1. ## Intersection of subgroups

Show that the intersection of 2 subgroups of finite index also has finite index.

Proof.

Let $\{S_{a} \} _{a \in I}$ and $\{ T_{b} \} _{b \in J}$

The intersection of them, well, since the index of both are finite, isn't the intersection $\bigcap _{c \in I \cup J } \{U _{c} \}$ well, then, the index has to be finite.

But I'm not sure if I write this up correctly.

Thanks.

2. Let $H,K$ be subgroups.
Define $(G:H)$ to be the cosets of $H$.
Define $(G:K)$ to be the cosets of $K$.
Define $[G:H] = |(G:H)|$ and $[G:K] = |(G:K)|$.

Note that $H\cap K$ is a subgroup.
Define a mapping from $(G:H\cap K)$ to $(G:H)\times (G:K)$ by $a(H\cap K)\mapsto (aH,aK)$.
We need to show this mapping is well-defined.
To show this we use a fact that if $aN = bN$ then $ab^{-1} \in N$ for $N$ some subgroup.

The next thing we notice is that this mapping is one-to-one.
For if $(aH,aK) = (bH,bK) \implies ab^{-1} \in H \text{ and }ab^{-1} \in K \implies a(H\cap K) = b(H\cap K)$.
Since this mapping is one-to-one it means $[G:H\cap K] \leq [G:H]\times [G:K] < \infty$.

3. Thanks, perfecthacker, as you can see my season is starting again, and this modern alg is hard...

I understand the proof in general, but I do have two questions:

How do we conclude from $aN=bN$ implies $ab^{-1} \in N$? (sorry, I know should be easy, it is just that I been out of touch from Algebra for the summer... I really should know the answer for this.)

And how does this fact show that this mapping is well-defined?

Thank you so much!

How do we conclude from $aN=bN$ implies $ab^{-1} \in N$?
Since $aN = bN$ and $a\in aN$ it means $bh=a$ for some $h\in H$. Thus, $h = ab^{-1} \in H$.
Say $a(H\cap K) = b(H\cap K)$.
This means $ab^{-1} \in H \text{ and }ab^{-1} \in K$.
Now $a(H\cap K)\mapsto (aH,aK) \text{ and }b(H\cap K)\mapsto (bH,bK)$.
By above $aH = bH \text{ and }aK=bK$.
Thus, $(aH,aK) = (bH,bK)$.