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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. $12$
in Set Theory & Algebra
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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.

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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 $\left \{ a_{1},a_{2},a_{3} \right \}$

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

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
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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