# How to verify the group operation?

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?

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What 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? –  Alex Becker May 10 '12 at 17:30
The 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. –  Arturo Magidin May 10 '12 at 17:30
Yes, I am assuming G is a subset of some other group with "addition" already defined for. –  CasterT May 10 '12 at 17:31
If 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 $K$? –  Arturo Magidin May 10 '12 at 17:33
If $G$ 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. –  Michael Joyce May 10 '12 at 17:50

You have a group $(K,+)$ (presumably abelian, since you are using $+$ for the operation). And you have a subset $G\subseteq K$.
To check whether $G$ is a subgroup of $K$, you need to check if the restriction of the operation $+$ from $K$ to $G$ makes G$into a group. Formally, this would require checking that: 1. For all$a,b\in G$,$a+b\in G$; (that$+$is an operation on$G$); 2. For all$a,b,c\in G$,$(a+b)+c = a+(b+c)$(the operation is associative); 3. There exists$0\in G$such that$a+0=0+a= a$for all$a\in G$; and 4. For every$a\in G$there exists$b\in G$such that$a+b=b+a=0$. In fact, if 1 holds, then 2 holds "for free" because the equality is true in$K$; and 3 holds if and only if the identity of$K$is in$G$, and 4 holds if and only if the inverse of$a$in$K$happens to be in$G$. So we can verify that$G$is a subgroup under$+$by checking only that: 1.$0\in G$; 2. If$a,b\in G$then$a+b\in G$; and 3. If$a\in G$, then$-a\in G$. Alternatively, one can also verify instead that: a.$G\neq\varnothing$; b. If$a,b\in G$, then$a-b\in G$. Indeed, if$G$satisfies (1), (2), and (3), then since$0\in G$then$G\neq\varnothing$; and if$a,b\in G$, then$-b\in G$by (3) applied to$b$, and therefore$a-b = a+(-b)\in G$by (2) applied to$a$and$-b$. So if$G$satisfies (1), (2), and (3), then it satisfies (a) and (b). Conversely, suppose that$G$satisfies (a) and (b). Let$x\in G$(possible by (a)); then$x-x=0\in G$, by applying (b) to$x$and$x$, so$G$satisfies (1). If$a\in G$, then since$0,a\in G$then by (b) we have$0-a = -a\in G$, so$G$satisfies (3). And if$a,b\in G$, then$-b\in G$(since we have established that (3) holds), so applying (b) to$a$and$-b$we get$a-(-b) = a+b\in G$, proving that (2) holds in$G$. So if$G$satisfies (a) and (b), then it satisfies (1), (2), and (3). So you can either: check that$0\in G$, that if$a,b\in G$then$a+b\in G$, and that if$a\in G$then$-a\in G$; or that$G\neq\varnothing$and if$a,b\in G$then$a-b\in G$. In particular, it is not enough to check that$a,b\in G$implies$a+b\in G$; and it is not enough to check that$a,b\in G$implies$a-b\in G$; in order to verify that$G$is a subgroup of$K$. If, on the other hand, you are asking: Suppose$(K,+)$is a group, and$G\subseteq K$is a group under some operation$(G,*)$. Is it enough to check that if$a,b\in G$then$a+b\in G$to conclude that$*$is actually$+$? Or check that$a-b\in G$? The answer is no. It's entirely possible for$G$to be a subgroup, and yet be a group under a completely different operation that has nothing to do with the operation$+$of$K$. Or it could be that$G$is closed under$+$, but the operation$*$has nothing to do with$+$. For example, take$K=\mathbb{R}$under the usual addition, and let$G$be the positive rationals under multiplication. Then$(G,*)$is a group,$G$is contained in$K$, and for every$a,b\in G$we have$a+b\in G\$, but multiplication of rationals is not the same as addition of reals.