Say you are given a group $G$. How can you show that the group operation of this group is addition? What I have in mind is $\forall (a,b) \in G$ if I can show $(a+b) \in G$, this will prove the above. Does $\forall (a,b) \in G, (a-b) \in G$ prove the same thing?
How to verify the group operation?
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2What do you mean? Are you assuming $G$ is a subset of some other group with "addition" already defined for it, and trying to show that the operation on $G$ is the same as this operation? – 2012-05-10
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2The question makes no sense. A group is a set $G$ together with a binary operation $G\times G\to G$. What you call the operation is irrelevant; it makes no sense to ask "is the group operation of this group 'addition'", because "addition" doesn't have an absolute meaning. – 2012-05-10
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0Yes, I am assuming G is a subset of some other group with "addition" already defined for. – 2012-05-10
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0The meaning of the question is not clear, one would need something more explicit. But if for any pair of elements of $G$, $a-b$ is in $G$, then it follows that the additive inverse of any element of $G$ is in $G$, and therefore that the sum of any two elements of $G$ is in $G$. – 2012-05-10
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2If your $G$ is contained in some group $(K,+)$, then you are really asking whether $(G,*)$ is actually a *subgroup* of $(K,+)$, as opposed to some random group structure that has nothing to do with that of $K$? – 2012-05-10
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0Sorry if my question makes no sense. What I have is a group $G$ that is given as a subgroup of a group $(K, +)$, as you mentioned. What I am trying to do is, verifying that the group operation is the same in $G$. – 2012-05-10
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1If $G$ is a subgroup of $K$, with the operation of $K$ given by $+$, then *by definition*, the operation on $G$ is the restriction of $+$ to $G$. (The fact that $G$ must be closed under this operation insures that $g_1 + g_2 \in G$ whenever $g_1, g_2 \in G$.) FWIW, people usually use $+$ for the group action when the group is abelian (commutative), but generally use $\cdot$ otherwise. – 2012-05-10
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0@CasterT: Does this have something to do with your question about my answer to [this question](http://math.stackexchange.com/questions/139339/how-to-prove-a-group-has-a-basis-with-exactly-one-element)? I just wondered because there's a superficial similarity, although this question doesn't seem to ask the same thing as you asked there. – 2012-05-10
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0See also the [answers here](http://math.stackexchange.com/q/75371/242) on [subgroup testing](http://en.wikipedia.org/wiki/Subgroup_test) – 2012-05-10