+1 vote
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Prove that :-

Every infinite cyclic group is isomorphic to the infinite cyclic group of integers under addition.
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+2
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Thank you !
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Do we have these in our Current GATE syllabus?

## 1 Answer

+3 votes
Best answer

I hope that you know the definitions of Homomorphism and Monomorphism of groups

To, solve the above problem you have to know one theorem

Theorem  : A homomorphism f : G→G1 of groups is a monomorphism iff Ker f = {e}

Now, come to the problem

Let, G = <a> be an infinite cyclic group. Then o(a) is not finite and this implies an =e iff n=0.

Define f:Z→G by f(n) = an for all n∈Z.

Clearly f is a surjective function and f(n+m) = an+m = anam = f(n)f(m) (Definition of Homomorphism of groups). for all n,m∈Z

Now, Ker f = { n∈Z | f(n) = e } = { n∈Z | an = e } = {0}

This implies f is injective (Theorem written above).

Thus f is an isomorphism.

So, the infinite cyclic group is isomorphic to the set of integers under addition.

by Loyal (9.6k points)
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bro ,could you please explain this thing "Ker f = {e}"
+1

Explanation of the statement ker f = {e}

ker f means kernal of f

where  f: G → G1

Let e = identity element of G and e1= identity element of G1

So, ker f = { n ∈ Z | f(n) = e1 ( identity element of G1) } = {e} (a set containing only the identity element of G)

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I felt that I didn't explained it properly before thats why I edited my comment.
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Define f:Z→G by f(n) = an for all n∈Z.

Clearly f is a surjective function and f(n+m) = an+m = anam = f(n)f(m) (Definition of Homomorphism of groups). for all n,m∈Z

@kushagra , according to this group G contains multiplication operation . right ?

because according to homomorphism definition ,

if we have 2 groups (G,*) and (H,o)  and mapping f: G --> H

then f(a*b) = f(a)o(b) for all a,b belongs to G.

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it is a representation like power set

not multiplication

rt?
+1
Here the operator on G need not to be multiplication

It can be anything say # defined in such a way that G forms an infinite cyclic group

In group theory a^m means a is operated on itself m times. i.e. a # a # a ....(m times)

Now, f(n+m) = a^(n+m) (i.e. a#a#a....(m+n times)) = a^m # a^n = f(m) # f(n).
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got it... thank you !

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