Question #120777
1. Build up the operation tables for group G with orders 1, 2, 3 and 4 using the elements a, b, c, and e as the identity element in an appropriate way.
2. i. State the Lagrange’s theorem of group theory.
ii. For a subgroup H of a group G, prove the Lagrange’s theorem.
iii. Discuss whether a group H with order 6 can be a subgroup of a group with order 13 or not. Clearly state the reasons.
1
Expert's answer
2020-06-11T20:48:31-0400
  1. We denote by * an operation of the group GG . We point out that the group can not have elements of order 3. I.e., if b3=eb^3=e then the operation is wrongly defined. We point out that the general number of possible definitions of operation * on the group GG is less than 9. As an example, * can be defined in the following way:


eabceeabcaaecbbbcaeccbea\begin{matrix} * && e & a& b& c\\ \\ e &&e & a& b& c\\ a& & a & e&c& b\\ b && b & c& a& e\\ c && c& b& e& a\\ \end{matrix}

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:

eabeeabaabebbea\begin{matrix} * && e & a& b\\ \\ e &&e & a& b\\ a& & a & b&e\\ b && b & e& a\\ \end{matrix}


2.(i) Lagrange’s theorem.

For every finite group GG the order of every subgroup HH divides the order of GG .


(ii) It is enough to prove that groups of order 3 do not exist. Suppose HH is a subgroup of order 3 with elements e,a,be,a,b . Let us try to extend the group operation on element cc . Suppose that ca=xc*a=x , where x is from the set of elements e,a,b,c.e, a, b, c. We multiply by x1x^{-1} and receive that c=xa1c=x*a^{-1} . Thus, cc has to belong to HH . 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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