Answer to Question #297042 in Abstract Algebra for John

Question

[DMe] Define group. Show that the set P3 of all permutations on three symbols 1,2,3 is a finite non-abelian group of order six with respect to permutation multiplication as composition.

Answer 1

Solution:

A group is a set G, combined with an operation *, such that:

1. The group contains an identity
2. The group contains inverses
3. The operation is associative
4. The group is closed under the operation.

Here $P_3=\{I,(1\ \ 2), (2 \ \ 3), (3 \ \ 1), (1 \ \ 2\ \ 3), (1 \ \ 3\ \ 2)\}$  where $\mathrm{I}$  is the identity permutation.

There are six elements in the set P₃ .

Let $\quad f_{1}=I, f_{2}=\left(\begin{array}{ll}1 & 2\end{array}\right), f_{3}=\left(\begin{array}{ll}2 & 3\end{array}\right), f_{4}=\left(\begin{array}{ll}3 & 1\end{array}\right), f_{5}=\left(\begin{array}{lll}1 & 2 & 3\end{array}\right)\ and\ f_{6}=\left(\begin{array}{lll}1 & 3 & 2\end{array}\right)$

Let us prepare a composition table for P₃ .

For completing the entries in the above table we have actually multiplied the permutations. Thus

$\begin{aligned} &f_{2} f_{3}=\left(\begin{array}{lll} 1 & 2 \end{array}\right)\left(\begin{array}{ll} 2 & 3 \end{array}\right)=\left(\begin{array}{lll} 1 & 3 & 2 \end{array}\right)=f_{6} \\ &f_{2} f_{4}=\left(\begin{array}{lll} 1 & 2 \end{array}\right)\left(\begin{array}{ll} 3 & 1 \end{array}\right)=\left(\begin{array}{lll} 1 & 2 & 3 \end{array}\right)=f_{5} \\ &f_{3} f_{3}=\left(\begin{array}{lll} 2 & 3 \end{array}\right)\left(\begin{array}{ll} 2 & 3 \end{array}\right)=f_{1}, \text { the identity permutation. } \\ &f_{5} f_{5}=\left(\begin{array}{llll} 1 & 2 & 3 \end{array}\right)\left(\begin{array}{lll} 1 & 2 & 3 \end{array}\right)=\left(\begin{array}{lll} 1 & 3 & 2 \end{array}\right)=f_{6} \text { etc. } \end{aligned}$

Now we observe that

$\left(G_{1}\right)$  All entries in the table belong to $P_{3}$  and therefore $P_{3}$  is closed with respect to product of permutations.

$\left(G_{2}\right)$ Product of permutations is an associative operation.

 $\left(G_{3}\right)$ The identity permutation f₁ is the identity of multiplication.

 $\left(G_{4}\right)$  Every element of P₃ possesses inverse in P₃ because

inverse of f₁ is f₁ ; inverse of f₂ is f₂ ; inverse of f₃ is f₃

inverse of f₄ is f₄ ; inverse of f₅ is f₅ ; and inverse of f₆ is f₆ .

$\left(G_{5}\right)$  The product of permutations in P₃ is not commutative as

$f_{2} f_{3}=f_{6}\ and\ f_{3} f_{2}=f_{5}\ and\ f_{2} f_{3} \neq f_{3} f_{2}$ .

Therefore P₃  is a finite non-abelian group of order six with respect to permutation multiplication.