Show the dual space is a reprensentation of a finite group G.
attempt: Suppose $V^{*} = {\{\phi^*: V → \mathbb{K}}\}$ , where $V^*$ is the set of all linear functions called, functionals. Then the dual space is $V^*$
Then let $\phi:G → GL(V)$ be a linear representation of $ V$. Then let $\phi^* : V → \mathbb{K} \in V^*$. Then we need to show the group action of $G$ on $V$.
$g \dot\ \phi^* : V → \mathbb{K}$ . defined by $g \dot\ \phi^*(v) = \phi^*(g^{-1} \dot\ v)$.
Let $g_1, g_2 \in G, v\in V$ , then
$(g_1g_2 \dot\ \phi^*) = \phi^*((g_1g_2^{-1})v) = \phi^*(g_2^{-1}g_1^{-1} \dot\ v) = \phi^*(g_2^{-1} \dot\ (g_1^{-1} \dot\ v)$ .
And $(g_1(g_2\phi^*)(v) = (g_2\phi^*)(g_1^{-1}v) = \phi(g_2^{-1}g_1^{-1}v)$
I am not sure how I have to prove the dual space is a representation. Or if I am even setting up the problem correctly. Can someone please help me?
If I understand what you are trying to do, you are trying to show that the dual representation of a representation $\rho:G \to \operatorname{GL}(V)$ really is a group homomorphism $\rho^*:G \to \operatorname{GL}(V^*)$. (Defined by $\rho^*_g(\phi^*)(v) = \phi^*(\rho_{g^{-1}}(v))$.) Indeed we have $$\rho^*_{g_1 g_2}(\phi^*)(v) = \phi^*(\rho_{g_2^{-1}g_1^{-1}}(v)) = \phi^*(\rho_{g_2^{-1}}(\rho_{g_1^{-1}}(v))) = \rho^*_{g_1}(\rho^*_{g_2}(\phi^*))(v).$$