Let’s do an example. Take a subgroup \(H= \{(1), (1234), (13)(24), (1432)\}\) of \(S_4\text{.}\) What are the cosets of the subgroup?
Write each coset as a left coset: \(aH\) for \(a \in S_4\text{.}\) For example, \((123)H\text{.}\) What is this? Note that this is NOT the same set as the corresponding right coset.
Let \(G\) be a finite group and let \(H\) be a subgroup of \(G\text{.}\) Then \(|G|/|H| = [G : H]\) is the number of distinct left cosets of \(H\) in \(G\text{.}\) In particular, the number of elements in \(H\) must divide the number of elements in \(G\text{.}\)
A cool consequence of this theorem is that it also tells us something about elements. The order of an element \(a\) is the least \(k\) such that \(a^k = e\) (the identity). Since \(\langle a \rangle\) is a subgroup, the order of the subgroup must be a divisor of the size of the group. But the order of the cyclic subgroup is the same as the order of its generator. So orders of elements must also divide the size of the group.
In particular, suppose a group \(G\) has an element \(a\) such that \(a^7 = e\text{.}\) Now this means that either \(a = e\) or the order of \(a\) is 7. That means that \(G\) must have size that is a multiple of 7.
