0

You know that map where G goes to H and then we also know a surjective group homomorphism G to G/N exists. How do I connect the G/N to H? As in, how do I show that there is a surjective map between them? Also, how come showing that a map is well-defined does not imply injective?

cakey
  • 245
  • What do you need to show exactly? – UNM Oct 31 '13 at 21:25
  • Try to be more specific about the groups here. It will help you more than it helps us. – Shaun Oct 31 '13 at 21:27
  • I need to connect the "canonical" group homomorphism maps to the image called H. I also know that such mapping isn't injective just from doing problems, but I don't really know how to prove it. – cakey Oct 31 '13 at 21:27

1 Answers1

4

The first isomorphism theorem for groups says that if $\phi:G\to H$ is a homomorphism of groups, then $G/\ker\phi\cong \phi(G)\subseteq H$. I assume that in your question, $N$ is meant to be the kernel of your homomorphism. If the original map is not surjective, there is not necessarily a surjective map from $G/\ker\phi\to H$. However, if the original $\phi$ is surjective, then the first isomorphism theorem tells you there is a surjection (as there is an isomorphism).

As for the second part of your question, a function $f:X\to Y$ is well-defined if you obtain the same element of the codomain no matter how you represent an element in the domain (different representations of the same element give the same result under the mapping). This has nothing to do with injectivity. For example, the function $f : \Bbb Q\to\Bbb Z$ given by $f(q) = n - m$ (with $q = n/m$) is not well-defined, because $2 = 2/1 = 4/2$, but $2 - 1 = 1\neq 2 = 4 - 2$, while the function $f : \Bbb Q\to \Bbb Z$ given by $f(q) = 0$ is well-defined (writing $q = n/m = n'/m'$ in two different ways will not change $f(q)$), but very far from being injective.

Stahl
  • 23,212
  • Okay, I think in my example it isn't isomorphic because I accidentally said maps to image but the image is mapping to something else I believe since in the problem, the map is only surjective. I forgot the term of when the kernal gets mapped to the image and then to something else. – cakey Oct 31 '13 at 21:34
  • for example, I get confused with proving injective because I prove "well-defined" by saying let a=b and then I come out with f(a)=f(b) and then I say injective, unless this is wrong? Is this the same way to prove well-definedness? – cakey Oct 31 '13 at 21:35
  • I assume you're trying to show that multiplication within the quotient group is well-defined. By definition, multiplication in $G/N$ is given by $aN\cdot bN = (ab)N$. You must show that if $aN = a'N$ and $bN = b'N$, then $(ab)N = (a'b')N$. That is, the multiplication does not depend on the choice of coset representative. Your statement $a = b\implies f(a) = f(b)$ doesn't show that $f$ is injective. If $b$ is an arbitrary way of writing $a$, then showing that $f(a) = f(b)$ is showing that $f$ is well-defined. – Stahl Oct 31 '13 at 21:40
  • To show that $f$ is injective, you would need to go the other way: take $f(a) = f(b)$ and show that $a = b$ (or take $a\neq b$ and show $f(a)\neq f(b)$). – Stahl Oct 31 '13 at 21:43
  • Oh okay, then the only other way I see to show injectivity is giving an inverse map? But i think I understand the question they are asking me. The want me to show that the kernal which is isomorphic to the image and gets mapped to H is surjective. So I have to show that all elements of G/N get mapped into H. – cakey Oct 31 '13 at 21:45
  • Just saw your comment about injectivity.I get the injectivity part now though thanks for that good explanation. But I suppose the hard part is showing surjectivity. How do I even know all elements of the image get mapped onto H? – cakey Oct 31 '13 at 21:46
  • For example I know since the image must map to H, how can I know that it is a surjective map, unless that is not even true? – cakey Oct 31 '13 at 22:03
  • Actually I reread the question and it says that G maps surjectively onto H, therefore...the image must map surjectively onto H, but I don't know how to formally prove that. – cakey Oct 31 '13 at 22:04
  • If $G$ maps surjectively onto $H$, then $Im(G) = H$ so the identity is a surjection $Im(G)\to H$. – Stahl Oct 31 '13 at 23:02
  • Okay, well I worked on connecting the kernal to the Im(G) be showing "well-definedness" there along with a homomorphism there, is that all I need to connect my kernal to the image?Do I need these both? If so, how come I can't just show well-definedness alone? – cakey Nov 01 '13 at 13:36
  • I'm not sure exactly what issue you're having but this might answer some of your questions (specifically Arturo's answer). – Stahl Nov 01 '13 at 15:55
  • This is what I'm asked to do. I think I'm also confused about what I'm being asked. "Show that φ is surjective, and determine its kernel. In general, φ is not an isomorphism. (Try to construct examples where it is not, and think about how this applies in the situation of group presentations!)" The φ is the map from G/N to H. I don't know what I need to show the surjection here. – cakey Nov 02 '13 at 02:41