## Tuesday, June 15, 2010

### A Group Theory Exercise

Here is nice exercise that uses all three isomorphism theorem (see here) at once. Let f be a group homomorphism from a finite group G and let H be a normal subgroup (we can extend this to non-finite groups but let us rather stay with finite group so that everything we deal with is are natural numbers). Prove that the relationship between (G:H) and (f(G):f(H)) is given by, $$(f(G):f(H))\cdot |\ker f| = (G:H)\cdot |\ker f \cap H|$$.

Solution: Define the function $$\hat f : G\to f(G)/f(H)$$ by $$g\equiv f(g) (\bmod f(H))$$. Clearly this function is a group homomorphism. We claim that its kernel is the product $$H\cdot \ker f$$. To see this let $$x\in \ker \hat f$$. Then $$f(x) \in f(H)$$ so f(x) = f(h) for some h in H. Thus, $$f(h^{-1}x)=1$$ which means x=hk where h is in H and k is in the kernel of f. Conversely, if x=hk where h is in H and k is some element of the kernel of f then it is clear that f(x) = f(h) which is in f(H). Thus, by the first isomorphism theorem we have $$f(G)/f(H) \simeq G/(H\cdot \ker f)$$. Now by the third isomorphism theorem we have $$f(G)/f(H)\simeq (G/H)/(H\cdot \ker f/H)$$. However, using the second isomorphism theorem we quickly see that $$f(G)/f(H)\simeq (G/H)/(\ker f / (\ker f\cap H))$$. From here the equation that we were supposed to prove easily follows.

What is the lesson of all of this? That the kernel measures in some way the wrapping of the morphism around itself. If the kernel is trivial then the morphism is injective and so there is no wrapping. The larger the kernel is the more the morphism wraps around itself. It is obvious that (f(G):f(H)) = (G:H) if the f is injective, we do not even need the proof above. However, in the case when f is not injective we have a certain wrapping factor, this is why we expect to find the kernel in the solution to the more general problem.