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# Answer to Question #16071 in Abstract Algebra for Sone

Question #16071
Let &empty;: G &rarr; H be a group homomorphism. Show that &empty; is one-to-one if and only if &empty;^(-1)(e) = {e}
1
2012-10-09T08:47:51-0400
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.

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