Prove or disprove that the set of even integers forms a group under addition possible duplicate but want work checked and verified

let $l =$ any even number and $m =$ any even number

$l+m = w$ where $w$ is any element in the even integers thus it is closed under addition

$2k+0=2k$ and $0-2k=2k$ so there is an identity both left and right

every even number can have its negative added to get 0 so there is an inverse for every element in the even integers i.e $2+(-2)=0$

Its associative
Your proof is okay, but it can be improved. 

Closed: For any $l,m \in \mathbb{E}$ (where $\mathbb{E}$ is the set of even integers) $l = 2a$ and $m = 2b$ for some $a,b \in \mathbb{Z}$. Thus, $l + m = 2a + 2b = 2(a+b) \in \mathbb{E}$. 

Identity: $0 \in \mathbb{E}$ and $0 + m = m + 0 = m$ for all $m \in \mathbb{E}$.

Inverse: For any $2k \in \mathbb{E}$, the number $2(-k) \in \mathbb{E}$ and $2k + 2(-k) = 2k - 2k = -2k + 2k = 0$.

Associative: The integers under addition are associative, so $\mathbb{E} \subset \mathbb{Z}$ inherits that property.
It's $2\Bbb Z$.  It's an infinite abelian group, and  is cyclic, so isomorphic to $\Bbb Z$.  

$\phi:\Bbb Z\to 2\Bbb Z$ by $\phi(k)=2k$ is an isomorphism.