I got stuck on this syllabus about abstract algebra:
Let $J$ be a set and $\{ \pi_J \ : \ j \in J\}$ be a collection of profinite groups. Show that $\prod_{j \in J} \pi_j$ is again profinite.
Own attempts
For any $j \in J$ we have a projective with its projective limit equal to $\pi_j$. The directed system is is composed of a directed poset $I_j$, a collection of groups $\{ G^j_i \ : \ j \in J\}$, and a collection of maps $\{ f^j_{lk} \ : \ G^i_l \to G^i_k \ | \ l,k \in I_j \text{ and } l \geq k \}$.
We thought of interchanging the projective limit and the product: $$ \prod_{j \in J} \pi_j \ = \ \prod_{j \in J} \varprojlim_{i \in I_j} G_i^j \ ``=" \ \varprojlim_{i \in I_j} \prod_{j \in J} G_i^j $$ We wanted to justify this interchange using the universal property. However, the interchange as presented above is meaningless. We should create a new projective system to make this work. our first thought was to define $I := \sqcup_{j \in J} I_j$, with the obvious partial ordering, but that set is not directed.
I think we just need to apply a convenient operation to the sets $\{I_j : j \in J \}$ to obtain the right limit. Is this right? And if so, could you tell me what kind of operation I could think of?
You can try the topological approach, show that the product is compact Hausdorff totally disconnected. (note that a product of compact sets is compact is equivalent to the axiom of choice so even if you choose the alebraic approach you will most likely need to use the axiom of choice).