined by $x / y = x y^*$ is commutative if and only if $x = x^3$ for all $x$, if and only if $x = x^*$ for all $x$.

In particular, in every semilattice (a semilattice being a special case of inverse semigroup) the operation $/$ is commutative.

Edit. One can also put the result the following way. The operation $/$ is commutative if and only if the multiplication on $S$ is commutative and coincides with $/$.

Bonus. The operation $/$ is commutative if and only if it is associative.

polmath
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  • Note that the question was now changed to be only about groups... This answer is (now) more relevant to the (new) question about quasigroups: http://math.stackexchange.com/questions/1145034/what-are-the-commutative-quasigroups-satisfying-a-b-b-a – the gods from engineering Feb 15 '15 at 14:04
  • Also, there's a name for such a quasigroup; it is called a "totally symmetric quasigroup". – the gods from engineering Feb 23 '15 at 05:16
0

Part of the problem here is the difference between a function and an operation.

A function f(x) is defined as the mapping from objects x to their counterparts f(x). The inverse function f-1(x) is the mapping that takes the objects back to where they started.

So f(f-1(x))=x

Under this definition, “adding” is not a function, but “adding two” is a function.

I believe that what you want is what I would describe as a binary operation. It is effectively a function of two values.

So the adding function can be written as add(x, y).

In this description commutativity means that add(x, y) = add(y, x)

I'm not sure that a binary operation can have an inverse in the proper sense, because how can we undo the addition to get the original pair of values? Given that add(x, y) = 6, for example, we can never determine which values x and y were added together.

So let's try to rephrase the question. I started thinking about addition and its “inverse” (or perhaps better “reverse”) subtraction. Subtraction can also be written as subtract(x, y) and it is not commutative; subtract(x, y) is not equal to subtract(y, x).

The effect of subtraction is really to reverse the addition process.

So subtract(add(x,y), y) = x

The question is therefore if there are a pair of functions f and g that act on two values that have this property:

f(g(x, y), y) = f(g(x, y), x)

But if f is the reverse of g, that would mean that f(g(a, b), b) = a

So we would end up with y = x

Therefore the answer to your question is, “No.”

tomi
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  • First, the OP is using "inverse" in the sense used in early school (division is inverse to multiplication, etc., as he says), while you're using inverse in the standard mathematical sense. Both senses exist - in my native language and (I believe) in English.

    Second, have you read the other answers? They have a different result, so there must be an error somewhere.

    Third, I'm convinced the bug is in your proof, and I think that f(g(x, y), y) = f(g(x, y), x) is wrong — how does it relate to the statement? I only see tempting but wrong ways to get there.

     – Blaisorblade Feb 15 '15 at 11:12
  • @Blaisorblade: tomi's inverse is not in any "standard mathematical sense" I know of. I'm not aware of any such standard sense for inverting a function of two arguments other than considering them as a tuple, which would give you non-unique inverses here. What tomi is talking about is in quasigroup sense. His formula is wrong though, as you noticed. The correct one for inverse would be (in his notation) $g(f(x,y),y) = f(g(x,y),y)$. And commutativity for both $f$ and $g$ would mean $g(f(x,y),y) = g(y, f(x,y)) = f(g(x,y),y) = f(y, g(x,y))$, but this still does not imply his formula. – the gods from engineering Feb 15 '15 at 13:51
  • @RespawnedFluff: A function of a tuple is just a unary function from the Cartesian product; as such, the standard definition of inverse (as used by tomi at the beginning of his answer) applies — and there are invertible functions from e.g. N × N → N (as used in the proof that Q ≅ N). The second part of the answer is indeed about the quasigroup sense — and we agree there. – Blaisorblade Feb 16 '15 at 03:45