# ISI2019-MMA-19

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

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

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

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

$1,2,3,6,12$