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Definition of **homomorphism of groups**

Let (G,*) and (G_{1},o) be two groups and f be a function from G to G_{1}. Then f is called a homomorphism of G into G_{1} 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 G _{1} and e,e_{1} are the identity elements of G and G_{1} respectively ,then**

**(i) f(e) = e _{1}**

**(ii) f(a ^{-1}) = f(a)^{-1 }for all a ∈**

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 e_{1} be the identity element of G_{1}.

Hence, f(a) = e_{1} ( From the definition of Ker f)

Again f(e) = e_{1 } ( 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} = e_{1}

=> f(x)f(y^{-1}) = e_{1} (From the theorem written above)

=> f(xy^{-1}) = e_{1} (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 →G_{1} of groups is a monomorphism iff Ker f = {e}.

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}.

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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