- #1

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they would be isomorphisms.

Are there situations in wich the kernel of such a homomorphism would reduce to the identity? I'm thinking of situations where the groups act on different sp

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- Thread starter TrickyDicky
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- #1

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they would be isomorphisms.

Are there situations in wich the kernel of such a homomorphism would reduce to the identity? I'm thinking of situations where the groups act on different sp

- #2

HallsofIvy

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If that is your question, then, strictly speaking, yes. In order to be an isomorphism, a homomorphism must be both "one to one" and "onto". A homomorphism is "one to one" if and only if its kernel is the identity. But that leaves "onto". It is possible for a homomorphism to be "one to one" yet not be "onto".

However, in such a case, the homomorphism

- #3

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Hi, that was not what I meant, I had some problem with the editor and the question appeared cut.

If that is your question, then, strictly speaking, yes. In order to be an isomorphism, a homomorphism must be both "one to one" and "onto". A homomorphism is "one to one" if and only if its kernel is the identity. But that leaves "onto". It is possible for a homomorphism to be "one to one" yet not be "onto".

However, in such a case, the homomorphismis"onto" a subgroup of the original range group and we normally then focus on the homomorphism being an isomorphism to that subgroup.

I was gonna give some example: consider the well known homomorphism SU(2)->SO(3), Since SU(2) is the Universal cover of SO(3) the only reason it is not an isomorphism is that its kernel has one more element besides the identity, -I, and my question is if there are situations where the kernel

of this homomorphism is reduced to the identity, amd therefore turned into an isomorphism. Like maybe wnen acting on spaces that are not vector spaces that habe isometries that reduce the kernel.

- #4

Fredrik

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I suspect that whatever the exact question is, the best answer you will get is that there's a theorem that says that if ##\phi:G\to H## is a homomorphism, then the quotient group ##G/\ker\phi## is isomorphic to ##\phi(G)##.

- #5

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/

I suspect that whatever the exact question is, the best answer you will get is that there's a theorem that says that if ##\phi:G\to H## is a homomorphism, then the quotient group ##G/\ker\phi## is isomorphic to ##\phi(G)##.

That was just an example that fulfills the conditions I stablished. By situations I mean there are group actions on manifolds other than the usual Rn that modify the kernel so that the quotient trivially becomes G/Identity.

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