Order of a product of subgroups.

Hi:
Let G be a finite group and let $H_1,H_2, \ldots H_n$ be subgroups of G such that:
(a) $G=H_1H_2 \ldots H_n$ but $G \neq H_1H_2 \ldots H_{i-1}H_{i+1} \ldots H_n$ for all $i$
(b) The $H_i$ have parewise trivial intersections.
(c) The orders of the $H_i$ are prime numbers.
Show that $|G| = |H_1||H_2| \ldots |H_n|$.

I have tried induction. So, $H_1\ldots H_{n+1} = HH_{n+1}$, where $H = H_1\ldots H_n$. Now, if I could prove that H is a subgroup of G, it is trivial that H, H1, ..., Hn satisfy (a), (b), (c). Therefore, $|H| = |H_1|\ldots |H_n|$. Furthermore,
using $|K_1K_2| = |K_1||K_2| / |K_1\cap K_2|$, if I could prove that $H \cap H_{n+1} = (e)$, (but this I think I could prove it using the fact that H is a subgroup), I would have $|G| = |H \ldots H_{n+1}| = |HH_{n+1}|= |H||H_{n+1}| = |H_1|\ldots |H_n||H_{n+1}|$.

But unfortunately, I cannot prove that H is a group. Any suggestion will be welcome. Thanks for reading.