ISI2019-MMA-19

Question

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

• A. $1$
• B. $2$
• C. $6$
• D. $12$

Answer 1

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.

Comments

@Shikha Mallick

what about this?

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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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Your answer is partially correct. For the full answer, see my answer below.

Answer 2

If the order of group is X and order of element is Y then X mod Y =0  always.

so let the element of group be “g” such that its order is “y” i.e. g^y = identity

thus 12 mod y=0,  Now “y” cannot be “1” as it means that the element is identity, thus another smallest value is “2”

Thus option B is correct

Comments

Your argument is partially incorrect, what you are saying is that there will be an element of order 2  because 2 divides 12, you are apply the converse of the statement. And secondly, what about 6 ? Why can't it be ? After all 6 is also a divisor of 12.

1 is reserved for identity element, so it is exempted  and now we are left with 11 elements, we can pair an element with its inverse , because it is commutative ( by shifting the brackets as required) ,and eventually everyone will be paired up except 1, that guy will be it's own inverse.  So the order is 2

12 can't be order of an element because then it would mean the group is cyclic and an Abelian group need not cyclic.

 

Answer 3

Answer: A, B (both)

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 Since it is given that $G$ is an abelian group of order $12$. It implies that $G\cong \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3 $ or $G\cong\mathbb{Z}_4\times \mathbb{Z}_3$.

In case of $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3$ the product has order $1$

In case of $\mathbb{Z}_4\times \mathbb{Z}_3$ the product has order $2$