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?
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?
The axioms for a group $G$ are
So the existence of a multiplicative identity is required by the axioms and isn't something that you have to prove.
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 $1$.
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 $F$.