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.

FizeauV 2021-02-25 Answered
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.

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Expert Answer

2abehn
Answered 2021-02-26 Author has 13452 answers

To prove any two groups are isomorphic:
The map \(\displaystyle\phi:Г\rightarrowΓ′\) is called an isomorphism \(Γ\) and \(Γ'\) and are
said to be isomorphic if
i) \(\displaystyle\phi\) is a homomorphism.
ii) \(\displaystyle\phi\) is bijective.
\(\displaystyleΓ\stackrel{\sim}{=}Γ′\) denotes \(Γ\) is isomorphic to \(Γ′\)
Let \(\displaystyle\phi:{Z}\rightarrow{2}{Z}\) be defined by \(\displaystyle{x}\mapsto{x}+{x}={2}{x}\)
i) To prove \(\phi\) is homomorphism:
\(\displaystyle\phi{\left({x}+{y}\right)}=\phi{\left({x}\right)}+\phi{\left({y}\right)}\)
\(\displaystyle\phi{\left({x}+{y}\right)}={2}{\left({x}+{y}\right)}\)
\(\displaystyle\phi{\left({x}+{y}\right)}={2}{\left({x}\right)}+{2}{\left({y}\right)}\)
\(\displaystyle\phi{\left({x}+{y}\right)}=\phi{\left({x}\right)}+\phi{\left({y}\right)}\)
Therefore, \(\displaystyle\phi\) is homomorphism
ii) To prove \(\displaystyle\phi\) 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,
\(\displaystyle\phi{\left({x}\right)}=\phi{\left({y}\right)}\)
\(\displaystyle{2}{x}={2}{y}\)
\(\displaystyle\Rightarrow{x}={y}\)
Therefore, \(\displaystyle\phi\) is one-one.
To prove \(\displaystyle\phi\) 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 \(\displaystyle{y}\in{2}{Z}\)
Then y=2k for some \(\displaystyle{k}\in{Z}\)
Since \(\displaystyle{k}\in{Z}\) and
\(\displaystyle\phi{\left({k}\right)}={2}{k}\)
\(\displaystyle\Rightarrow{y}\)
\(\displaystyle\phi\) is onto
\(\displaystyle\Rightarrow\phi\) is bijective.
Since, it is homomorphism and bijective.
it is isomorphic.
\(\displaystyleΓ\stackrel{\sim}{=}Γ′\)
Therefore,The group (Z,+) and (E,+) are isomorphic.

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