Alternative definitions of "internal direct product"

Question

Let be $H_1,\cdots,H_n$ subgroups of the group $G$, then we say that $G$ is an internal direct product of $H_1,\cdots,H_n$ iff

1. $H_j\triangleleft G$ for $1\le j\le n$, and
2. $G=H_1H_2\cdots H_n$, and
3. $H_k\cap H_1H_2\cdots \hat{H_k}\cdots H_n=\{e\}$.

Can property (3) be replaced with the following property?

4. $H_j\cap H_k=\{e\}$ for all $j\ne k$.

I proved 1+2+3 imply 1+2+3'. Now I have to prove the converse.

By contradiction suppose that there is a $k$ and a $h_k\in H_k$ with $h_k\ne e$ and $h_k=h_1\cdots h_{k-1}h_{k+1}\cdots h_n$.

I dont't need necessarily a proof. To know that the 1+2+3' is a rightfully alternative would suffice.

EDIT It seams that it is not correct according to this post.

Could it be an alternative in case of an abelian group?

Answer 1

Your property (4) is strictly weaker than (3). For example, take $G = \Bbb{Z}\times \Bbb{Z}$ (pairs of integers) under pointwise addition and let us write the group operation as $+$: $(a, b) + (c, d) = (a + c, b + d)$. Now take $n = 3$ and take $H_1$, $H_2$ and $H_3$ to be the subgroups generated by $(1, 0)$, $(0, 1)$ and $(1, 1)$ respectively. Then your properties (1), (2) and (4) hold, but (3) does not: $H_3 \subseteq H_1 + H_2$.