Prove that the general linear group, GL(V) is a group If V is a vector space and GL(V) is the set of all linear transformations from V to V that are bijections, prove that GL(V) is a group with operation composition. 

I am out of practice with algebra, and perhaps this is too abstract for me, but isn't the fact that the linear transformations are bijections, isn't associativity and inverse proved? But to prove the existence of an identity element?
Let $T \in GL(V)$. We wish to show that $S := T^{-1}$ is a linear map.

Proof: Let $w, w' \in V$. Then, since $T$ is a bijection, there exists $v, v' \in V$ such that $w = T(v)$ and $w' = T(v')$. Then, $S(w) * S(w') = S(T(v)) * S(T(v')) = v*v' = S(T(v*v')) = S(T(v)*T(v')) = S(w*w')$.
Let $T \in GL(V)$. We wish to show that $S := T^{-1}$ is a linear map.

Proof: Let $w, w' \in V$. Then, since $T$ is a bijection, there exists $v, v' \in V$ such that $w = T(v)$ and $w' = T(v')$. Then, $S(w) * S(w') = S(T(v)) * S(T(v')) = v*v' = S(T(v*v')) = S(T(v)*T(v')) = S(w*w')$.