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I'm working a rather simple problem on whether the natural numbers satisfy the conditions for a group over the binary operations addition, subtraction, and/or multiplication. For the sake of consistency with the text's notation, I will exclude $0$ from the naturals, though I do think this actually makes a difference in terms of the final answer: for example, we would have an identity under addition and subtraction if $0$ were included, but the absence of $0$ means $\mathbb{N}$ no longer satisfies these properties.

With that said, my main confusion is in whether $\mathbb{N}$ satisfies the requirement of containing an inverse. Clearly, under multiplication, it does not, as, for example, the inverse of $2$ is $\frac{1}{2}$, which is a rational, but not a natural, number. Under addition, it seems that no such inverse would exist, as this element would be an integer, but not a natural. Subtraction is a bit confusing: the answer key states that this property does not hold. But, can't any natural number simply act as its own inverse? For example, for any $x \in \mathbb{N}$, since we've excluded $0$ from the set, isn't $x - x = 0$, the identity?

So, I suppose my question boils down to this final bit on whether subtraction has an inverse in $\mathbb{N}$, assuming we exclude $0$, which obviously has no inverse, from $\mathbb{N}$. It's also very possible I'm defining the inverse incorrectly, as it seems that $e$ or $1$ are the most common representations and it varies by operation. (For example, I think we're aiming to find a product of $1$ with multiplication but a sum/difference of $0$ under addition and subtraction.)

Thanks.

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    $\begingroup$ is it that $1-2\in\Bbb N$? $\endgroup$ – janmarqz Jul 11 '18 at 23:34
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Subtraction is not even an operation in $\Bbb N$, whether or not $0$ is included. There is no result to $1-2$, for example. In the integers it is an operation, but $0$ is still not an identity. It is true $a-0=a$, but you also need $0-a=a$ for $0$ to be an identity. The integers do form a group under addition with $0$ as the identity.

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As Ross Millikan says, subtraction isn't an operation on the natural numbers, and in the integers it is an operation but lacks an inverse. It's worth noting that there is yet another way it fails to yield a group: subtraction is not associative. In general we have $(a-b)-c\not=a-(b-c)$; indeed, the two expressions are equal only when $c=0$.

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If you exclude $0$ from natural numbers, then $\mathbb{N}$ has no identity element under addition (as you said). When an algebraic structure, namely a set with a binary operation, does not have an identity element talking about inverses becomes completely meaningless.

And as you figured out on your own, no element in $\mathbb{N}$ except $1$ has an inverse under multiplication. So, it's not a group under multiplication either.

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