+1 vote
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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. $1$
2. $2$
3. $6$
4. $12$
edited ago | 162 views

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.
selected
0
will it not be 6??

It is told summation of all elements
0
$\prod_{i=1}^{12}a_i$ is product symbol. How summation will be 6? Didn't understand. :)
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 $\left \{ a_{1},a_{2},a_{3} \right \}$

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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 \}........$$\left \{ a_{1},a_{2},a_{3},............,a_{12} \right \} So, order will be 1,2,3,6,12 right?? 0 Shouldn't the product be a_1.a_2.a_3...a_{12}? I don't understand how you are calculating the product. Explain a bit more. 0 0 Why did you write this:$$a_1.\{a_1, a_2\}.\{a_1, a_2, a_3\}.....\{a_1, a_2, a_3, ..., a_{12}\}$$0 U mean$G$same as$\prod a_{i} ??$answer will be$12$right? 0 @Shikha Mallick what about this? 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.

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