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Topic: Epimorphic groups
Replies: 10   Last Post: Apr 5, 2013 3:37 PM

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Butch Malahide

Posts: 894
Registered: 6/29/05
Re: Epimorphic groups
Posted: Apr 4, 2013 8:56 PM
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On Apr 4, 6:11 pm, Kaba <k...@nowhere.com> wrote:
>
> Let G, G', and H be groups with G a sub-group of G'. Let f : G --> H and
> f' : G' --> H be surjective homomorphisms such that f = f'|G, the
> restriction of f' to G. The kernels of f and f' are equal. Does it then
> follow that G = G'?


Yes. Assume for a contradiction that G is a proper subgroup of G'.
Choose x in G'\G; then f'(x) is in H. Since f is surjective, f'(x) =
f(y) for some y in G. Since f is the restriction of f' to G, f'(x) =
f(y) = f'(y). Since f'(x) = f'(y), x/y is in the kernel of f', which
is equal to the kernel of f, so x/y is in G. Since x/y is in G and y
is in G, it follows that x = (x/y)y is in G, contradicting the
assumption that x is in G'\G.



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