Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics I / 6 / b (Finite-index subgroups)

For every $g\in G$, left multiplication gives a bijectionwhose inverse is $x\mapsto g^{-1}x$. Hence every left coset has the same cardinality as $H$. If $H$ is an infinite set, every left coset is infinite; if $H$ is a finite set, every left coset is finite. These are the only possibilities.

Suppose $H$ were finite, say $|H|=r$. There are at most $n$ left cosets, each with $r$ elements by part (i), and they partition $G$. Hence $G$ would be a finite set with at most $nr$ elements. This contradicts the assumption that $G$ is an infinite group. Therefore $\boxed{H\text{ is infinite}}$.

Every left coset $kH$ with $k\in K$ is also a left coset of $H$ in $G$. Moreover equality of two such cosets means the same thing in $K$ and in $G$:Thus the map from the set of left cosets of $H$ in $K$ to the set of left cosets of $H$ in $G$, taking $kH$ to itself, is injective. Consequently the index of a subgroup satisfiesThese are sets of cosets; no normal subgroup assumption is needed.

Define a map between sets of left cosets byIt is well defined: replacing $g$ by $gu$ for $u\in H\cap K$ changes neither component. It is injective becauseThe Cartesian product on the right has at most $nm$ elements. ThereforeThis proves the basic bound for a finite intersection of finite-index subgroups.

Put $L={\bigcap }_{i=1}^k{H}_i$. Repeated application of part (iv), or mathematical induction on $k$, givesPart (ii) therefore implies that $L$ is an infinite set. If $G\setminus U$ is nonempty, choose $g\notin U$. We claim its left coset $gL$ is wholly outside $U$. If $g\ell \in H_i$ for some $\ell \in L$, then $\ell \in H_i$ and closure under products and inverses would give $g=(g\ell ){\ell }^{-1}\in H_i$, a contradiction.

Thus $gL\subseteq G\setminus U$. Part (i) makes $gL$ infinite, so the complement is infinite. If there is no such $g$, the complement is empty. This proves the complement of a finite union of finite-index subgroups property: