Prove that G = { 1,2,3,4,5,6} is a finite abelian group of order 6 multiplication module 7.

**Solution:** Algebraic system $\rightarrow$ $(G_1 \times 7)$

**1] Closure Axiam:**

V $a_1$ b E G If $AX_7$ b E G then $(G_1 \times \ 7)$ is closed.

prepare composition table for ( $G_1 \times \ 7)$

$X_7$ | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|

1 | 1 | 2 | 3 | 4 | 5 | 6 |

2 | 2 | 4 | 6 | 1 | 3 | 5 |

3 | 3 | 6 | 2 | 5 | 1 | 4 |

4 | 4 | 1 | 5 | 2 | 6 | 3 |

5 | 5 | 3 | 1 | 6 | 4 | 2 |

6 | 6 | 5 | 4 | 3 | 2 | 1 |

From above composition table $(G_1 \times 7)$ is closed.

**2] Associative prop:**

V a,b,c E G if $a \times 7(b X_7 \ c) = (a X_7 \ b) X_7 \ C$

then $(G_1 X_7)$ is associative.

Let a = 2, b = 3, c = 5

$2X_7 (3 X_7 \ 5) = (2 X_7 \ 3) X_7 \ 5$

$2 X_7^1 = 6 \times 5$

2 = 2

$\therefore$ $(G_1 X_7)$ is associative.

**3] Identity element property:**

V a E G 7 an ele e E G such that

$a X_7^e = a = e x_7 \ a$

$\therefore$ e = 1 E G

$\therefore$ Identity element e=1 exist in $(G_1 X_7)$

**4] Inverse element property:**

V a E G 7 an ele $\alpha$ E G such that

$a x_7 \alpha = e = \alpha x_7 \ a$

$\therefore$ $1^{-1} = 1$ $4^{-1} = 2$

$2^{-1} = 4$ $5^{-1} = 3$

$3^{-1} = 5$ $6^{-1} = 6$

$\therefore$ Inverse element exist in $(G_1 X_7)$

$\therefore$ $(G_1 X_7)$ is a group.

**5] Commutative property:**

$V \ a_1$ b E G if a $x_7$ b = b $x_7$ a then

$(G_1 X_7)$ is commutative.

say a = 3 , b = 4

AX7b = bX7a

3$x_7^4=4 x_7 ^3$

5 = 5

$\therefore$ $(G_1 X_7)$ is commutative.

$\therefore$ $(G_1 X_7)$ is an abelian group