Is the cardinality of a Group is equal to the number of cosets of one of it's subgroups?

Question

Assume I have $H \subseteq G$

It seems to me that since the right cosets of $H$ from a partition of $G$ then the cardinality of the set: $\{Hg_1, Hg_2,\dots, Hg_n\}$ (not necessarily finite)

Must be that of $G$.

is this correct?

Answer 1

You are right that the set of cosets of a subgroup $H$ of $G$ are a partition of $G$. However, two cosets $H x$ and $H y$ can be the same even if $x \not= y$ since we could have $x = h y$ where $h \in H$.

What is true in general is that the cardinality of $G$ is equal to the cardinality of $(G / H) \times H$ (where $G / H$ is the set of cosets of $H$ in $G$).

Answer 2

Except in special cases, no. The obvious special case is when $H = \{1\}$.

Here is an easy counter-example. Take $G = \mathbb{Z}$ and $H = 2\mathbb{Z}$. We have that $G/H \cong \mathbb{Z}_{2}$.

Note that as $G = \mathbb{Z}$ is abelian, the left and right cosets of $G$ with respect to $H$ are the same.

Answer 3

The mistake you are making is counting the whole subgroup $H$ as one element. Imagine you have many cartons and packed all your books in the cartons all having equal number of books. The number of books in your possession is not the number of cartons (Unless you pack one book in each carton, and this is the point in the answer by m10105).