Multiplicative group of prime order greater than 2 For any multiplicative group $\mathbb{Z}_n$, the order is $\phi(n)$. We know that $\phi(n)$ is always even for $n\ge3$.

Then how can I get a prime order multiplicative group.

Is the order of a multiplicative group means the multiplicatiive order of the generator element in that group?

In Builinear pairing, it is given that $e: \mathbb{G} \times \mathbb{G}_T$ , where both are of same prime order.

Please provide an example for $\mathbb{G}$ and $\mathbb{G}_T$.
I think you're confusing two senses of the phrase "multiplicative group". Often that phrase is used to mean any group, where the group operation is written using product notation: the identity is written $1$, inverses are written $x^{-1}$, products are written using concatenation $xy$, and powers are written using superscripted exponents like $xx=x^2$.


One particular kind of multiplicative group is the group of units modulo $n$, written $(\mathbb Z/n\mathbb Z)^*$. As you note, for $n\geq 3$, the order of this group is even. When discussing this group, one often emphasizes the word multiplicative to distinguish it from the additive group $\mathbb Z/n\mathbb Z$.
There are many other groups that are written multiplicatively. These include the multiplicative group of complex numbers $\mathbb C^*$ and its subgroups. As Adelafif notes in the comments, the group of third roots of unity is such a group of order $3$.


When a group is cyclic, yes, its order is the same as the order of any of its generator elements. When a group is not cyclic, it doesn't have a generator element, so of course this doesn't apply.
$Z_n$ is not a multiplicative group, inverse does not exist for 0. The order is not \phi(n), please dont get confused, multiplicative group means "a set together with multiplication operation becomes group".for example $U(n)=\{x \in N | x &lt; n$ and $gcd(x,n)=1\}$ is a multiplicative group.