I want to understand the concept of groups that have a finite index intuitively.
Suppose $H$ and $K$ are infinite subgroups, $H \subset K$, and $[ K:H ]$ is finite.
Intuitivelly, does this mean there are a finite set of elements in $H$ that are not in $K$? Since each element of $H$ not in $K$ will map to a coset of $K/H$?
Or does it mean there could be infinite elements in $H$ that are not in $K$, but quotient map $ K \rightarrow K/H$ is not injective, and there are only a finite number of cosets?
Best Answer
When $H$ is a subgroup (not necessarily a normal subgroup) of $K$, we can form a partition of $K$ by left cosets $kH = \{kh \mid h\in H \}$ using elements $k\in K$. Counting the number of distinct such left cosets gives us the index $[K:H]$ of $H$ in $K$.
[The Reader is encouraged to prove that two such left cosets of $kH$ and $k'H$ are either equal or disjoint, which is why they partition $K$. The same applies to right cosets, and the Reader is encouraged to find a bijection between these left cosets and right cosets, so that the count $[K:H]$ is the same whether we use one or the other choice to partition $K$.]
What it would mean for $n=[K:H]$ to be finite is that there are $n$ elements $k_1,\ldots,k_n \in K$ such that $K$ is the disjoint union $k_1H \cup \ldots \cup k_nH$.
Note that one of these parts must be the coset $H$, distinguished by being the coset that contains the identity $e$ of $K$ (and of $H$). One easily shows that $H$ is a normal subgroup of $K$ exactly when each left coset is also a right coset of $H$. It follows that subgroups of index $2$ will always be normal subgroups.