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This question already has an answer here:

I know that a Group G is defined as a set of elements with a binary operation with these three properties:

1) Associatively

2) Inverses Exist

3) Identity Exists

However there are many definitions such as the one found on wikipedia that say closure is also a property and what not. However I have seen some other things about groups and it said closure is not included as a property, mainly Harvard Abstract Algebra videos on Youtube. I know that closure can come from those three properties sometimes, but not all the times. So my question is, is closure a property of a group?

I believe it is because there has been more information supporting it being one, then it not being included.

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marked as duplicate by Yuval Filmus, apnorton, user147263, Claude Leibovici, Najib Idrissi Oct 3 '14 at 7:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Closure is part of being a binary operation. Some people like to emphasize it by adding it as a fourth axiom but it is actually redundant. $\endgroup$ – Seth Oct 3 '14 at 2:33
  • $\begingroup$ oh thank you very much. I didnt even think of looking for that combination. $\endgroup$ – Jack Armstrong Oct 3 '14 at 2:35
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    $\begingroup$ One reason that closure is traditionally included as a group axiom is that you need to check it when trying to show that a subset of a group is a subgroup. $\endgroup$ – Derek Holt Oct 3 '14 at 2:50
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Closure is an important property of groups. However, sometimes the binary operation is defined as an operation $G^2 \to G$, in which case closure is part of the definition of the binary operation. (This is probably the "right" way to go about it anyhow.)

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The idea of a group doesn't make sense without closure, really; to be sure, most definitions say a group is a pair $(G,\cdot)$ where $G$ is a set and $\cdot$ is a map $G\times G \rightarrow G$ (i.e. a binary operation) which is associative, has inverses, and has an identity. The fact that we define $\cdot:G\times G\rightarrow G$ implies closure, and none of the other properties would make sense without it, because it is assumed that for any $c$ in the codomain of $\cdot$, we can apply $\cdot$ again. That is, $ab$ must be in $G$ because we are demanding certain properties of applications of $\cdot$ to $ab$, like that $(ab)c = a(bc)$.

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