let s be a binary set with a binary operation * and identity element e. let $a^{-1}$ be an inverse then $a^{-1}$ is the only element in S

Question

Question:

Let be a set with a binary operation ∗ and identity element . Let $^{−1}$ ∈ be an inverse for some element ∈ . Then $^{−1}$ is the only such element of with this property in relation to

My answer

Suppose $x \in S$ and $y_1$ and $y_2$ are both inverses of $x$

$y_1=y_1*e$ because e is the identity

$=y_1*(x*y_2)$ because $y_2$ is the inverse of $x$

$=(y_1*x)*y_2$ bu associativity

$=e*y_2$ because $y_1$ is the inverse of $x$

$=y_2$

so $y_1 = y_2$

I assumed this is a uniqueness question. Correct me if I am wrong

Answer 1

This looks good! It is worth noting that you're assuming $y_1$ is an inverse on the left, and $y_2$ is an inverse on the right. In general, these might be different, and it's worth watching out for that! Since you said "inverse", we can assume that $y_1$ and $y_2$ are inverses on both sides, but in the interest of completeness, here is a condition for left and right inverses agreeing:

Whenever every element has an inverse, then left and right inverses agree. In what follows, write $x^{-1}$ for the left inverse of an element:

$(x * x^{-1})$ we first cleverly multiply by $e$

$= e * (x * x^{-1})$ now we expand $e$ as something times its left inverse

$= ((x * x^{-1})^{-1} * (x * x^{-1})) * (x * x^{-1})$ now we apply associativity, and simplify some

$= (x*x^{-1})^{-1} * (x * x^{-1})$ but notice this is $y^{-1} * y$, so w can condense again.

$= e$

Thus left inverses are also right inverses, as long as we can find a left inverse for everything. A similar proof shows that right inverses are also left inverses, provided we can find a right inverse for everything.

---

Your proof was fine, but I wanted to include something more in the answer, and I know I didn't realize left/right inverses could be different for a long time, so it felt worthwhile to bring this up!

I hope this helps ^_^

Answer 2

At one step in your calculation, you invoked associativity. But nothing in the question says that $*$ is associative. If $*$ were associative (and if, as others have pointed out, "inverse" means "two-sided inverse) then your proof is correct. If, on the other hand, $*$ can be non-associative, then not only your proof but the result you're trying to prove fails. The simplest counterexample seems to be a $3$-element set, $\{e,a,b\}$, with $e$ being the identity element and $a*a=a*b=b*a=b*b=e$.