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Suppose $G$ is a set and $\cdot$ is a binary operation on $G$ such that there exists an $e\in G$ such that $a\cdot e=a$ for a in $G$ and given $a\in G$, there is a $y(a)\in G$ such that $y(a)\cdot a=e$. I wish to find a $G$ such that $G$ is not a group under $\cdot$.

An obvious place to start looking would be sets with some non-commutative operation. An example that comes to my mind would be that of square matrices under multiplication but it seems that if a square matrix is left invertible it has a right inverse as well, so this example doesn't work.

Martin Sleziak
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    see http://math.stackexchange.com/questions/433546/is-a-semigroup-g-with-left-identity-and-right-inverses-a-group – M.U. Dec 16 '15 at 13:34
  • The question is not so clear. Are you just looking for a set $G$ and an arbitrary binary operation on $G$ such that both do not form a group? Or are you looking for a set $G$ for which there is no binary operation on $G$ such that they form a group? – Björn Friedrich Dec 16 '15 at 14:38
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    I was looking for a set $G$ and an arbitrary binary operation on $G$ such that both do not form a group. –  Dec 16 '15 at 14:40
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    The class of binary operations which yield group structures is infinitessimal among all binary operations. All you'd need to do is take a random stab in the dark and you'd surely hit one that's not a group structure. In fact you already did: the set of all square matrices under multiplication does not form a group: there are noninvertible matrices. – anon Dec 17 '15 at 15:53

3 Answers3

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A group doesn't need to be commutative, but we do require associativity. One example would be the integers under subtraction.

You can take $e=0$, and $y(a)=a$, so $\langle \mathbb{Z}, - \rangle$ satisfies your requirements, but is not a group.

mrf
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    Ah, thank you. I was only trying to look for examples where there is a left inverse but no right inverse. I completely forgot about associativity. –  Dec 16 '15 at 13:41
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As an addendum, note that your question asks for an example of a binary operation which has an identity (two-sided) and left inverses for every element. The two axioms which are missing here are the existence of two-sided inverses and associativity. The following fact shows that associativity must fail in such an example:

Let $(G, \cdot)$ be a set and binary operation, such that $\cdot$ is associative, and $G$ contains a two-sided identity for $\cdot$, and every $a \in G$ has a left inverse. Then $G$ is a group.

Proof: Let $a \in G$. Let $b$ be a left inverse for $a$. Let $c$ be a left inverse for $b$. Then $c(ba) = c(e) = c$, but $(cb)a = (e)a = a$. By associativity, we therefore have $c=a$, which makes $b$ a two-sided inverse for $a$.

Dustan Levenstein
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For an example that is commutative (but not associative), let $G$ be any set with more than $2$ elements and define $a\cdot e=e\cdot a=a$ for all $a\in G$ as usual and

$$a\cdot b=e\quad\text{if }a,b\not=e$$

Note that inverses here are not unique.

Another non-associative (but in this case non-commutative as well) example: Let $G$ be the positive integers, let $y(a)=e=1$ for all $a\in G$, and let $a\cdot b=a^b$.

Barry Cipra
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    So, for $a,b,c$ distinct and not $e$, $(a\cdot b)\cdot c=e\cdot c=c$ while $a\cdot (b\cdot c)=a\cdot e=a$ and so the product is not associative. –  Dec 16 '15 at 14:32