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$\begingroup$

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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    It's called the [trivial group](http://en.wikipedia.org/wiki/Trivial_group).2012-08-06
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    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?2012-08-06
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    @qiaochu yuan Exactly!2012-08-06
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    I'd like to ask, what is the argument against it?2012-08-06
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    That "a group cannot contain only one element, just like a Field".2012-08-06
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    It is too small a thing to argue about. But you are right.2012-08-06
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    I've added a small question to the main question.2012-08-06
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    **Hint**: How do you check whether or not something is a group?2012-08-06
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    Checking the axioms one by one. My qustion is if 0 is the **only** subset of Z to form a group with $\triangle$2012-08-06
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    Did you first check if $G$ itself is a group? If not, what axiom(s) are contradicted? This might give you a hint as to why ${0}$ is the only subset that makes a group under this operation.2012-08-06
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    I have a clarifying remark/question about your original question: when you write $G=\{0, \cdot\}$, do you mean that the underlying set for $G$ is $\lbrace 0 \rbrace$ or is it empty?2012-08-06
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    I meant that the set for $G$ is the single element set of $\{ 0 \}$. I know that with Z it's not a group since there is no inverse (since $\frac{-x}{1+x} \notin \mathbb{Z}$ ). I asked whether there are more subsets of Z to form a group with $\triangle$.2012-08-07
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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.