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I was wondering how one would go about proving the inverse of an inverse of a group element is the element itself, but without being able to use either $a + a^{-1} = a^{-1} + a$ or $0+a=a+0$?

If we only have associativity and $a + a^{-1} = 0$ (that is, if we know $a^{-1}$ is the inverse of a), I just don't see how I'd show $a = (a^{-1})^{-1}$. Any suggestions?

I'm really puzzled by this one, I have to say. I can show the equality if we have $0+a=a+0$ for all a in the group, but not without.

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In fact, I'm not quite sure what your assumptions even are. You've mentioned some things you don't want to use; can you tell us exactly what things you are using? –  Chris Eagle Jan 9 '13 at 20:41
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I think one can prove that the second inverse of an element is the same as the element under fairly weak sounding hypotheses (I'm thinking only being allowed same side identity and inverse), but I agree with @ChrisEagle that we need more information about what is available. –  peoplepower Jan 9 '13 at 20:45
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Why are you mixing up notations, using $+$ for the group operation but $a^{-1}$ for inverse? Usually, we either go with $(-a)$ and $+$ and $a^{-1}$ and $\cdot$ –  Thomas Andrews Jan 9 '13 at 21:00
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Then @Ryker, whoever did it either tried to make some very particular point or else he loves very messy notation...or he's simply wrong. As Thomas told you, the notation $\,a^{-1}\,$ is usually reserved for multiplicative inverse, whereas the additive inverse is almost always denoted $\,-a\,$ . Is this from a book? Because if it is you can try to gives us a link to it and to the page, or copy and paste the particular page. –  DonAntonio Jan 9 '13 at 21:05
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@Ryker Do you see how that too is not helpful? By saying where it did not come from just makes the whole scenario more fantastic –  peoplepower Jan 9 '13 at 21:12

3 Answers 3

up vote 2 down vote accepted

Note that we need to assume that $0 + a = a$ for all $a$ and that for all $a$ there is an $a^{-1}$ such that $a^{-1} + a = 0$ (or of course in the other order, but we cannot just assume those two conditions with the order switched in one of them).

Under these assumptions, let us check that $a + a^{-1} = 0$. We get $a^{-1} + a + a^{-1} = 0 + a^{-1} = a^{-1}$ but we can cancel $a^{-1}$ on the left dues to our assumptions, so we get $a + a^{-1} = 0$ as we wanted.

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Why would we be able to cancel $a^{-1}$ on the left? –  Ryker Jan 9 '13 at 21:08
    
Because we are assuming that each element has a left inverse. –  Tobias Kildetoft Jan 10 '13 at 0:01
    
But of course :) I got confused because my assumptions in regards to "handedness" were the opposite of yours, and it just didn't click. I hang my head in shame... –  Ryker Jan 10 '13 at 1:50

The claim does not follow from the assumptions you are using. For example, suppose $X$ is any set with at least two elements (one of which is called 0), and let $+$ be defined by $x+y=0$ for every $x$ and $y$, and $^{-1}$ be defined by $x^{-1}=0$ for every $x$. Then $+$ is associative, and $x+x^{-1}=0$ clearly holds for every $x$, but $(x^{-1})^{-1}=0$.

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Under the hypotheses that the operation (which I will refer to as juxtaposition) is closed, associative, and satisfies the properties that there is a right identity (I will call this 1) and a right inverse is guaranteed for each element of the set $G$, it is known that $G$ is a group. We are interested in the seemingly smaller result that $a=(a^{-1})^{-1}$ for all $a\in G$.

For any $a$, we compute $a^{-1}a=a^{-1}a1=a^{-1}a(a^{-1}(a^{-1})^{-1})=(a^{-1}1)(a^{-1})^{-1}=a^{-1}(a^{-1})^{-1}=1$ where the associative law was used to cancel middle terms. This proves that the inverse $a^{-1}$ applies to both sides of a. In particular, $a^{-1}(a^{-1})^{-1}=1=(a^{-1})^{-1}a^{-1}$, so we can now prove the desired statement.$$a=a1=a(a^{-1}a)=1a=((a^{-1})^{-1}a^{-1})a=(a^{-1})^{-1}$$

Notice that in doing so, we prove that $a1=a=1a$; all we did was show that $G$ is a group and do the usual proof of idempotence of the inverse in a group.

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Of course, in the last sentence I mean that "the inverse has order 2 in a group" instead of "idempotent". –  peoplepower Jan 10 '13 at 1:46
    
Nice! Thanks! Sorry I didn't choose your answer as accepted, I realized Tobias was also right, and since he posted first I felt obliged to choose his. I actually like your approach better (i.e. it aligns better with what I'd have done), so please don't be offended. Anyway, thanks for all the help, I guess I was really confused that the things you guys proved first were those that were given to us as needing proof in later parts of the question, with my original question being the first one. But I guess we've established you can't prove that without proving what you guys did, no? –  Ryker Jan 10 '13 at 1:54
    
@Ryker Thanks for your kind remarks. I don't get worked up over accepts so don't worry. :) As for the final remark I do not believe a single sequence of equalities will prove more than what we did. In particular, I would be surprised if there was a sequence of equalities linking $a$ and $(a^{-1})^{-1}$ without needing to draw intermediate conclusions along the way. –  peoplepower Jan 10 '13 at 2:05

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