Surjection between kernel and image of a homomorphism

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Discussion Overview

The discussion revolves around the existence of surjective homomorphisms related to the kernel and image of a homomorphism, particularly in the context of groups and vector spaces. Participants explore implications of the first isomorphism theorem and related questions about isomorphisms between normal subgroups and their implications for the groups themselves.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether the first isomorphism theorem implies a surjective homomorphism from the kernel to the image or vice versa.
  • Another participant clarifies that the first isomorphism theorem establishes an isomorphism between the quotient of the group by the kernel and the image, but does not directly imply a surjection between the kernel and the image.
  • A participant expresses uncertainty about the implications of having an isomorphism between normal subgroups of equally sized finite groups and whether it implies the groups themselves are isomorphic.
  • Another participant provides a counterexample involving non-isomorphic groups that share a normal subgroup isomorphic to a common group.
  • There is a discussion about the possibility of guaranteeing a surjection from a group to its normal subgroups.
  • One participant shifts the focus to vector spaces, questioning the existence of surjections from a vector space to its subspaces.
  • Another participant suggests that surjections can be established by comparing dimensions of the vector spaces involved.
  • Further elaboration is provided on how to construct surjections using projections based on bases of the vector spaces.
  • One participant concludes that a surjection from the kernel to the image may exist if the dimension of the kernel exceeds that of the image, but notes that this surjection may not relate to the original homomorphism.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the first isomorphism theorem and the existence of surjective mappings, indicating that multiple competing views remain. The discussion about isomorphisms between normal subgroups also reveals disagreement, particularly regarding whether such isomorphisms imply group isomorphism.

Contextual Notes

Participants acknowledge limitations in their reasoning, particularly regarding assumptions about dimensions and the nature of mappings between groups and vector spaces.

Who May Find This Useful

Readers interested in group theory, homomorphisms, vector spaces, and the implications of the first isomorphism theorem may find this discussion relevant.

burritoloco
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Hi, I was wondering whether the following is true at all. The first isomorphism theorem gives us a relation between a group, the kernel, and image of a homomorphism acting on the group. Could this possibly also imply that there exists a surjective homomorphism either mapping the previous kernel to the image or the image to the kernel? It's not a homework question per se, just a question of mine. Cheers.
 
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The first isomorphism theorem gives an isomorphism between ##\frac{G}{\textrm{Ker}(f)}## and ##f(G)##. I'm not sure how you would use this to find a surjection between the kernel and ##f(G)##. Can you explain why you think the first isomorphism theorem would imply this?

Also, notice that ##f(\textrm{Ker}(f)) = \{0\}##, so ##f## will not be the surjection you want.
 
Thanks! I'm not sure it would imply it at all, but it would definitely come in handy for what I'm trying to do :). Another related question: If I have a group isomorphism between two normal subgroups of two equally sized finite groups, then would the two groups also be isomorphic? And if so, would the same mapping of the normal subgroups (showing their isomorphism) also imply the group isomorphism?
 
No. Consider the non-isomorphic groups ##D_6## (dihedral group of order ##6##) and ##\mathbb{Z}_2\times \mathbb{Z}_3##. Both groups contain a normal subgroup isomorphic to ##\mathbb{Z}_3##.
 
I forgot to mention that for the first question the image is also contained within the group, if this is of any use. Also, I'm wondering whether we could even guarantee a surjection from a group to any of its normal subgroups?
 
Thank you for that :)
 
Ahh, now I see that I should have mentioned that I'm actually dealing with vector spaces instead of groups lol. D_6 is not abelian..
 
Vector space over what? Some kind of finite field?
 
Rightly guessed!
 
  • #10
"Another related question: If I have a group isomorphism between two normal subgroups of two equally sized finite groups, then would the two groups also be isomorphic? And if so, would the same mapping of the normal subgroups (showing their isomorphism) also imply the group isomorphism?"

I just realized this is not true at all even for vector spaces. Scratch this. But I'm still curious about the 1st question. Even about a guarantee that we can find surjections from a vector space to any of its subspaces.
 
  • #11
Intuitively I think it makes sense a surjection exists.
 
  • #12
Then to find a surjection, you only need to compare the dimensions of the spaces in question. So in particular, if we have a vector space ##V## and a subspace ##W## we can always find a surjection from ##V## to ##W##.
How? Well, we can find a basis ##\{e_1,...,e_n\}## of ##V## such that ##\{e_1,...,e_k\}## is a basis of ##W##. Then we can use projections as a surjection, that is, take the function

f(\alpha^1e_1 + ... + \alpha^ne_n) \rightarrow \alpha^1e_1 + ... + \alpha^ke_k

Similarly, if ##V## and ##V^\prime## have the same dimension, and if there is an isomorphism ##f:W\rightarrow W^\prime## between subspaces, then we can always extend this isomorphism to an isomorphism between ##V## and ##V^\prime##. We again take right bases for ##V## and ##V^\prime## and we use them to find an isomorphism.

Now, for your question in the OP. There will be a surjection from ##\textrm{Ker}(f)## to ##f(G)## if the dimension of the kernel is greater than the dimension of the image. But even then, I don't think the surjection that we will find will necessarily have anything to do with ##f##.
 
  • #13
Perfect, thank you. The vectors that are not in the span of the subspace basis are just sent to 0!
 

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