Let $G=H_{1} \times H_{2} \dots \times H_{n}$ if and only if $ \; \forall g \in G$, $g= h_{1}h_{2}\ldots h_n$ where $h_{i} \in H_{i}$ is unique.

Question

Let $G=H_{1} \times H_{2} \dots \times H_{n}$ if and only if $ \; > \forall g \in G$, $g= h_{1}h_{2}\ldots h_n$ where $h_{i} \in H_{i}$ is unique. Here, $H_{i}\lhd G \; \forall i$

I have proved this for the case when we have only $2$ normal groups, $(n=2)$ but I am finding it difficult to prove the general result. I feel like I am missing some important intermediate results.

Answer 1

Hint: As you told you have done this for 2 normal subgroup then you can tackle other by mathematical induction.

Take base case for n=2 . which you had already proved

Assume this result for n such normal group direct product

then to prove for n+1 take assumed normal group as single piece normal subgroup and this problem reduces to 2 now