LAGRANGE THEOREM

LAGRANGE THEOREM. GROUP 7. MKG3013.

GROUP MEMBERS. LEI WING YAN (S63165) NURIN IMAN NAJIHAH BINTI MD ADNAN (S62313 ) NUR LIYANA FARAIN BT MOHD NOR (S63150).

Lagrange Theorem statement. As per the statement, the order of the subgroup H divides the order of the group G. This can be represented as; |G| = |H| Before proving the Lagrange theorem, let us discuss the important terminologies and three lemmas that help to prove this theorem..

Cosets. 01. Proof. 02. Corollary. 03. What will we be discussing?.

What is coset?. 01.

If G is a finite group, and H is a subgroup of G, and if g is an element of G, then;.

02. What is lagrange theorem proof ?.

Let H be any subgroup of the order n of a finite group G of order m. Let us consider the cost breakdown of G related to H..

03. What is lagrange theorem corollary?.

Corollary 1. If G is a group of finite order m, then the order of any a∈G divides the order of G and in particular a m = e..

Whoa!. Proof corollary 1:. Let p be the order of a, which is the least positive integer, so, ap = e Then we can say, a, a2, a3, …., ap-1,ap = e, the elements of group G are all distinct and forms a subgroup. Since the subgroup is of order p, thus p the order of a divides the group G. So, we can write, m = np, where n is a positive integer. So, am = anp = (ap)n = e Hence, proved.

Proof corollary 2:. Let us consider, the prime order of the group G is m. Now, m has only two divisors 1 and m (prime numbers property). Therefore, the subgroups of G will be and G itself. So, there are no proper subgroups. Hence, proved..

Proof corollary 3:. Suppose, G is the group of prime order of m and a ≠ e∈G. Since the order of a is a divisor of m, it is either 1 or m. But the order of a, o(a) ≠ 1, since a ≠ e. Therefore, the order of o(a) = p, and the cyclic subgroup of G generated by a are also of order m. It proves that G is the same as the cyclic subgroup.

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