Let $G$ be a topological group, and $H$ be subgroup of $G$. Is it true that:
$H$ is open subgroup of $G$・・・①
is equivalent to:
$H$ is of finite index in $G$・・・②
I have seen ① implies ②, but I don't have confidence the converse holds. Only results and backgrounds are welcome.
No, it is not generally true.
A counter-example would be the following: let $\mathbb{Z}$ be the additive group of integers equipped with discrete topology. Then the trivial subgroup $\{0\}$ is open but with infinite index - the index of trivial subgroup in $\mathbb{Z}$ is the cardinality of $\mathbb{Z}$.
But if $G$ is compact, then $H$ being open does imply that $H$ is of finite index. This can be seen if you consider the cosets of $H$, which is a minimal open covering by translations (or cosets) of $H$ and must be finite.
Also, the converse is not true even if $G$ is compact. A conter-example is given by Dietrich in his comment: the profinite group $\widehat{\mathbb{Z}}^\times$ admits non-open finite index subgroup.
As noted in the comments, the result is not true in general.
It is easy to come up with groups which have open subgroups which are not of finite index. For example, any infinite discrete group works, and any product of two groups, one of which has an open subgroup of infinite index, also has the same property.
It is a bit more difficult to find profinite groups which have finite index subgroups which are not open.
For example, if $G=C_2^{\mathbf N}$ (a product of countably many copies of the cyclic group of order 2), then $V=G$ is a vector space over the two-element field of dimension continuum. It follows that the dual space $V^*$ has dimension (and hence also cardinality) $2^{\mathfrak c}$; kernels of distinct linear functionals are distinct subgroups of index 2 in $G$, so there are $2^{\mathfrak c}$ many such groups. On the other hand, $G$ is second countable as a topological space, so there are only continuum many open subsets, let alone open subgroups (in fact, I'm pretty sure there are only countably many open subgroups, but we do not even need that).
There are some positive results, though.
Firstly, if $G$ is a compact group, then every open subgroup has finite index. This is easy by compactness.
Finally, a celebrated result of Nikolov and Segal says that this equivalence actually is true if $G$ is both profinite and (topologically) finitely generated.
