I am new to group theory and I have come across the theorem
The orders of all classes must be integral factors of the order of the group.
on page 15 of Chemical Applications of Group Theory (3rd edition) by F. Cotton (I'm not a math major), which doesn't give a proof but states that a proof is "similar to that used in connection with the orders of subgroups".
On that the order of any subgroup divides the order of the group, in pages 12–13 the book provides this proof:
- Suppose that the set of $g$ elements $A_1, A_2, A_3, \ldots, A_g$ forms a subgroup.
- Now take another element $B$ in the group but not in the subgroup, and form all of its products with elements in the subgroups: $BA_1, BA_2, BA_3, \ldots, BA_g$.
- None of these products can be in the subgroup. Proof by contradiction: if $BA_i = A_j$, then $BA_iA_i^{-1} = B = A_jA_i^{-1}$ and thus $B$ is in the subgroup by closure, contradicting that $B$ is not in the subgroup.
- These elements are also different, therefore there are now at least $2g$ elements in the group. If we can find another element that is not in $A_i$ nor $BA_i$, then there must be $3g$ elements in the group, and so on. Thus there are integral multiples of $g$ elements in the group. QED
I tried to formulate a similar proof for the theorem on classes. Using the same notation as above (except the $A$'s are now elements of the class), I could prove $A_i^{-1}BA_i$ is not in the class, but then all the different similarity transforms of $B$ are not necessarily the same. Also, $BA_i$ can be not in the class because classes are not necessarily closed under multiplication. Thus, I cannot proceed with Step 4 as above.
I've found proofs using the class equation, which I've not studied yet. But is it possible to formulate a simple proof as described above? (Or, did I interpret anything wrong/missed out anything?)