Proof of Correspondence theorem

[Gaussian97]

Summary:: I'm reading Adkins' book "Algebra. An approach via Module Theory" and I'm trying to prove theorem 3.15

In theorem 3.15 of Adkins' book says:
Let $N \triangleleft G$. The 1-1 correspondence $H \mapsto H/N$ has the property
$$H_1 \subseteq H_2 \Longleftrightarrow H_1/N \subseteq H_2/N$$
and in this case
$$[H_2: H_1] = [H_2/N: H_1/N].$$
The first part is proven by the book, but it left to prove $[H_2: H_1] = [H_2/N: H_1/N]$ to the reader. The book says that one should try to find a 1-1 correspondence between the sets $S_1 = \{aH_1 : a\in H_2\}$ and $S_2 = \{\bar{a}H_1/N : \bar{a}\in H_2/N\}$

I think I have the proof, but I'm not sure if it's correct/rigorous, so I would appreciate if anyone can look at it. My proof goes like this:
I think that the most natural correspondence is to define
$$\begin{array}? f: &H_2/H_1 &\rightarrow &(H_2/N)/(H_1/N) \\ & h_2H_1 & \mapsto & (h_2N)(H_1/N)\end{array}$$

First I need to prove that is a well-behaved function, i.e. that fulfils
$$h_2H_1 = h'_2H_1 \Longrightarrow (h_2N)(H_1/N) = (h'_2N)(H_1/N).$$
$$h_2H_1 = h'_2H_1 \text{ means that } \exists h_1, h'_1 \in H_1 : h_2h_1 = h'_2h'_1.$$ Therefore $h_2h_1N = h'_2h'_1N$, and because $N\triangleleft G$ I know (from Proposition 3.6 of the book) that the cosets of $N$, with the multiplication $(aN)(bN)=(ab)N$ form a group, and therefore
$$h_2h_1N = h'_2h'_1N \Longrightarrow (h_2N)(h_1N) = (h'_2N)(h'_1N) \Longrightarrow (h_2N)(H_1/N) = (h'_2N)(H_1/N).$$ So the function $f$ is well-defined.

Also, because any element of $(H_2/N)/(H_1/N)$ can be written as $(h_2N)(H_1/N)$, $f$ is surjective.

The only thing left to prove is injection, i.e.
$$(h_2N)(H_1/N) = (h'_2N)(H_1/N) \Longrightarrow h_2H_1 = h'_2H_1.$$
$(h_2N)(H_1/N) = (h'_2N)(H_1/N)$ means that
$$\exists h_1, h'_1 \in H_1 : (h_2N)(h_1N) = (h'_2N)(h'_1N) \Longrightarrow (h_2h_1)N = (h'_2h'_1)N$$
Because $G/N$ is a group, with neutral element $N$ and inverse $(a^{-1})N$ we can multiply by $(h_2h_1)^{-1}N$ to get
$$N = ((h_2h_1)^{-1}h'_2h'_1)N \Longrightarrow ((h_2h_1)^{-1}h'_2h'_1)\in N \Longrightarrow h_2h_1n= h'_2h'_1, \qquad n\in N$$
Finally, because $N \subseteq H_1$ implies $h_1 n \in H_1$ and that proves $h_2 H_1 = h'_2 H_1$.

With that, $f$ is a 1-1 correspondence and therefore the two groups must have the same number of elements, proving $[H_2: H_1] = [H_2/N: H_1/N]$.

If you can tell me if this prof is correct, or point me where it fails or where I need to be more rigorous I would appreciate.
Thank you

 

[member 587159]

I quickly read through your proof and the idea looks ok to me but some things are written somewhat awkwardly, but I guess that is part of the learning process. For example, when proving in the injectivity part that $h_2H_1= h_2' H_1$ it would be easier to show both the inclusions $\subseteq, \supseteq$ directly from the assumption and not take such a detour. But this might be matter of taste.

What is important here is that you wrote down a natural candidate for the correspondence and showed that it is well-defined. This is key, the rest are details that need checking but are less important. Throughout your work, I was also convinced that you understood the concept of quotient groups. Well done!