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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
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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
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3 Answers

up vote 5 down vote accepted

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
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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
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@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
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@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
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If a group has only one element which acts as the inverse of every element in the group, then that group is the trivial group. Proof: the identity must have an inverse, which must be the identity, hence the identity times any element in the group is the identity, hence the identity is the only element of the group.

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.

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