# Can there be "identity-like" mappings that do not involve the identity element in a group?

I'm working with the definition of an identity from Pinter's intro book on abstract algebra. He writes as follows:

I understand that if an operation has an identity it must be unique, which can be seen letting $$e$$ and $$f$$ be identities and noting that $$e = e*f$$ and $$e*f=f$$, so $$e = f$$. However — and I'm not sure how to word this — can there be specific non-identity elements on which another element has identity-like action?

Consider a later problem in Pinter's book where we identify that the operation

$$(a, b) * (c, d) = (ac, bc + d), \quad \text{on the set } \left(\mathbb{R}-\{0\}\right) \times \mathbb{R}$$

is a group with the identity element $$e = (1, 0)$$. But if we add the deleted zero back into the first coordinate's domain and do the same checks on that reformed $$\mathbb{R}^2$$ we run into the motivation for my question. I am not concerned with whether the operation still defines a group — it definitely does not since the inverse breaks — I am concerned with whether the identity still holds. It's still clear that $$(1, 0)$$ works for all elements in the new domain but it's not clear if it's acceptable that a specific element now has other identity-like relationships, like

$$(0, 0) * (x, 0) = (0, 0) = (x, 0) * (0, 0) \tag{\star}$$

Because there is still only one element as Pinter describes — one that works for every element in $$\mathbb{R}^2$$. My instinct is that the identity still holds, but I was reviewing an answer here: https://www.math.wisc.edu/~mstemper2/Math/Pinter/Chapter03B, problem 2 & 3, and the author states that $$(\star)$$ breaks uniqueness. So, to what part of the definition is uniqueness applied? It sounds like author of the linked answer has taken the converse as the definition.

Edit: Due to comments let me try to restate my problem more succinctly:

For a set and and a binary operation on that set, if there is/are some element(s) $$a$$ such that $$xa=ax=x$$ for many, possibly infinite choices of $$x$$, but only one element $$e$$ such $$xe=ex=x$$ for every $$x$$, is the property of the uniqueness of $$e$$ in the context of groups violated? What about for general sets with binary operations?

Edit2: I am utterly convinced that for a group

1. There is only one identity element $$e$$
2. There is no other element $$e'$$ in that group such that $$xe'=e'x=x$$

But I am not convinced that these statements are equivalent for operations on sets, only that they happen to coincide for a group. Based on Pinter's definition, I think that an identity for an operation on a set can be unique (statement 1), yet there may still be elements $$e'$$ in such that $$xe'=e'x=x$$ for some value of $$x$$ but not all. So in the linked answer when the author states

So this operation does not have a unique identity element.

I become confused because it isn't true. I don't see how it could ever be true. Of course this is probably just a quickly typed answer and the answerer meant to say group instead of operation, but the comments below are making me second guess this.

• $(0,0)$ is what is called a "zero element" (unsurprisingly). In a semigroup $S$, a zero element is an element $0$ such that $0x=x0=0$ for all $x\in S$. That does not look like an "identity-like relationship" to me... it looks like a relationship where $0$ overwhelms everything it comes in contact with... the very opposite of an identity. Commented Apr 27, 2020 at 1:20
• Agreed, so then the linked answer incorrect in its reasoning? Commented Apr 27, 2020 at 16:36
• There is no incorrect reasoning in either problem 2 or 3. There is no lack of uniqueness in problems 2. Problem 3 is noting that in that set, there are many possible solutions to the equation $x_1e_1=x_1$. So if you try to proceed as in the previous problem, you will find that for some elements $\mathbf{x}$, there are many possible elements $\mathbf{e}$ that satisfy $\mathbf{x}\mathbf{e}=\mathbf{x}$. If this happens for even a single element, then the resulting structure is not a group. What exactly is it you think is wrong about that reasoning? Your post seems to wander all over the place Commented Apr 27, 2020 at 17:28
• I find the reasoning to be off in the sense that, as you say, "for some elements $x$, there are many possible elements $e$ that satisfy $xe=x$" which does not contradict that there may be a particular unique element $e$ such that $xe=x$ for all elements $x$. Commented Apr 27, 2020 at 21:29
• In a group, any equation of the form $ax=b$ has a unique solution namely $x=a^{-1}b$. If $xf=x$, then $x^{-1}xf = x^{-1}x$, the left hand side is $f$, and the right hand side is $e$. So the only possible solution is $e$. If you can find more than one solution for any particular $e$, then what you have is not a group. In a group, what you are positing is, in fact, utterly impossible. Commented Apr 27, 2020 at 21:39

From comments, I finally figured out where you were going... (or so I thought, my confidence undermined by subsequent comments while I was writing this...)

In a semigroup (a set with a binary associative operation), it is possible for you to have elements that act as identities for some, but not all, elements. This shows up in rings (whose multiplicative structure is just a semigroup or a monoid (a semigroup with an identity element) and not groups. For example, the ring $$\mathbb{Z}\times\mathbb{Z}$$ contains a subring $$\mathbb{Z}\times\{0\}$$ that has an “internal multiplicative identity” $$(1,0)$$ that is however not the identity for the whole ring (which is $$(1,1)$$).

However, in a group this just doesn’t happen. There are many ways to that fact, but basically, if you have a group $$G$$, and you have an element $$a\in G$$, then the only element of the group that satisfies the equation $$ax=a$$ is $$x=e$$, the identity. No element can act as an “identity” for any particular element except the identity itself.

This is a consequence of the following well-known observation:

Theorem. Let $$G$$ be a group. Then for each $$a,b\in G$$, there exist unique solutions $$x,y\in G$$ to the equations \begin{align*} ax &= b\\ \text{and }ya &=b. \end{align*}

Proof. Existence: let $$x=a^{-1}b$$, and let $$y=ba^{-1}$$. Then $$ax = a(a^{-1}b) = (aa^{-1})b = eb = b$$, and $$ya=(ba^{-1})a = b(a^{-1}a) = be = b$$.

Uniqueness: Suppose that $$x_0$$ and $$x_0’$$ are both solutions to $$ax=b$$. Then $$ax_0=b=ax_0’$$. So $$ax_0=ax_0’$$. Multiplying on the left by $$a^{-1}$$, we get $$x_0 = ex_0 = (a^{-1}a)x_0 = a^{-1}(ax_0) = a^{-1}(ax_0’) = (a^{-1}a)x_0’ = ex_0’ = x_0’.$$ Therefore, $$x_0=x_0’$$. The same argument shows the uniqueness of $$y$$, by multiplying on the right by $$a^{-1}$$. $$\Box$$

In fact, this is a defining characteristic of groups:

Theorem. Let $$G$$ be a nonempty set with a binary associative operation such that for all $$a,b\in G$$, there exist unique solutions to the equations \begin{align*} ax & = b\\ ya &= b. \end{align*} Then $$G$$ is a group under the binary associative operation.

Note: In fact, you may drop the “uniqueness clause.

Proof. Pick $$a\in G$$. Let $$e_a$$ be a solution to $$ax=a$$; I use the subscript $$a$$ because at this point we are not sure that it works for every element of $$G$$. Similarly, for every $$g\in G$$, there is an element $$e_g\in G$$ such that $$ge_g=g$$. I claim that $$e_g=e_a$$ for all $$a$$.

To see that, let $$g\in G$$ be arbitrary. Then there is a solution to $$ya=g$$. Then $$ge_a = (ya)e_a = y(ae_a) = ya = g$$. Since both $$e_a$$ and $$e_g$$ are solutions, it follows from the uniqueness hypotheses that $$e_a=e_g$$. So there is a unique element $$e\in G$$ such that $$ae=e$$ for all $$a\in G$$.

Likewise, for each $$a\in G$$ there exists $$a’\in G$$ such that $$aa’=e$$, by solving $$ax=e$$.

I claim that $$a’’=(a’)’=a$$. Indeed, we have $$a’ = a’e = a’(aa’) = (a’a)a’.$$ Therefore, $$e = a’(a’’) = ((a’a)a’)a’’ = (a’a)(a’a’’) = (a’a)e = a’a.$$ But the only solution to $$a’x=e$$ is $$a’’$$, so $$a=a’’$$.

In particular, for every $$a$$, $$aa’=a’a=e$$.

Finally, we just need to show that $$ea=a$$ for all $$a$$. But $$ea = (aa’)a = a(a’a) = ae = a,$$ so this follows. Thus, $$G$$ is a group. $$\Box$$

Now, in a semigroup/monoid it is possible for there to be elements $$e’$$ such that $$xe’=x$$ for some, but not all, elements $$x$$.

In fact, given any semigroup $$S$$, one can construct a new semigroup $$S^1$$ as follows: let the $$e_0$$ be something that is not an element of $$S$$. Let $$S^1 = S\cup\{e_0\}$$. Define an operation $$\cdot$$ on $$S^1$$ as follows: $$x\cdot y = \left\{\begin{array}{ll} xy &\text{if }x,y\in S;\\ x &\text{if }y=e_0;\\ y &\text{if }x=e_0. \end{array}\right.$$ It is easy to verify that $$S^1$$ is a monoid, and that $$e_0$$ is the identity of the monoid. This, even if $$S$$ already had an identity. If you perform this construction multiple times, you end up with a sequence $$S\subsetneq S_1\subsetneq S_2\subsetneq S_3\subsetneq\cdots$$ of monoids such that for each $$j$$ there is an element $$e_j$$ in $$S_j$$ that is the identity of $$S_j$$, but such that $$e_j$$ is not an identity for $$S_{j+1}$$.

In a monoid this can happen; but only if it is not a group. If it is a group, then this situation just cannot happen.

(A monoid is a set $$M$$, together with a binary associative operation such that there exists $$e\in M$$ such that $$ae=ea=a$$ for all $$a\in M$$; from the definition, it follows that this element $$e$$, called the identity of $$M$$, is unique.)

This is the reason why when we define “submonoid” of $$M$$ we require the subset $$N$$ to be a monoid and contain the identity of $$M$$ (and why a “subring-with-identity” is required to contain the identity of the ring), and why a monoid homomorphism $$f\colon M\to N$$ is required to take the identity of $$M$$ to the identity of $$N$$.

Coming back to Pinter, the exercise in question is correct. Even though the semigroup with underlying set $$\mathbb{R}\times\mathbb{R}$$ and operation $$(a,b)(c,d) = (ac,bc+d)$$ is a monoid (and thus has a unique identity element), the fact that it fails to satisfy the theorem above about uniqueness of solution to the equations $$ax=a$$ establishes immediately that it cannot be a group.

The monoid in question is not a group because not every element has an inverse. In fact, the elements that have no inverses are precisely the elements $$a$$ for which the equation $$ax=a$$ has multiple solutions, namely the elements $$(a,b)$$ with $$a=0$$. (If $$ax=a$$ and $$a$$ has an element $$a’$$ such that $$a’a=e$$ where $$e$$ is the monoid identity, then $$ax=a$$ implies $$a’ax =a’a$$, which implies $$x=e$$, giving the uniqueness. This is because the set of invertible elements of a monoid forms a group).

• Many proofs about groups commonly skip over this very important step in the proof without mentioning it anywhere else in the text. I came across this in proofs about subgroups and the identity element, and your proof was extremely helpful for filling in the gaps. Commented Mar 13, 2023 at 21:36

No, there cannot be other "identity-like" elements for a group. In other words for a group that has an identity element $$e$$ there can't be even one $$e'$$ such that $$xe'=x$$, but there can in other structures, as pointed out by Arturo Magidin in the comments.

So while the linked answer may be interpreted as correct for subtler reasons (again, see Magidin's comments), it does state something incorrect which is that the operation as defined on $$\mathbb{R}^2$$ has no identity element. The opeation has the identity element pointed out in the post, and it is unique since no other elements $$R^2$$ satisfy Pinter's definition.