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In this statement and example

The set of all nonnegative integers (including 0) under addition is not a group. There is an identity element 0, but no inverse for 2

This is my confusion. Isn't 0, under addition, an inverse for all nonnegative integers? That is

a * a' = a'*a = a + a' = a' + a

Let $a = 2$ and $a' = 0$ and then $2 + 0 = 0 + 2$

Which is also doing the identity element (does this have to be unique for a group G to be a group?)

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    This is kind of silly but when I first learned these, I viewed the inverse of an element as what I used to "kill" it, and the identity is what it became when it was "dead." Group elements, like people, are all destined for the same place - the *grave* - but we all get there by different methods. That's why it's *the* identity, but every element has its own, unique inverse.2012-10-06

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An inverse $b$ of $a$ is such that $ab = ba = e$ where $e$ is the identity element of the group. Can you find a positive integer $b$ such that $2+b = b+2 = 0$?

In the comment OP asked about the uniqueness of inverses. Yes, they must be unique: suppose there is another inverse $b'$ of $a$, and try to get to a contradiction.

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    Oh okay, I overlooked that part. Throwing in something extra, does the inverse have to be unique?2012-10-06
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No, 0 isn't an inverse for all integers; it's the identity . Recall that the inverse $a'$ of element $a\in G$ has the property that $a+a' = a'+a=$ identity, which is in this case 0. For $a=2$, the inverse would be $-2$. Since $-2 \notin G$, $G$ isn't a group under addition.

The property you showed : $2+0=0+2=2$, illustrates that 0 is the identity element, not the inverse of 2. Hope this helps.