Let $K\le H\le G$ with $[G:H]<\infty$. Is $[G:K]<\infty$ necessarily?

Question

While reading this proof for the Tower Law for Subgroups, the data for the theorem only stipulates that H is a subgroup of G with finite index, and that K is a subgroup of H. Later in the proof however they write "..By hypothesis these numbers are finite." when referring to the indexes of both K and H in G.

Is this simply an omission error (they should've explicitly mention K having finite index as well), or is there a theorem which states that when a subgroup H has finite index in G, so does every subgroup K of H?

Would appreciate any counter-examples otherwise!

Answer 1

No, that's false as stated. A counterexample: $G = \mathbb{Z}$, $H = 2\mathbb{Z}$, and $K = \{0\}$. Clearly, $[G: H] = 2 < \infty$, but $[H: K] = \infty$. Of course, $[G: K] = \infty$ as well, so as long as you're comfortable with an "equation" like $\infty = 2 \cdot \infty$, the index tower law still holds.

One easy fix is to just start with the hypothesis that $[G: K] < \infty$. In that case, it does follow that $[G: H] < \infty$ and $[H: K] < \infty$, and the formula holds.

Answer 2

Observe: $[2\Bbb Z:\{0\}]=\infty$ with $[\Bbb Z:2\Bbb Z]=2,$ but $$[\Bbb Z:\{0\}]=\infty.$$

---

For a proof of the Tower Law for infinite indexes, see here.