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$\begingroup$

I have $G =\Bbb Z_+$ and $ \mathbf{a}\cdot\mathbf{b} = ab $ , I have to prove that is a group and the answer is no, but I don't understand why. Is because the neutral element has no inverse? Please help

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    Actually, the neutral element is the *only* element having an inverse ... – 2017-01-28
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    yes that is what I thought too, then why this isn't a group? – 2017-01-28
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    With the identity element $\mathbf{1}$ but otherwise no inverses, what you have is an example of a [monoid](https://en.wikipedia.org/wiki/Monoid). It can be extended to a group of positive rational numbers. – 2017-01-28
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    @hardmath Do you think this is a useful point of view for someone struggling to show this is a group? – 2017-01-28
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    That's why. To be a group, inverses are required for *all* elements – 2017-01-28
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    @MorganRodgers: Sometimes it helps to have a name for things. The example is *not* a group, but the OP's Question and Comment indicate a difficulty in letting go of the notion that it might be possible to prove it is a group. So calling such a thing by a name (monoid) and explaining that it is part of a group might be helpful, putting some context around the failure to be a group. – 2017-01-28

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The neutral element is always the inverse of itself. Here the pathology is that every element different to the neutral element has no inverse.

Try to check that for every $a \in \Bbb Z_+\space a\neq 1$ you don't find an element $b$ such that $ab=1$. Since there are elements without inverse, the set $\Bbb Z_+$ with that operation is not a group.

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    Yes, but he consider the set with only positive integer. Clearly also $-1$ has inverse in $\Bbb Z$. – 2017-01-28