Let $G$ be a group and $R$ be a commutative ring. Let $X$ be the set of all mappings $\phi : G \rightarrow R$ with $\phi(g) \neq 0$ for finitely many $g \in G$. For all $g \in G$ define $$(\phi_1 + \phi_2)(g) := \phi_1(g) + \phi_2(g)$$ $$(\phi_1 \cdot \phi_2)(g) := \sum_{h \in G} \phi_1(h) \phi_2(h^{-1}g)$$
Show that $(\phi_1 \cdot \phi_2) \in X \ \forall \phi_1,\phi_2 \in X$.
The problem for the existence seems to be the condition with the finitely many g, so that $\phi(g) \neq 0$. If I took some $\phi_1, \phi_2 \in X$, $(\phi_1 + \phi_2) \in X$, as the condition applies for $\phi_1$ and $\phi_2$, so I cannot generate a mapping with infinite mapped zeros from them.
But what about the multiplication? G is not necessarily finite. Can you please help me to go on?
Of course, for a fixed $g \in G,$ $(\Phi_1 \cdot\Phi_2)(g)$ is well defined. We have to show that $(\Phi_1 \cdot\Phi_2)(g) \neq 0$ for only finitely many $g \in G.$ To this end, put $$ M = \{h \in G | \Phi_1(h) \neq 0\} $$ and $$ N = \{h \in G | \Phi_2(h) \neq 0 \}. $$ Note that $M$ and $N$ are finite, by assumption. Also note that for $g,h \in G,$ we have $\Phi_1(h)\Phi_2(h^{-1}g) \neq 0$ iff ($h \in M$ and $h^{-1}g \in N$) iff ($h \in M$ and $h \in gN^{-1}$) iff $h \in M \cap gN^{-1}.$ From this, we see that $$ (\Phi_1\cdot\Phi_2)(g) \neq 0 \Longrightarrow M \cap gN^{-1} \neq \emptyset. $$ Now we make the following
Claim: $M \cap gN^{-1} \neq \emptyset$ holds only for finitely many $g \in G.$
We prove the claim by contradiction. Suppose there are inifinitely many $g \in G$ such that $M \cap gN^{-1} \neq \emptyset.$ Then we can pick sequences $m_i \in M$, $n_i \in N$, and $g_i \in G$ for $i \in \mathbb N$ such that $$ m_i = g_in_i^{-1} $$ and the $g_i$ are pairwise different. Since $M$ is finite, by passing to a subsequence, we can assume WLOG that $m_i$ is constant, i.e. there is an $m \in M$ such that $$ m = g_in_i^{-1}\qquad for\ i \in \mathbb N. $$ We can repeat this argument: since $N$ is finite, by passing to a subsequence, we can assume WLOG that $n_i$ is constant, i.e. there is an $n \in N$ such that $$ m = g_in^{-1}\qquad for\ i \in \mathbb N. $$ But now we get that $$ g_i = mn\qquad for\ i \in \mathbb N $$ is constant. This is a contradiction to the fact that the $g_i$ are pairwise distinct. So the claim must be true.
This in turn shows that $(\Phi_1\cdot\Phi_2)(g)\neq 0$ for only finitely many $g\in G,$ as desired.