Question: Show that if G is the internal direct product of H$_1$ H$_2$, . . . , H$_n$ and i $\neq$ j with 1 $\leq$ i $\leq$ n, 1 $\leq$ j $\leq$ n, then H$_i$ $\cap$ H$_j$ = {e}.
Def of internal direct product from course: Let H$_1$, H$_2$, . . . , H$_n$ be a finite collection of normal subgroups of G. We say that G is the internal direct product of H$_1$, H$_2$, . . . , H$_n$ and write
G = H$_1$ $\times$ H$_2$ $\times$ H$_n$, if
- G = H$_1$H$_2$...H$_n$ = {h$_1$h$_2$ ... h$_n$ | h$_i$ $\in$ H$_i$},
- (H$_1$H$_2$ ... H$_i$) $\cap$ H$_{i+1}$ = {e} for i= 1,2, ..., n-1
I approached this problem by trying to find the elements in H$_i$ $\cap$ H$_j$ and show that it is only the identity. This is what I tried so far but I am stuck.
Let h$_i$ $\in$ H$_i$ and h$_j$ $\in$ H$_j$, then h$_i^{-1}$ $\in$ H$_i$ and h$_j^{-1}$ $\in$ H$_j$
Since H$_i$ and H$_j$ are normal, h$_j$h$_i^{-1}$h$_j^{-1}$ $\in$ H$_i$ and h$_i$h$_j$h$_i^{-1}$ $\in$ H$_j$
Thus h$_i$h$_j$h$_i^{-1}$h$_j^{-1}$ $\in$ H$_i$ and h$_i$h$_j$h$_i^{-1}$h$_j^{-1}$ $\in$ H$_j$
So h$_i$h$_j$h$_i^{-1}$h$_j^{-1}$ $\in$ H$_i$ $\cap$ H$_j$
i want to show that h$_i$h$_j$h$_i^{-1}$h$_j^{-1}$ = e.
I'm not sure what to do next, or if this is even the correct method to approach it.
If $a \in H_i \cap H_j \text{ and } a \neq e $ then $a \in H_1H_2\ldots H_i\ldots H_{j-1}$ so $H_1H_2\ldots H_i\ldots H_{j-1} \cap H_j \neq \{e\}$, contradiction.