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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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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

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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 $2$

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