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As part of a larger problem I am working on, I need to come up with an example as follows: A set of exactly 4 elements with a binary operation. The set plus the operation have to comply with all the requirements of a group with the exception of the last one. I.e. Invertibility. It can be a known operation or a made up one * with a table. It is important though that invertibilty will not exist for a least one of the elements. In other words the 4 elements set, exhibits the first 3 qualities of a group but not the last one.

I tried thinking of a known operation such as addition but could not find an example that will maintain closure. With a made up operation and a table it is very difficult to check for associativity so I'm lost there as well.

Any help would be appreciated. Cheers

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Consider a set $A=\{1,2\}$ with two elements, now consider it's power set $$P(A)=\{\emptyset, \{1\}, \{2\}, \{1,2\}\}.$$ with four elements and the binary operation as intersection of two sets. $(P(A), \cap)$ is a monoid (group without requiring existence of inverses) with identity element as $A$ but inverse need not exist.

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  • $\begingroup$ will I be able to just say that associativity exists in this set with ∩ or there needs to be prof provided. What I mean is that it is fairly known and acceptable that the intersection operation is assosiative but not sure about showing that it also exists in this particular set. $\endgroup$ – eventhorizon02 Jun 14 '17 at 7:24
  • $\begingroup$ @eventhorizon02 the associativity of the intersection operation is valid for any type of sets and is very well-known (a standard result) hence can be used without a need of proof (however the exact requirement will depend on the context in which this question was asked). $\endgroup$ – Anurag A Jun 14 '17 at 7:28
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A fairly obvious example is $Z_n=\{0,1,\ldots,n-1\}$ under the operation of multiplication modulo $n$. (Here $n=4$.)

My other example (now deleted) was $\{0,1,\ldots,n\}$ under the operation $a*b=\min(a+b,n)$. (Here $n=3$.)

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  • $\begingroup$ Your first example looks good. Is there a quick easy way to prove that associativity exists? I mean even for such a small set of 4 elements it's a lot of combinations to go through. $\endgroup$ – eventhorizon02 Jun 14 '17 at 6:48
  • $\begingroup$ @eventhorizon02 It's surely well-known that multiplication of integers is associative? $\endgroup$ – Lord Shark the Unknown Jun 14 '17 at 18:02
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Ordinary multiplication, over the set $\left\{0,1,\frac{-1+\sqrt{-3}}{2},\frac{-1-\sqrt{-3}}{2}\right\}$.

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