Does an inverse function always give an identity element?

Question

So for multiplication the inverse element is 1 over the number and gives 1 which is the identity element for the multiplication operation. Are there examples of operations where there is an inverse function which gives the same result for each element in the group but said result is not an identity element (e.g. an "inverse" which when multiplied by a number gives a number not equal to 1). If these inverses don't exist then why?

Answer 1

I could be wrong, but I think the question comes from the idea that for all $g$, there's some $g^{-1}$ such that $g \circ g^{-1} = e$, where $e$ is the identity element of the group.

Maybe we could call $g^{-1}$ "the (right-)inverse of $g$ with respect to $e$" because right-multiplying $g$ by $g^{-1}$ gives $e$ (I don't know if there's a phrase for this kind of thing, since the identity element is baked into the definition of inverse elements). I think your question is asking if there are "(right-)inverses of $g$ with respect to any fixed group element $b$", that is, some $x$ with $g \circ x = b$ for any $g$ and fixed $b$.

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Thus, as I understand the question:

Suppose $G$ is a group with operation $\circ$, and fix $b \in G$.

For each $g \in G$, can we find some $x$ so that $g \circ x = b?$

Yes, this is possible. If you've talked about Cayley tables, this is basically the observation that each group element shows up in each row and each column exactly once (in particular, $b$ shows up in each row exactly once).

In any group, you can solve the equation $g \circ x = b$ by left-multiplying each side by $g^{-1}$ to get the desired value $x = g^{-1} \circ b$. A similar technique will show that $x \circ g = b$ always has a solution too.

Answer 2

The identity element $e$ of a group $(G,\circ)$ is defined by $$\forall\; g\in G: g\circ e = g = e\circ g$$ The inverse element $g^{-1}$ of an element $g$ of a group $(G,\circ)$ is defined by $$g^{-1}\circ g = e = g\circ g^{-1}$$ For a group, the above statement is true for all $g\in G$. However, if given some other operation $\star\neq\circ$ on $G$, we could have $g\star g^{-1}\neq e$, because $e$ means the identity element in respect to $\circ$ and $g^{-1}$ means the inverse in respect to $\circ$. For example, $2\cdot\frac{1}{2}=1$, but $2+\frac{1}{2}=\frac{5}{2}\neq 1$.

Note that "Inverse" needs "identity element", not the other way around. A set with a binary operation that has an identity element but not an inverse for every element is no group but a monoid.

The inverse $g^{-1}$ is unique for each $g\in G$ (this can be proved). Therefore, given element $g,h,x\in G$, one can always solve the equation $h\circ x = g$ for $x$, it is $$h^{-1}\circ (h\circ x) = (h^{-1}\circ h)\circ x = e\circ x = x \stackrel{!}{=} h^{-1}\circ g$$