Question #297042

[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.


1
Expert's answer
2022-02-17T10:06:17-0500

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 P3={I,(1  2),(2  3),(3  1),(1  2  3),(1  3  2)}P_3=\{I,(1\ \ 2), (2 \ \ 3), (3 \ \ 1), (1 \ \ 2\ \ 3), (1 \ \ 3\ \ 2)\}  where I\mathrm{I}  is the identity permutation.

There are six elements in the set P3 .

Let f1=I,f2=(12),f3=(23),f4=(31),f5=(123) and f6=(132)\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

f2f3=(12)(23)=(132)=f6f2f4=(12)(31)=(123)=f5f3f3=(23)(23)=f1, the identity permutation. f5f5=(123)(123)=(132)=f6 etc. \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

(G1)\left(G_{1}\right)  All entries in the table belong to P3P_{3}  and therefore P3P_{3}  is closed with respect to product of permutations.

(G2)\left(G_{2}\right) Product of permutations is an associative operation.

 (G3)\left(G_{3}\right) The identity permutation f1 is the identity of multiplication.

 (G4)\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 .

(G5)\left(G_{5}\right)  The product of permutations in P3 is not commutative as

f2f3=f6 and f3f2=f5 and f2f3f3f2f_{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.


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