Let (Z,+) be a group of integers and (E,+) be a group of even integers. Find and prove if there exist an isomorphism between them.

To prove any two groups are isomorphic:
The map ${\displaystyle \varphi :Г\to \Gamma \prime }$ is called an isomorphism $\Gamma$ and ${\Gamma }^{\prime }$ and are
said to be isomorphic if
i) ${\displaystyle \varphi }$ is a homomorphism.
ii) ${\displaystyle \varphi }$ is bijective.
${\displaystyle \Gamma \stackrel{\sim }{=}\Gamma \prime }$ denotes $\Gamma$ is isomorphic to $\Gamma \prime$
Let ${\displaystyle \varphi :Z\to 2Z}$ be defined by ${\displaystyle x\mapsto x+x=2x}$
i) To prove $\varphi$ is homomorphism:
${\displaystyle \varphi (x+y)=\varphi \left(x\right)+\varphi \left(y\right)}$
${\displaystyle \varphi (x+y)=2(x+y)}$
${\displaystyle \varphi (x+y)=2\left(x\right)+2\left(y\right)}$
${\displaystyle \varphi (x+y)=\varphi \left(x\right)+\varphi \left(y\right)}$
Therefore, ${\displaystyle \varphi }$ is homomorphism
ii) To prove ${\displaystyle \varphi }$ is bijective:
Bijective is one-one and onto
To prove phi is one -one:
"Let us consider ‘f’ is a function whose domain is set A. The function is said to be injective(1-1) if for all x and y in A,
$f(x)=f(y)$, then $x=y^{″}$
Let,
${\displaystyle \varphi \left(x\right)=\varphi \left(y\right)}$
${\displaystyle 2x=2y}$
${\displaystyle \Rightarrow x=y}$
Therefore, ${\displaystyle \varphi }$ is one-one.
To prove ${\displaystyle \varphi }$ is onto:
"A function f from a set X to a set Y is surjective (also known as onto, or a surjection), if for every element y in the codomain Y of f, there is at least one element x in the domain X of f such that $f(x)=y^{″}$.
Let ${\displaystyle y\in 2Z}$
Then y=2k for some ${\displaystyle k\in Z}$
Since ${\displaystyle k\in Z}$ and
${\displaystyle \varphi \left(k\right)=2k}$
${\displaystyle \Rightarrow y}$
${\displaystyle \varphi }$ is onto
${\displaystyle \Rightarrow \varphi }$ is bijective.
Since, it is homomorphism and bijective.
it is isomorphic.
${\displaystyle \Gamma \stackrel{\sim }{=}\Gamma \prime }$
Therefore,The group (Z,+) and (E,+) are isomorphic.

i) $\varphi$ is a homomorphism.
ii) $\varphi$ is bijective.
$\Gamma \cong {\Gamma }^{\prime }$ denotes Γ is isomorphic to Γ'
Let $\varphi :Z\Rightarrow 2Z$ be defined by $x\Rightarrow x+x=2x$
1) To prove $\varphi$ is homomorphism:
$\varphi (x+y)=\varphi (x)+\varphi (y)$
$\varphi (x+y)=2(x+y)$
$\varphi (x+y)=2(x)+2(y)$
$\varphi (x+y)=\varphi (x)+\varphi (y)$
Therefore, $\varphi$ is homomorphism
ii) To prove $\varphi$ is bijective:
Bijective is one-one and onto
To prove phi is one -one:
"Let us consider ‘f’ is a function whose domain is set A. The function is said to be injective $(1-1)$K if for all x and y in A,
$f(x)=f(y),thenx=y"$
Let, $\varphi (x)=\varphi (y)$
$2x=2y$
$\Rightarrow x=y$
Therefore, $\varphi$ is one-one.
To prove $\varphi$ is onto:
"A function f from a set X to a set Y is surjective (also known as onto, or a surjection), if for every element y in the codomain Y of f, there is at least one element x in the domain X of f such that $f(x)=yZ$".
Let $y\in 2Z$
Then $y=2k$ for some $k\in Z$
Since $k\in Z$ and
$\varphi (k)=2k$
$\Rightarrow y$
$\varphi$ is onto
$\Rightarrow \varphi$ is bijective.
Since, it is homomorphism and bijective.
it is isomorphic.
$\Gamma \cong \Gamma \prime$
NSK
Therefore,The group $(Z,+)and(E,+)$ are isomorphic.