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A homomorphism $f:G$ to $G1$ of groups is a monomorphism iff Ker $f = \{e\}$.
asked in Set Theory & Algebra by (475 points)
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Definition of homomorphism of groups

Let (G,*) and (G1,o) be two groups and f be a function from G to G1. Then f is called a homomorphism of G into G1 if for all a,b∈G

            f(a*b) = f(a) o f(b)

To do the above problem you have to know the following theorem

If f is a homomorphism from a group G into a group G1 and e,e1 are the identity elements of G and G1 respectively ,then

(i) f(e) = e1

(ii) f(a-1) = f(a)-1 for all a ∈ G

Now, come to the given problem

Suppose f is a monomorphism. Then f is injective.

Let a∈ ker f and also assume e be the identity element of G and e1 be the identity element of G1

Hence, f(a) = e1 ( From the definition of Ker f)

Again f(e) = e ( From the theorem written above)

Since f is injective we have a=e.

Hence ,ker f = {e}. 

Now, to prove the converse 

Conversely, suppose ker f = {e}.

Let x,y ∈ G such that f(x) = f(y)

Then f(x)f(y)-1 = e1

=> f(x)f(y-1) = e1 (From the theorem written above)

=> f(xy-1) = e1 (From the definition of homomorphism of groups).

=> xy-1 ∈ ker f = {e} and so xy-1 = e.

Hence, x=y.

Therefore f is injective and so f is a monomorphism.

Hence the problem 

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

answered by Loyal (9.5k points)
selected by
Why  u took f as monomorphism?
In the question we have to prove

f is monomorphism if and only if ker f = {e}

This means we have to prove two statements

1) If f is a monomorphism then ker f={e}

2) If ker f = {e} then f is a monomorphism.

So, in the 1st part of the problem I have proved ker f= {e} assuming f is a monomorphism.

And in the 2nd part (converse) I have proved f is a monomorphism assuming ker f = {e}.

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