# Answer to Question #16071 in Abstract Algebra for Sone

Question #16071

Let ∅: G → H be a group homomorphism. Show that ∅ is one-to-one if and only if ∅^(-1)(e) = {e}

Expert's answer

Suppose f is one-to-one, that is if

f(a)=f(b)

then

a=b.

We

should prove that

f^(-1)(e) = {e}.

Indeed, suppose that f^{-1}(e)

contains some other element a distinct from e, so

f(a)=f(e)=e.

Since

f is one-to-one, we get that a=e, which contradicts to the assumption that

a<>e.

Thus

f^(-1)(e) = {e}.

Conversely, suppose

f^(-1)(e) = {e}.

Suppose also that there are two elements a,b from G such

that f(a)=f(b).

We should prove that a=b.

Indeed, since f(a)=f(b), it

follows that

e = f(a) f(b)^{-1} = f(ab^{-1}),

that is

ab^{-1}

belongs to f^(-1)(e) = {e}.

This means that

ab^{-1} = e

and so

a=b.

f(a)=f(b)

then

a=b.

We

should prove that

f^(-1)(e) = {e}.

Indeed, suppose that f^{-1}(e)

contains some other element a distinct from e, so

f(a)=f(e)=e.

Since

f is one-to-one, we get that a=e, which contradicts to the assumption that

a<>e.

Thus

f^(-1)(e) = {e}.

Conversely, suppose

f^(-1)(e) = {e}.

Suppose also that there are two elements a,b from G such

that f(a)=f(b).

We should prove that a=b.

Indeed, since f(a)=f(b), it

follows that

e = f(a) f(b)^{-1} = f(ab^{-1}),

that is

ab^{-1}

belongs to f^(-1)(e) = {e}.

This means that

ab^{-1} = e

and so

a=b.

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