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How to make existence proof for Abelian Group condition 3( unit element e), when ($\mathbb{N}. \cdot )$, where $\cdot$ is natural multiplication? Is it by example? Is it done by constructive proof method?

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Are you asking how to show that $(\mathbb N,\cdot)$ has an identity? What do you think the identity is? Note that while it does have an identity, it's not actually a group. –  Devlin Mallory Aug 17 '13 at 17:58
    
Are you looking for a unit element for multiplication in $\mathbb N$? Note: $\mathbb N$ is not a group under multiplication, as multiplicative inverses don't always exist. –  Mark Bennet Aug 17 '13 at 18:01
    
@DevlinMallory: I think they're just checking the existence of an identity condition, not trying to prove that it's a group. –  Jim Aug 17 '13 at 18:03
    
@Devlin: yes, I'm both asking for a proof that this "set" has indentity element and how to make existence proof in this case, because i'm not sure about my proof that it is right. –  laovultai Aug 17 '13 at 18:11
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"i'm not sure about my proof that it is right"... Then showing said proof would be more direct, don't you think? –  Did Aug 17 '13 at 18:15
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Note that $(\mathbb N, \cdot)$ has identity $e = 1$. Take any arbitrary $n \in \mathbb N$ and note that by the definition of multiplication, and by commutativity, we have $1\cdot n = n \cdot 1 = n$. This gives us that $\forall n \in \mathbb N,$ since $1 \cdot n = n \cdot 1 = n$, then $1$ is by definition the multiplicative identity for the set of natural numbers.

However, just to be clear, $(\mathbb N, \cdot)$ fails to be a group, hence cannot be an abelian group. While it's true that multiplication on $\mathbb N$ is commutative, $(\mathbb N,\cdot)$ is not a group. It fails to be a group since it is not closed under taking inverses.

For example, with identity $e = 1$, there exists no $2^{-1} \in \mathbb N$ such that $2 \cdot 2^{-1} = 2^{-1} \cdot 2 = 1$.

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@Clayton Dahhh...yes, you're right...I was thinking of the problem with inverses...will edit accordingly. –  amWhy Aug 17 '13 at 18:41
    
I was pretty sure :) –  Clayton Aug 17 '13 at 18:41
    
@Clayton thanks for pointing it out! ;-) –  amWhy Aug 17 '13 at 18:43
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If you want to show that $(\mathbb N, \cdot)$ has an identity you have to pick an element and prove that it satisfies the condition of being an identity element. So pick $1 \in \mathbb N$ and note that $1 \cdot a = a \cdot 1 = a$ for all $a \in \mathbb N$.

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