The latter table is for group of order 4. We point out that if we consider only first row and first column, we obtain a table for the group of order 1 and if we remain first two rows and columns we get a table of order 2. It is not possible to obtain a table for a group of order 3 from the table for the group of order 4 (see Lagrange’s theorem). Therefore we present a table for a group of third order separately:
2.(i) Lagrange’s theorem.
For every finite group the order of every subgroup divides the order of .
(ii) It is enough to prove that groups of order 3 do not exist. Suppose is a subgroup of order 3 with elements . Let us try to extend the group operation on element . Suppose that , where x is from the set of elements We multiply by and receive that . Thus, has to belong to . It shows that it is not possible to extend the group property from 3 to 4 elements. It completes the proof.
(iii) Due to the Lagrange's theorem, a group of order 6 can not be a subgroup, since 6 does not divide 13.
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