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To show that a set $G$ is a group under addition, do we first need to show that $G$ is closed under addition, or is that implied by proving the three properties of a group, namely

  1. there exists an identity element $e \in G$ such that $a \cdot e = e \cdot a = a$ for every $a \in G$

  2. for every $a \in G$, there exists an $a^{-1} \in G$ such that $aa^{-1} = a^{-1}a = e$

  3. the operation $\cdot$ is associative

(Here, $\cdot$ is some binary operation.)

For context, I'm doing the following exercise:

enter image description here

and want to know if I should first prove that sets are closed under addition for those that turn out to be groups under addition.

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  • $\begingroup$ Not implied, you need to show that $\endgroup$ – Snufsan Sep 10 '14 at 6:52
  • $\begingroup$ Please see my revision to the question $\endgroup$ – James Cameron Sep 10 '14 at 6:53
  • $\begingroup$ You need to prove the closure property. $\endgroup$ – Shodharthi Sep 10 '14 at 7:00
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Indeed we have to show that the set is closed under addition in each case. For example, consider $2/3$ and $1/3$. Both of them satisfy $|.|<1$ but $|2/3+1/3|$ is not strictly smaller than $1$. So c doesn't make a groupid either. The similar story can be seen for d for example. In fact $|3/2+(-4/2)|<1$ while $|3/2|\ge1,~~|-4/2| \ge1$.

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Consider the set $G=$ {-1,0,1} under addition.

It satisfies 3 properties mentioned by you, namely
1.) $\exists$ 0∈G such that a+0=0+a=a , $\forall$ a∈G.
2.) $\forall$ a∈G, $\exists$ -a∈G such that a+(-a)=(-a)+a=0.
3.) the operation + is associative.

But it doesn't form a group under addition since it is not closed.
i.e. 1+1=2 $\notin G$.
Hence G serves as a counter example.

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If $\cdot$ is a binary operation from $G\times G\to G$, then closure is trivially statisfied. In these cases however, you have the binary operation $+$ from $\mathbb{Q}\times\mathbb{Q}\to\mathbb{Q}$, which you are now restricting to some set $G\subset\mathbb{Q}$, considering it as a map from $G\times G\to\mathbb{Q}$. You need to verify the group axioms, as well as showing that the image of this restriction of the binary operation is actually contained in $G$, i.e. that $+$ defines a binary operations from $G\times G\to G$.

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