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Take the following operation $*$ on the set $\{a, b\}$:

  • $a * b = a$
  • $b * a = a$
  • $a * a = b$
  • $b * b = b$

$b$ is the neutral element. Can $a$ also be its own inverse, even though it's not the neutral element? Or does the inverse property require that only the neutral element may be its own inverse but all other elements must have another element be the inverse.

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5 Answers 5

up vote 18 down vote accepted

Yes, an element other than the identity can be its own inverse. A simple example is the numbers $0,1,2,3$ under addition modulo 4, where 0 is the identity, and 2 is its own inverse.

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Sorry for my lack of understanding, but how is 0 the identity? –  Matt Ellen Jun 17 '11 at 9:48
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@Matt: $0 + a = a$ for all $a$. –  Qiaochu Yuan Jun 17 '11 at 10:01
    
@Qiaochu: Thanks for the help. –  Matt Ellen Jun 17 '11 at 10:49
    
@MikeRand: Yes. Let's rename your $b$, call it $1$, and your $a$, call it $-1$, and think of your $\ast$ as ordinary multiplication. Then we have a group, and $-1$ (your $a$) is its own inverse, but $-1$ is not the identity element. –  André Nicolas Jun 17 '11 at 12:24
    
Qiaochu means 0+a=a and a+0=a, he probably just didn't say it, since one can derive one from the other in the context of modulo addition (either via knowing modulo addition on a set as a group, or that modulo addition commutes), even if we didn't know 0 as the neutral. After all, one of the binary operations F on {0, 1} goes 00F=0, 01F=1, 10F=0, 11F=0, so 0+a=a in this case, but a+0 does not equal a. –  Doug Spoonwood Jun 17 '11 at 15:39
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Your set is isomorphic to the two-element group: $b=1$, $a=-1$, $*=$multiplication. So yes, $a$ can very well be its own inverse.

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Denoting 1 for the identity element, we have for every group element $a$:

$a=a^{-1}\Leftrightarrow a^2=1$.

So a non-identity element is its own inverse iff it has order 2. This is perfectly possible, as Gerry Myerson showed. Or look at the group of automorphisms of $\mathbb{C}-\{0\}$ (invertible ring homomorphisms), the identity element being the identity map, and group multiplication being composition of automorpisms. Then complex conjugation has order 2: it sends $x+iy$ to $x-iy$, and the latter is sent back to $x+iy$.

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Technical: the OP didn't say he was talking about groups. Just a binary operation. –  GEdgar Jun 17 '11 at 13:51
    
Yes, you are right. –  wildildildlife Jun 17 '11 at 14:02
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All the elements that are its own inverses constitute a special example of the so-called elements of finite orders; in the finitely generated commutative groups, these elements constitute a subgroup, called the torsion group, of finite order itself. Of course then, these elements need not be neutral elements.

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First off, let's get some definitions straight.

Definition 1: A neutral n on a set S with a binary operation * we define as an element n of S such that for all s belonging to S, (s*n)=s and (n*s)=s.

Definition 2: An inverse i on a set S of an element j of S which has neutral n and binary operation * we define as an element i of S such that (j*i)=n and (i*j)=n. j and i need not come as distinct, texts just use different letters so that they may come as distinct, as they often do.

Now for the operation * on {a, b} described in the original post, we have b as the neutral. We have (a*a)=b also. Thus, by definition 2 element "a" of {a, b} qualifies as its own inverse. So, you have a simple example in the original post.

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My confusion was based on the fact that, as you noted, texts often use different letters or define the inverse as "another" element. –  MikeRand Jun 17 '11 at 18:08
    
@MikeRand: I agree it's a bit confusing to write 'another element' rather than just 'an element'. Using different letter is on the other hand quite sensible; if i and j are elements it is of course possible that i=j. –  wildildildlife Jun 18 '11 at 11:00
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