Is there a name for this type of groups?

A group having more than one elements with only one element as inverse of each element in the group.

Is there any name for that? Let me explain my question: $(\{0\}, +)$ is a trivial group with only one element (additive identity). Additive inverse of element $0$ is $0$ in this group. There is only one element and thus an inverse.

Similarly, if we define a group $(G,\wedge)$ with elements $\{0,1,2\}$ and a binary operation $\wedge$ defined as $$a \wedge b =a^b \ \forall a,b \in G$$. Then it is clear that the identity element in this group is $1$ and inverse of each element is $0$. What should I call this group? I think that there might be a special name for this kind of groups.

edit: Unfortunately I gave a wrong example. But not considering the example here, does there exist such a group with more than one elements where one element works as inverse of all elements?

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I don't understand? What the heck is $2 \wedge 2$ supposed to be? $a^b$ can't be regular exponentiation because then $2^2$ wouldn't be in $G$. –  kahen Jan 2 '12 at 7:55
In any group, if the inverses of $a$ and $b$ are equal, then $a$ and $b$ are equal. So the name for such a group is "trivial". Your $(G,\wedge)$ is not a group: it's not closed under the operation ($2\wedge 2 = 2^2\notin \{0,1,2\}$), the operation is not associative ($(2\wedge 1)\wedge 2 = (2^1)^2 = 2^2$, $2\wedge(1\wedge 2) = 2^{(1^2)} = 2^1\neq 2^2$), and $1$ is not the identity, since $1\wedge 2 \neq 2$. –  Arturo Magidin Jan 2 '12 at 7:55

There is no such thing as a group with more than one element in which the same element works as the inverse of everything.

Since the inverse of the identity is the identity, if $x^{-1}=e$ for all $x\in G$, then $e = xx^{-1} = xe = x$, so $G=\{e\}$. That is, the only group with this property is the trivial group.

More generally, if $G$ is a group, and $x,y\in G$, then $x^{-1}=y^{-1}$ if and only if $x=y$ (different elements have different inverses): if $x^{-1}=y^{-1}$, then $$x = xe = x(y^{-1}y) = (xy^{-1})y = (xx^{-1})y = ey = y.$$

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Confusion solved. Thanks. –  gaurav Jan 2 '12 at 8:11

Given an x in some group G, the inverse of x is unique. So whatever you have is not a group (an obvious stab at what fails is associatity, as this is also the crucial step used in proving uniqueness of inverses in an actual group).

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I think the question is not about having each element with a unique inverse, but rather having a single element "work" as the inverse of everything. Rather than "the inverse of x is unique", the relevant property is that different elements have different inverses. –  Arturo Magidin Jan 2 '12 at 7:58
Oops!! I presented a wrong example. This operation is associative but there are two inverses for element 1. –  gaurav Jan 2 '12 at 8:00
@gaurav: No, the operation is not associative, the set is not closed under the operation, there is no two-sided neutral element, and there are no two-sided inverses. This is not a group, because it fails all the requirements. –  Arturo Magidin Jan 2 '12 at 8:01
@gaurav Furthermore, the statement "This operation is associative but there are two inverses for element 1" is false not just in this example but in any possible example of anything. –  Alex Becker Jan 2 '12 at 8:12
That follows fromy my comment. If x was an inverse for everything, then given a and b; a + x = e = b + x and inverses are unqie so a = b. –  Adam Jan 2 '12 at 17:26
What you have is not a group, since $2\wedge 2 = 2^2\notin G$. But even if we forget the element $2$, we still have $$0 = 0^1 = 0^{(1^0)} = (0^1)^0 = 0^0 = 1$$ a contradiction, hence associativity fails.