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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
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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
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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
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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

1 Answer 1

From what I seem to understand of your question:

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.

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