Answer to Question #120777 in Abstract Algebra for Roshanthi

Question

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.

Answer 1

1. We denote by $*$ an operation of the group $G$ . We point out that the group can not have elements of order 3. I.e., if $b^3=e$ then the operation is wrongly defined. We point out that the general number of possible definitions of operation $*$ on the group $G$ is less than 9. As an example, $*$ can be defined in the following way:

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

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:

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

2.(i) Lagrange’s theorem.

For every finite group $G$ the order of every subgroup $H$ divides the order of $G$ .

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