I want to prove the following:
Let $G$ be a finite group, $ V $ be a finite dimensional vector space over $\Bbb C$ and $V^{**}$ be the double dual of $V$. Then $(\pi,V) $ is an irreducible representation of $G$$\iff$ $(\pi^{**},V^{**})$ is an irreducible representation of $G$.
In the above statement $(\pi^{**},V^{**})$ is the usual representation induced by $(\pi,V)$ which can be obtained as follows:
Firstly, let $\pi^{*}:G\to Aut(V^*)$ defined by $(\pi^*(g).\lambda)(v)=\lambda(\pi(g)^{-1}v)$.
From this we get a representation $(\pi^*,V^*)$ of $ G$.
Next, let $\pi^{**}:G\to Aut(V^{**})$ defined by $(\pi^{**}(g).\psi)(f)=\psi(\pi^*(g)^{-1}f)$.
So we get the representation $(\pi^{**},V^{**})$ of $ G$.
Now coming to the actual problem, I was able to show that $(\pi,V)$ and $(\pi^{**},V^{**})$ are both equivalent representations of $G$ via the bijective intertwining operator given by, $$\eta:V\to V^{**}$$
$$v\mapsto\eta_v$$
where, $$\eta_v:V^*\to \Bbb C$$
$$f\mapsto f(v)$$
I have no clue how to proceed from here.
Hints Please!
This equivalence you showed also preserve subreps. So one of them is reducible, it means that the other is reducible as well and vice versa. Therefore, irred. iff irred. is also concluded.