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Let $G=\{0, \cdot\}$.

I'm arguing with someone over if $G$ is a group with the regular multiplication since I don't see why it isn't.

Addition:

Now, $G=\{\mathbb{Z},\triangle \}$ with $x \triangle y=x+y+xy$. Is it true that $G$ is not a group and the only subset of $\mathbb{Z}$ to form a group with $\triangle$ is $\{0\}$?

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5  
It's called the trivial group. –  anon Aug 6 '12 at 22:59
7  
How can you argue with someone over whether something is a group? There is a small finite list of axioms to check. Is the operation associative? Does it have an identity? Does it have inverses? –  Qiaochu Yuan Aug 6 '12 at 23:01
    
@qiaochu yuan Exactly! –  Amihai Zivan Aug 6 '12 at 23:05
    
I'd like to ask, what is the argument against it? –  Cocopuffs Aug 6 '12 at 23:12
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It is too small a thing to argue about. But you are right. –  André Nicolas Aug 6 '12 at 23:18
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1 Answer

up vote 3 down vote accepted

Yes.

  1. Closure $0\cdot0=0 \in G$

2.Associativity $(0\cdot0)\cdot 0 = 0(\cdot0\cdot0)$

3.Identity element is $0$.

4.Inverse element holds, because if not, eist $x\neq0 \in G$ that don't have Inverse element. Absurd.

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