A finite group $G\neq \{1\}$ whose only characteristic subgroups are $\{1\}$ and $G$ is either simple or a direct product of isomorphic simple groups. Okay I don't want an entire proof, but more of a clearifcation of this part of the proof that I am reading in the book.

Choose a normal subgroup H of G with $H\neq \{1\}$ and whose order is minimal among such subgroups. Write $H=H_1$ and consider all subgroups of the form $H_1\times H_2\times ...\times H_n$ $(n\geq 1)$, where $H_i\lhd G$ and $H_i\cong H$ for each $i;$ choose one such direct product of maximum order and call it M. Note that $M\lhd G$ for it is generated by normal subgroups of G....

I cannot for the life of me figure out why $M\lhd G$ or rather why $M\subset G$. Are they saying M is isomorphic to some normal subgroup of $G$? Another thing is the assumption that there is a list $H_1,H_2,...,H_n$ such that $M=H_1\times H_2\times ...\times H_n$ of maximum order. What stops me from just adding $H_{n+1}=H$ to the list and making the order larger?  
 A: I assume $M = H_1 \times H_2 \times \cdots \times H_n$ is intended to mean the internal direct product of these subgroups, so $M$ is a subgroup of $G$ rather than merely isomorphic to such a subgroup. In other words, $M = H_1 H_2 \cdots H_n$, where the product is direct.
At some point, if you try to add another $H_{n+1}$ to the product, it won't get you any more elements, in other words, $M = H_1 H_2 \cdots H_n$ will equal $M = H_1 H_2 \cdots H_n H_{n+1}$. Or, the proposed product $M = H_1 H_2 \cdots H_n H_{n+1}$ will not be direct so it won't qualify.
Finally, $M \lhd G$ because $gMg^{-1} = gH_1g^{-1}gH_2g^{-1}\cdots gH_{n}g^{-1}$.
A: In this case, the subgroups isomorphic to $H$ are all pairwise disjoint, since we would have $H \cap H^{\prime} \triangleleft G$, and $H$ has minimal order among the normal subgroups.  Also, you know that you have more than one normal subgroup isomorphic to $H$ since $H$ cannot be characteristic (so there is an automorphism of $G$ that does not fix $H$), but there are a set number of these.
The big product is formed by taking a collection of $H_{i}$ which pairwise commute (the collection may contain only $H_{1}$) (*thanks zibadawa timmy), that is, for every $h_{i} \in H_{i}$ and $h_{j} \in H_{j}$, $h_{i}h_{j} = h_{j}h_{i}$ (for every pair $(i,j)$ with $i \neq j$) so that we can form the internal direct product in a way that makes sense. This will let you make sense of the big product they give (essentially, it lets you think of $H_{i}H_{j} = \{h_{i}h_{j} \ |\  h_{i} \in H_{i},\ h_{j} \in H_{j}\}$ as the direct product $H_{i} \times H_{j}$).
