Abstract algebra: Take a group G, and a group H, with a group operation *. G is isomorphic to H if there exist a map f: G->H such that:
1: f is injective
2: f is surjective
3: f(a*b)=f(a)*f(b) for all a,b in G
If f satisfies these three properties, f is called an isomorphism.
1: f is injective
2: f is surjective
3: f(a*b)=f(a)*f(b) for all a,b in G
If f satisfies these three properties, f is called an isomorphism.
The map f: Z -> E given by f(a)=2a where Z is the integers and E is the even integers is an isomorphism.
Proof:
Showing injectivity
f(b)=f(a) => 2a=2b (from the given function) <=> a=b
Showing surjectivity
Suppose n is in E. n is an even integer hence n=2k for some integer k.
f(k)=2k=n, hence f is surjective.
Homomorphism:
f(a+b)=2(a+b)=2a+2b=f(a)+f(b)
Hence f is an isomorphism. Q. E. D.
Proof:
Showing injectivity
f(b)=f(a) => 2a=2b (from the given function) <=> a=b
Showing surjectivity
Suppose n is in E. n is an even integer hence n=2k for some integer k.
f(k)=2k=n, hence f is surjective.
Homomorphism:
f(a+b)=2(a+b)=2a+2b=f(a)+f(b)
Hence f is an isomorphism. Q. E. D.
by qsqazxcvfrew March 29, 2018
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