Broadly given an operator $*$ the term identity is used for an element $e$ such that $x * e = x$ for all elements. However is there a term for a value $ x * O = O$ for all values? This was brought to mind by this question What is the identity in the power set of $\Sigma^*$ as a monoid? that shows that the empty language has this property under concatenation. False has this property under the and operator.
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Can such an element $O$ exist necessarily? – Manas Dogra Jun 24 '20 at 02:52
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3If you are in a semigroup, though, an element $0$ such that $x0 = 0x = 0$ for all $x$ is called, perhaps unsurprisingly, a “zero element”. However, you can’t have zero elements in nontrivial monoids or groups. Note that your two examples are semigroups. – Arturo Magidin Jun 24 '20 at 03:07
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Thanks I've changed that. – Q the Platypus Jun 24 '20 at 05:46
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5@ArturoMagidin What makes you say you can't have a zero element in a nontrivial monoid? Consider for example the set $\Bbb{Z}{\geq 0}$ of nonnegative integers with binary operation given by multiplication. This operation is clearly associative, and $x\cdot 1 = 1\cdot x = x$ for all $x\in\Bbb{Z}{\geq 0}$, but $0\cdot x = x\cdot 0 = 0$ for all $x$ as well. In fact, given any ring $R,$ $R$ is a monoid under multiplication, and will be a nontrivial monoid with $0$ element so long as $R$ itself is a nontrivial ring. – Stahl Jun 24 '20 at 05:52
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3@Stahl: A momentary moment of silliness. Thank you. – Arturo Magidin Jun 24 '20 at 13:58
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There is no such element in a nontrivial group. Every element of a group has an inverse.
Let $G$ be a group and suppose (for purpose of contradiction) $G$ contains your proposed element $O \neq e$. Then there is a $p = O^{-1} \in G$ such that $pO = e \neq O$. But this contradicts the definition of $O$. Therefore, there is no nontrivial group, $G$, containing an $O \neq e$ as described.
Another way to get at this, using the required existence of inverses, is that from $$ x O = O \text{,} $$ we have $$ x = xO O^{-1} = O O^{-1} = e \text{.} $$ So the assumed multiplication properties of $O$ are incompatible with its membership in a group unless the only element of the group is $e$ (in which case $e = O$ does satisfy the properties of both the multiplicative identity in a multiplicative group and the properties of the $O$ element you describe). (This is why I wrote "$O \neq e$" in the second paragraph: to avoid the case that we were secretly only talking about the group with one element.)
Eric Towers
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Shouldn’t your first line say, There’s no such element in a ‘nontrivial’ group – Jac Frall Jun 27 '20 at 04:10
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Excepting the special case of a group with only one element, groups cannot have $0$ as an element, and multiplication as its operation, as it is required that every element have an inverse, and what is the multiplicative inverse of $0$?
It sounds like you are interested in Rings.
Rings take a set with two binary operations, one operation is analogous to addition and the other is analogous to multiplication. A ring has an additive identity ($0$) and an multiplicative identity ($1$) and requires that multiplication distribute over addition. As consequence $a\cdot 0 = 0.$
(Yes, there is also the special case here, where the ring has one element.)
Rings have a generalization of $0$, which is called an "ideal." An ideal is a subset of the ring such that for every element in the ideal multiplied by a member of the ring gives an element in the ideal.
Doug M
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