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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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    In the definition of a field this case is explicitly removed. You need two non-equal distinguished elements and this cannot happen if you only have one element! This isn't the case in a group.2012-08-07

1 Answers 1

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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.