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That is the statement of the third isomorphism theorem. Let K and B be normal subgroups of a group G with N $\subset$ K $\subset$ G. Then K / N is normal subgroup of G / N, and the quotient group $(G / N) / (K / N)$ is isomorphic to G / K.

I have read and understood the proof but I am trying to visualize more the situation to get better understanding.

(G / N) / (K / N) = Nk * Ng I don't immediately see how it is isomorphic Kg according to my intuition the reason it is isomorphic to Kg is that we know Nk will partition K and Ng will partition G hence there product should be somehow related to the Kg, but my intuition isn't carrying anymore further and isn't helping more if somehow have more insight that would be nice.

Andrés E. Caicedo
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1 Answers1

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It's easiest (but not most effective) to view it in an abelian setting.

Consider the integers. For any subgroup $k\mathbb{Z}$ of the integers that contains the subgroup $n\mathbb{Z}$ we have the direct relation $(k,n)>1$. If $k,n$ are the least representatives then $k\mid n$. This is the number theoretic standpoint of the theorem. The third isomorphism theorem gives $(\mathbb{Z}/n\mathbb{Z})/(k\mathbb{Z}/n\mathbb{Z})\cong \mathbb{Z}/k\mathbb{Z}$. What is $k\mathbb{Z}/n\mathbb{Z}$? It is the subgroup of $\mathbb{Z}/n\mathbb{Z}$ generated by the element $k$. Try writing out some examples if you need. The fact that we can take any of these quotients is a nice property of abelian groups.

Most of the examples I can come up with will be of the commutative type. Here is a geometric (still abelian) view.

Consider the space $\mathbb{R}^3$ with addition as vector addition. Any plane running through the origin will be a normal subgroup of $\mathbb{R}^3$, as the addition of two vectors in a plane will result in another in the plane and the additive inverse of $v$ is simply $-v$ which is still in the plane. Call the plane the set generated by two elements (linearly independent!) $\langle x,y \rangle$. Now consider a line contained in the plane that passes through the origin (we may call this $\langle z \rangle$). This is a subgroup of the plane, following a similar argument to the one given above. The quotient space $\mathbb{R}^3/\langle z \rangle$ is isomorphic to the real plane, $\mathbb{R}^2$ (what are its cosets?), and the space $\langle x,y\rangle/ \langle z\rangle$ is isomorphic to the the real line $\mathbb{R}$ (what are its cosets?). The third isomorphism gives $(\mathbb{R}^3/\langle z \rangle)/(\langle x,y\rangle/ \langle z\rangle)\cong \mathbb{R}^3/\langle x,y\rangle\cong \mathbb{R}$. But this makes sense, as the plane quotiented by a line should be a line.

Typically it is easier to view the theorem in the simplest backgrounds before finding examples in less obvious places. Perhaps keep a small collection of groups to test these on. Of course, I only provided two examples of abelian groups and nowhere in the proof of the third isomorphism theorem does it mention that the group is commutative. Good luck searching for more examples!

Eoin
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