Note "(* )" is an arbitrary binary operation The questions is:

Let S be a set with an associative binary operation (*) and assume that e $\in$ S is the unit for the operation.

a) Show that the inverse for the element $s_1$ (* ) $s_2$ is given by $s_2^{-1}$ (* ) $s_1^{-1}$

b) Show that every element has at most one inverse. Hint: Assume that there are two inverses and prove that they have to be the same.

I think the key of this problem these two definitions:

$s$ (* ) $e$ = $s$ and $s$ (* ) $s^{-1}$ = $e$

I literally spent hours trying to solve this equation I tried several things but at the end it looked like nonsense, basically saying

($s_1$ (* ) $s_2$) (* ) $x$ = $e$
where $x$ is the inverse we substitute $s_1^{-1}$ (* ) $s_2^{-1}$ for $x$ and we get the inverse and since we have the identity as the result. However that doesn't seem very logical and in the question it doesn't say its commutative so I can't just swap $s_1^{-1}$ and $s_2^{-1}$ to get $s_2^{-1}$ (* ) $s_1^{-1}$.


Note (* ) is an arbitrary binary operation


HINT: (I will write simply $s*t$.)

  • Use associativity repeatedly to simplify $(s_1*s_2)*(s_2^{-1}*s_1^{-1})$. First step: $$\color{crimson}(s_1*s_2\color{crimson})*(s_2^{-1}*s_1^{-1})=s_1*\color{crimson}{\big(}s_2*(s_2^{-1}*s_1^{-1}\color{crimson}{\big)}\;.$$

  • If $t_1$ and $t_2$ are both inverses of $s$, calculate $t_1*s*t_2$ in two different ways.

  • $\begingroup$ First of the all thanks for answering. I got the first one I kept simplifying until I got e which I think answers the first part. The second part if you could explain more on what I'm expecting to find, I have simplified it and eventually I got t_1 or t_2 depends on which I choose first but my question is does that prove that there is an inverse for every element of S? $\endgroup$ – Zed Oct 10 '16 at 19:17
  • $\begingroup$ @Z69: You’re welcome. It sounds as if you did indeed get the first part. You probably also got the second — you just don’t realize it. You’re not trying to prove that every element of $S$ has an inverse: you’re trying to prove that no element of $S$ has more than one inverse. If you succeeded in showing that $t_1=t_2$, you’ve done that: any two inverses of $s$ are in fact the same element, so either $s$ has no inverse, or $s$ has exactly one inverse. $\endgroup$ – Brian M. Scott Oct 10 '16 at 19:22
  • $\begingroup$ What i'm thinking is: $t_1 * (s * t_2) = t_1 * e = t_1$ and $(t_1 * s) * t_2 = e * t_2 = t_2$ and since $e$ is an identity the order does not matter. So the final result will be $ t_1 * e = t_1$ and $ t_2 * e = t_2$. However, I am not sure if I succeed showing that $t_1 = t_2$ $\endgroup$ – Zed Oct 10 '16 at 19:43
  • $\begingroup$ @Z69: Yes, you have: $$t_1=t_1*e=t_1*(s*t_2)=(t_1*s)*t_2=e*t_2=t_2$$ $\endgroup$ – Brian M. Scott Oct 10 '16 at 19:45
  • $\begingroup$ Ohhhhh I couldn't see it for some reason, now I completely get it, thank you for helping me =). $\endgroup$ – Zed Oct 10 '16 at 19:48

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