Is 1 always an element in multiplicative group? Let  $\mathbb{G}_T$ be a multiplicative group.
Is 1 $\in \mathbb{G}_T$ ?
I think its true, because if $a \in \mathbb{G}_T$ then $a^{-1} \in \mathbb{G}_T$, So $aa^{-1} =1  \in \mathbb{G}_T$. 
Is the above argument true?
 A: The axioms for a group $G$ are


*

*Associativity

*The existence of an element $e$ such that $ge = eg = g$ for all $g\in G$.

*For all $g\in G$ there is a $g^{-1}\in G$ such that $gg^{-1} = g^{-1}g = e$.


So the existence of a multiplicative identity is required by the axioms and isn't something that you have to prove.
A: It is always true that a group contains a neutral element. However, it is not necessarily "the" $1$ that is this element. Instead it is customary to denote the nuetral element in virtually any (multiplicatively written) group by the symbol $1$.
A: By definition every group $G$ contains an identity/neutral element, that is there is an element $h \in G $ such that $hg=gh = g$ for every $g \in G$.
It can be shown that this element is always unique. 
It is not uncommon to use the notation $1_G$ or just $1$ for this element when one uses multiplicative notation. 
However, this element, denoted $1$, is not necessarily related to the number $1$ in any way whatsoever. 
On your argument: you use at least that the group is non-empty. Indeed, if we were not  requiring the existence of a neutral element as an axiom the empty set would be a group. Moreover, note that $aa^{-1}$ is the neutral element. To define what an inverse is you need a neutral element to begin with. 
Let me add (based on a comment) that if your group is the subgroup of the multiplicative group of a finite field $F$ then, yes, this subgroup will always contain $1 \in F$, the multiplicative identity of $F$.
