The internal n-fold direct product theorem says:
A group $G$ is a $n$-fold product if and only if it has $n$ normal
subgroups $H_1,\ldots,H_n$ such that $G = H_1 \ldots H_n$ and $H_i
\cap (H_1 \ldots H_{i-1} H_{i+1} \ldots H_n) = \{1\}$.
Why is the independency condition for the subgroups written as $H_i \cap (H_1 \ldots H_{i-1} H_{i+1} \ldots H_n) = \{1\}$ rather than $H_1 \cap H_2 \ldots \cap H_n = \{1\}$? Is the latter condition stronger than the former? How? Could someone give me some examples? It's not really clear to me.
Best Answer
To see the conditions are not the same, consider the Klein $4$-group $G=C_2\times C_2$, and take $$\begin{align*} H_1 &= \{(1,1), (x,1)\}\\ H_2 &= \{(1,1), (1,x)\}\\ H_3 &= \{(1,1), (x,x)\} \end{align*}$$ where $x$ is the generator of the cyclic group of order $2$.
Note that $H_iH_j=G$ whenever $i\neq j$, so you never have $H_k\cap (H_iH_j)=\{(1,1)\}$ if $i\neq j$. However, $H_1\cap H_2\cap H_3 = \{(1,1)\}$.
Note that here the latter condition holds but the former does not, the latter cannot be stronger than the former.
To see that in fact the latter is weaker than the former, let us verify that the former condition implies the latter. That is: if $H_i\cap(H_1\cdots H_{i-1}H_{i+1}H_n = \{1\}$, then $H_1\cap H_2\cap\cdots \cap H_n=\{1\}$. This follows because each $H_j$ with $j\neq i$ is contained in the product $H_1\cdots H_{i-1}H_{i+1}\cdots H_n$ (by taking the identity in each of the other factors). So $$H_1\cap\cdots\cap H_n\subseteq H_1\cap H_2\subseteq H_1\cap(H_2\cdots H_n)=\{1\}.$$ Thus the former condition implies the latter, and so the former is stronger than the latter.