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  1. I have seen authors taking $G_1=G_2=G_T=G$ to be the same group of prime order $q$.

  2. What I know is that for pairing of type $$e:G_1\times G_2\rightarrow G_T,$$ size of the element in the target group is $kn$ where $n$ is the size of an element in $G_1$ and $k$ is the embedding degree.

Source: A New Family of Pairing-Friendly elliptic curves by Michael Scott and Aurore Guillevic. and this question

I am confused as it looks like these two points are contradicting each other.

Shweta Aggrawal
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    Where is $G_1 = G_T$ stated? Generally $G_1$ and $G_2$ are subgroups of elliptic curves (and may be equal, i.e. a Type 1 pairing), and $G_T$ is a subgroup of the multiplicative group of a finite field, so $G_1 \neq G_T$. – meshcollider Apr 25 '22 at 03:24
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    @meshcollider I know of some proposals to use pairings on singular curves. In such cases, there is a natural isomorphism between the curve group and $\mathbb F_p^\times$ which can then be taken as $G_1$, $G_2$ and $G_T$. I don't know if this is what OP refers to. – Daniel S Apr 25 '22 at 05:54

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