will it not be 6??

It is told summation of all elements

It is told summation of all elements

1 vote

Let $G =\{a_1,a_2, \dots ,a_{12}\}$ be an Abelian group of order $12$ . Then the order of the element $ ( \prod_{i=1}^{12} a_i)$ is

- $1$
- $2$
- $6$
- $12$

3 votes

Best answer

Every element is multiplied with its inverse. The identity element is its own inverse. That leaves us with $11$ elements other than identity. So, one element (except the identity element) is its own inverse. The other $10$ elements get cancelled out as they are multiplied with their respective inverse elements. So, the order of the element $\prod_{i=1}^{12}a_i$ is $2$.

Option $(B)$ is correct.

Option $(B)$ is correct.

0

yes, product

$i=1$, So elements are $a_{1}$

$i=2$, So elements are $\left \{ a_{1},a_{2} \right \}$

$i=3$, So elements are $\left \{ a_{1},a_{2},a_{3} \right \}$

.......................................

Now, $G=\left \{ a_{1},a_{2},a_{3},............,a_{12} \right \}$, here order of group is $12$, So order of it's subgroup

$1,2,3,6,12$

Now if we do product of these elements , it will be $a_{1}.$$\left \{ a_{1},a_{2} \right \}$$.\left \{ a_{1},a_{2},a_{3} \right \}........$$\left \{ a_{1},a_{2},a_{3},............,a_{12} \right \}$

So, order will be $1,2,3,6,12$

right??

$i=1$, So elements are $a_{1}$

$i=2$, So elements are $\left \{ a_{1},a_{2} \right \}$

$i=3$, So elements are $\left \{ a_{1},a_{2},a_{3} \right \}$

.......................................

Now, $G=\left \{ a_{1},a_{2},a_{3},............,a_{12} \right \}$, here order of group is $12$, So order of it's subgroup

$1,2,3,6,12$

Now if we do product of these elements , it will be $a_{1}.$$\left \{ a_{1},a_{2} \right \}$$.\left \{ a_{1},a_{2},a_{3} \right \}........$$\left \{ a_{1},a_{2},a_{3},............,a_{12} \right \}$

So, order will be $1,2,3,6,12$

right??

0

Shouldn't the product be $a_1.a_2.a_3...a_{12}$? I don't understand how you are calculating the product. Explain a bit more.

0

The order of an element is defined as "m" such that $a^{m}$ is equal to the identity.

So, the given product evaluates to some $(a')^{2}$ where $a' $ is the element which is its own inverse.

So that means that $(a')^{2}$ is equal to identity. Hence, order of $(a')$ is 2. Therefore, the order of the product is 2.

So, the given product evaluates to some $(a')^{2}$ where $a' $ is the element which is its own inverse.

So that means that $(a')^{2}$ is equal to identity. Hence, order of $(a')$ is 2. Therefore, the order of the product is 2.