
Re: Epimorphic groups
Posted:
Apr 4, 2013 8:56 PM


On Apr 4, 6:11 pm, Kaba <k...@nowhere.com> wrote: > > Let G, G', and H be groups with G a subgroup 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.

