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I am trying to prove the so-called "Tower law' for subgroups:

Given $G \supset H \supset K$, subgroups of $G$ such that $[G:H]$ and $[H:K]$ are finite, we have $$[G:K] = [G:H][H:K].$$

I do not want to make the assumption that $G$ is finite. If that were the case, then the result follows immediately from Lagrange's theorem. If I am not mistaken, if I assume that one of these indices is finite, then all of them are finite, but the group itself may still be infinite, even if it's partitioned into an infinite number of cosets. (I can't think of an example off-hand of this, so if anyone knows of a canonical one, that'd be very helpful.)

I've done some reading on this and have an attempt written down, but I can't understand a few of the steps.

By assumption, $[G:H]$ and $[H:K]$ are finite. Denote the former by $m$ and the latter by $n$. Since left cosets of $H$ in $G$ partition $G$ and the left cosets of $K$ in. $H$ partition $K$, we choose $g_1, \ldots, g_m$ and $h_1, \ldots, h_n$ such that \begin{align*} G = \bigsqcup\limits_{i=1}^m g_i H, \; H = \bigsqcup\limits_{j=1}^n h_j K. \end{align*} We therefore have: \begin{align*} G = \bigsqcup\limits_{i=1}^m g_i H = \bigsqcup\limits_{i=1}^m g_i \left(\bigsqcup\limits_{j=1}^n h_j K \right) = \bigsqcup\limits_{i=1}^m g_i \bigsqcup\limits_{j=1}^n g_i (h_j K) = \bigsqcup\limits_{i=1}^m \bigsqcup\limits_{j=1}^n (g_i h_j)K, \end{align*} so $G$ is a disjoint union of $mn$ left cosets of $K$ in $G$, so we have \begin{align*} [G:K] = mn = [G:H][H:K], \end{align*} as desired.

I approached this by "naive" substitution, treating these disjoint unions as if they were sums or products, but I'm not convinced that I fully understand the operation. When I put two unions together, I assume this is, in effect, a 'Cartesian product.' Is that right? When I "pull $g_i$ within the inner union," I can't say I'm using "linearity," but I don't know what the multiplication is, since I have coset multiplication within the union (and I'm multiplying $g_i$ by a coset) but also a group multiplication.

In a virtually identical proof, Artin comments that the last step is justified because multiplication by $g_i$ is an invertible operation. I do not understand where this comes from, unless its using the defining equivalence relation $a \sim b \iff a^{-1} b \in H$ that defines a coset.

Any help on understanding this would be grtand, greatly appreciated.

user861776
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  • The function that sends $x$ to $g_ix$ is the invertible function: it is invertible because the function that sends $y$ to $g_i^{-1}y$ is the inverse function. – Arturo Magidin Apr 12 '21 at 22:22
  • I am not sure this works. When you say “subgroups of finite index”, I interpret that to mean that $H$ and $K$ are both of finite index in $G$. Instead, you seem to be interpreting it to mean that $H$ is of finite index in $G$ and $K$ is of finite index in $H$. Of course, the tower law holds for cardinalities, finite or infinite, so the two assumptions end up being equivalent, but it is not clear to me what the assumption is supposed to be. – Arturo Magidin Apr 12 '21 at 22:25
  • This is a very good point. I was trying to say that both indices on the right-hand side are finite. I'll modify this. – user861776 Apr 12 '21 at 22:28
  • A different question, though: If $H$ has finite index in $G$, does it immediately follow $K$ has finite index in $H$? I'm trying to think of the best way to simplify the assumption. – user861776 Apr 12 '21 at 22:31
  • Think $G=\mathbb{Z}$, $H=2\mathbb{Z}$, $K=0$. Then $[G:H]$ is finite but $[H:K]$ isn't. – Alvaro Martinez Apr 12 '21 at 22:37

1 Answers1

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When I put two unions together, I assume this is, in effect, a 'Cartesian product.' Is that right?

It is a (disjoint) union of (disjoint) unions, therefore a (disjoint) union, nothing more.

When I "pull $g_i$ within the inner union", I can't say I'm using "linearity", but I don't know what the multiplication is.

In general, given a set $X\subset G$ and $g\in G$, $gX$ is a shorthand for $\{gx:x\in X\}$. Therefore $g\bigsqcup_{i=1}^n X_i=\{gx:x\in \bigsqcup_{i=1}^nX_i\}=\bigsqcup_{i=1}^n\{gx:x\in X_i\}=\bigsqcup_{i=1}^ngX_i$. This means that $g$ "distributes over unions".

Hopefully this clears up the proof for you!

Alvaro Martinez
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