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Take this mathematical structure for example:

(Z, #) where Z is the set of all integers and where a#b returns the larger of the two elements.

In order to prove this is a group I must prove these four properties: -Closure -Associative -Identity -Inverse

I am stuck on the Identity Element. As far as I know, the only element e that would prove to be true so that a#e = a = e#a would be -infinity. Is this possible, or would this mathematical structure just fail to be a group?

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  • $\begingroup$ Um, why do you assume the result will be that this is a group? $\endgroup$ – fleablood Nov 16 '16 at 2:11
  • $\begingroup$ I don't. I am trying to prove that this is a group. If Inverse fails then it means that it is not a group. @fleablood $\endgroup$ – Tom Nov 16 '16 at 10:42
  • $\begingroup$ Then prove identity fails. Neg infinity is not an integer so it is not an acceptable identity. Prove there is no identity. You will be done. $\endgroup$ – fleablood Nov 16 '16 at 18:26
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Your example doesn't work, but the answer to this question is an unequivocal yes. This is true for the same reason you can define a group of order two whose underlying set consists of a Cornish hen and my uncle Joe. The equations say it all: $$\text{Joe}\cdot \text{Joe}=\text{hen}$$ $$\text{hen}\cdot\text{hen}=\text{hen}$$ $$\text{hen}\cdot\text{Joe}=\text{Joe}\cdot\text{hen}=\text{Joe}$$ If instead of a Cornish hen we used the symbol $\infty$, we would still have a group of order two.

The moral of the story is that it doesn't matter what you call the elements. What matters is what the binary operation does to them, and you have lots of freedom in defining it.

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$-\infty \not\in \Bbb {Z} $ so that would only work if your algebraic structure were $( \Bbb{Z} \cup \{ - \infty \}, \#) $

Doing that, closure, associativity and identity are trivially true. But what you would you then define as the inverse? Can you define such a thing with this operation?

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Suppose $(\Bbb{Z},\#)$ is a group. Then the left and right cancellation hold. Note that $$2\#3=3=1\#3$$ But clearly $2\neq1$. This is a contradiction.

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$-\infty \not \in \mathbb Z $ so it is not an acceptable identity element.

If $e \in \mathbb Z $ and $e\#a=a $ for all $a \in \mathbb Z $ then $e\#(e-1)=e-1$ and therefore $e-1 \ge e $ which is impossible.

So this construct can not have an identity element and it is not a group. As there is no identity, the concept of "inverse" is meaningless.

Even if there were an identity, invertibility would be impossible. $a\#b $ is either $a $ or $b $ so $g\#a=e;$ means either $g$ or $a $ is $e $, so if inverses can only always exist if there are at most two distinct integers.

As a consequence, the construct has no "solvabilty". If $a\#b = c\#b $ it does not follow that $a=c $. That is a requirement of groups as a direct result of the existence of inverses. As invertibility is impossible (as well as meaningless) this does not hold.

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