Date: Apr 5, 2013 5:31 AM
Author: Kaba
Subject: Re: Epimorphic groups
5.4.2013 3:56, Butch Malahide wrote:

> 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.

Very nice, thanks!:)

For those interested, this proves in Clifford algebra that the group of

pinors is equal to the group of (unit) versors, while they look

different by their definition. Both are epimorphic in the above manner

to the group of orthogonal transformations O(V) in a bilinear space V,

and unit versors are a sub-group of pinors.

--

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