Surjection between kernel and image of a homomorphism

[burritoloco]

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.

 

[micromass]

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.

 

[burritoloco]

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?

 

[micromass]

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 $\mathbb{Z}_3$.

 

[burritoloco]

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?

 

[burritoloco]

Thank you for that :)

 

[burritoloco]

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..

 

[micromass]

Vector space over what? Some kind of finite field?

 

[burritoloco]

Rightly guessed!

 

[burritoloco]

"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.

 

[burritoloco]

Intuitively I think it makes sense a surjection exists.

 

[micromass]

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 $f$.

 

[burritoloco]

Perfect, thank you. The vectors that are not in the span of the subspace basis are just sent to 0!