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This is straight out of Fraleigh's book, "A first course in abstract algebra":

Let $\mathbb{R}^*$ be the set of all real numbers except $0$. Define $*$ on $\mathbb{R}^*$ by $a * b = |a|b.$

One of the problems is to show there is a left identity and right inverse, but I can't (obviously by design) get unique solutions out of the problem:

Suppose there's a left identity: $e * x = x \implies |e| = 1$, which gives us two possible identities. If I try to find a right identity, the solution is similar (though possibly different in sign): $e = \frac{x}{|x|}$.

For the right inverse ($x'$), the same problem comes up: $x * x' = e \implies |x|x' = e \implies x' = \frac{e}{|x|}$.

Not Legato
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    You mean left identity not right identity. And yes, there are no reason why left identity is unique, when you don't have a right identity. – user10354138 Jun 05 '19 at 20:42
  • Are there any constraints on $\Bbb{R}^*$? Is it meant to be a particular kind of algebraic structure? – R. Burton Jun 05 '19 at 20:50
  • user10354138: Does that mean the question is "show there is /an/ identity element", even if there are technically multiple, and these do not form a group?

    R. Burton: It's nothing but $\mathbb{R} - {0}$.

     – Not Legato Jun 05 '19 at 20:53
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    If it is not specified that $\Bbb{R}^*$ is any particular kind of structure (ex. a group, ring, etc.), then there is no reason that the identity or the inverse should be unique (unless otherwise specified). – R. Burton Jun 05 '19 at 20:56
  • Ah, so essentially the last question "Explain the significance of this exercise" probably refers to a lesson about the fact that the sheer existence of /an/ identity or inverse doesn't immediately give you a group structure, and one should be careful. – Not Legato Jun 05 '19 at 21:03
  • You can't have the "right identity" be $\frac{x}{|x|}$; because the right identity must work for every $x$, and here your $e$ changes depending on who $x$ is. So this doesn't work. – Arturo Magidin Jun 05 '19 at 21:25
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    @NPerhaps you should also notice that you have to be careful: One may define group (with associativity given) in terms of unique two-sided identity and unique two-sided inverse, but in fact existence of a left-identitiy and a left-inverse are enough; likewise right-identity and right-inverse a enough. But as this exercise shows, mixing these restrictions (left-identity and right-inverse) fails – Hagen von Eitzen Jun 06 '19 at 06:03

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