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Sorry for such a trivial question, but just wanted to check my understanding.

When proving a statement, for example, that the inverse of a group element is unique (in elementary group theory) one starts by supposing that there exists two inverses $h$ and $k$ for a given element $g \in G$, where $G$ is some group with binary operation $\ast:G\times G \rightarrow G$, such that $h\ast g = g\ast h = e$ and $k\ast g = g\ast k = e$ where $e \in G$ is the unique identity for the group $G$. From this, one can show that $h=k$ in the following manner: $$ h=h\ast e =h\ast\left( g\ast k\right) = \left(h\ast g\right)\ast k = e\ast k =k$$ and as such $h=k$. Now, is the reason we can from this state that the inverse of an element $g\in G$ is unique because $h$ and $k$ were chosen arbitrarily, apart from the requirement that they are inverses of $g$, and as such, if we know the value of one of the inverses, say $h$, then we know that for any other value $k$ to be an inverse of $g$ it must be equivalent to the known inverse $h$, and thus $h$ is the unique inverse of $g$. Is this reasoning correct?

Sorry for the wordiness of this question, but just wanted to check my understanding explicitly. Also, although I understand that, logically, I should have proven this first (but I'd already written out the inverses part before thinking of this, so apologies for that), but is the reasoning the same for arguing that the identity element of a group is unique? (i.e. If we assume that there are two identity elements $e,g \in G$ and subsequently show that $e=g$, then this implies that if we know the form of one of the identities, say $e$, then for any other value $g$ to be an identity of the group $G$ it must be equivalent to $e$ and thus the identity element of a group is unique).

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    $\begingroup$ Yes, the typical way to show something is unique is to assume that there are two, and then prove that they are the same element. If you're saying that $g$ is a group element, then you don't need to show that $g$ has an inverse - it is a consequence of being an element of the group. It's not that they're arbitrary, but they're arbitrary elements with that property - being inverses of some element. Also, I think you mean $e \in V$ is the unique $\textit{identity}$. $\endgroup$ – dannum Aug 27 '14 at 22:38
  • $\begingroup$ Yes, you're right, I meant that $e\in G$ is the unique identity. I'll change it now. $\endgroup$ – Will Aug 27 '14 at 22:46
  • $\begingroup$ I guess my issue is, I was trying to justify why it is enough to assume that there are two inverses and not more than that to imply that the inverse is unique (and the same for the identity)? $\endgroup$ – Will Aug 27 '14 at 22:53
  • $\begingroup$ Well, you're not assuming that there are exactly two, just more than one (I suppose this contradicts what I said above, but it suffices to assume there are two). The way that I like to think of it is that you're fixing one particular one, and then choosing ANY other one that is also an inverse. That is, choosing an arbitrary element with this property, and showing that it must be the same as the one you have fixed. This takes care of every possibility - any number of other inverses, not just two. $\endgroup$ – dannum Aug 27 '14 at 23:05
  • $\begingroup$ Ah, I see. So we assume that a given element $g$ has a known inverse $h$, then if $k$ is any other element in G that satisfies $k\ast g =g\ast k=e$, then we find that $h=k$, and as we didn't specify exactly which element $k$ was, just that it satisfies $g\ast k=e$ this must be true for any $k$. Something like that? $\endgroup$ – Will Aug 27 '14 at 23:31
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Group axioms tell you that an identity element must exist, and also that every element has an inverse. They don't tell you that there's only one identity element, and they don't tell you that an element can have only one inverse: these are things that you have to prove.

So, if for example you want to prove that there is only one identity element, suppose instead that there are more than one. If so, you can choose two of them that are not the same element: yet as the proof goes on you see that those elements are instead the same one, as you know. This means that supposing that there are lots of identity elements yields a contradiction: hence, since you can't have more than one of them, and since at least one has to exist because of the group axioms, you see that the only possible option is that there's exactly only one identity element.

The same happens when you're trying to show that every element of the group has only one inverse.

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  • $\begingroup$ So it's a proof by contradiction then: you assume that (using the inverses example again) there are two distinct inverses for a given element and then arrive at a contradictory statement that actually those two inverses are equivalent and hence must conclude that in fact the inverse of a given element must be unique (as like you said, at least one of them must exist initially from the group axioms). Correct? $\endgroup$ – Will Aug 28 '14 at 13:16
  • $\begingroup$ @WillEmond Yes, you may see it as a proof by contradiction. It's just that you don't assume that there are two of them, because the opposite of "there's only one" is not "there are two of them", but "there are several distinct ones" (or none, but this cannot be because of the group axioms): but then, if there are several elements you can choose two distinct ones among them... and then the proof goes on as you just described. $\endgroup$ – Labba Aug 28 '14 at 13:25
  • $\begingroup$ Thanks, appreciate your help on the matter. So basically you assume that there is more than one inverse (not specifying how many) and then take any two inverses and show that in fact they are the same. As we chose these two inverses arbitrarily (out of the set of inverses that we're assuming exist) this must be true for any two inverses and hence there must actually only be one (unique) inverse. Would this be a correct assessment? $\endgroup$ – Will Aug 28 '14 at 15:57
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Suppose $x\in G$ has two inverses $y$ and $z$ then $$zx=yx\Rightarrow (zx)y=(yx)y\Rightarrow z(xy)=y(xy)\Rightarrow ze=ye$$ So $y=z.$

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    $\begingroup$ I don't think the OP is asking how to prove the statement. $\endgroup$ – Git Gud Aug 27 '14 at 22:35

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