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I am trying to represent myself quotient groups and I'm having trouble seeing what the kernel of a homomorphism : $\Phi: G \rightarrow G/H$ is (be it a ring homomorphism or a group homomorphism).

I understand that $\mathbb{Z}/n\mathbb{Z}$ is a quotient group that is cyclic and that the kernel of the application: $\Pi : \mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}$ are all the elements of $n\mathbb{Z}$.

But in the more general case I have trouble truly visualizing the kernel of a quotient group:

Let us take the 2 following examples:

  • In the case of a group homomorphism, Let G be a group and H be a subgroup of G: $\Xi : G \rightarrow G/H$, $\Xi(g) \mapsto gH$ I know that its kernel is H but I don't understand why.

  • In the case of a ring homomorphism: If $(A,+,*)$ is a ring and $I$ is a two-sided ideal of $A$. Let the ring homomorphism $\chi: A \rightarrow A/I$, $\chi(a) \mapsto a+I$ I know that its kernel is I but I don't understand why.

Thank you for any clarification

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I'll answer for the case of a group homomorphism. Sometimes it helps to draw a picture in order to clarify this idea of a kernel for quotient groups. A big circle for G and G/H with circles in each for the kernel, image, ect, and a lines denoting the mapping between the groups. I'll try my best to clarify it in words. In any group homomorphism, Φ:G→G/H, you must have the property that, Φ: $e_G$--->$e_{G/H}$. So, the identity element of G must be sent to the identity element of G/H. In order to visualize the kernel, you have to ask yourself 1) What is the identity element of the coset gH? But, this is just H, since xH*H = xH for all xH in G/H. 2) What elements in G get mapped to this identity element? Well, this is all the elements h $\in$ G such that h $\in$ H. Since, if we take, Φ( h )---->hH=H. I hope this helps, I can try and find a good picture for you.

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  • $\begingroup$ This helped me visualize it thank you, just one question: $\Phi(h) \rightarrow hH = HI$ So is $HI = e_{G/H}$ $\endgroup$ – aribaldi May 27 '16 at 7:22
  • $\begingroup$ The identity element of G/H, is the whole set H. Don't worry to much about ideals in this case, you need Φ(h)→hH=Hh=e $\endgroup$ – wesssg May 27 '16 at 20:50

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