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If we let $G$ be a group with $n$ subgroups $N_i$ such that

  1. $\prod_{i=1}^n N_i= G$

  2. $N_i \cap N_j = \{e\}$ for all $i \ne j$ s.t. $1 \le i < j \le n$

  3. $N_i \unlhd G$ for any $1 \le i \le n$

then $G$ is not necessarily an internal direct product of the $N_i$, for there exist counter-examples that show $G$ could still not be isomorphic to the external direct product $N_1 \times \ldots \times N_n$.

But I'm curious if the converse is true: that is, if $G$ is an internal direct product of $N_1, \ldots, N_n$, then does this imply either that (i) $N_i \cap N_j = \{e\}$ for all $1 \le i < j \le n$ or (ii) any $N_i \unlhd G$?

NOTE: I'm assuming that if $G = \prod_{i=1}^n N_i$ and each $g \in G$ has a unique representation of form $h_1 \cdot \ldots \cdot h_n$ s.t. $h_i \in N_i$, then $\prod_{i=1}^n N_i$ forms an internal direct product of $G$.

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I'm confused. How are you defining internal direct product? I would have defined it as 1-3. – Alex Youcis Dec 10 '12 at 19:26
I just added the definition of an internal direct product above. My (1)-(3) I believe is equivalent to that definition only if the number of the $N_i$ is 2. – user1770201 Dec 10 '12 at 19:42
@Alex Youcis: Let $G = \{1,g_1,g_2,g_3\}$ be a Klein 4-group, $N_i = \{1,g_i\}$ for $i=1,2,3$. These $N_i$ satisfy 1-3 above, but the group is not the direct product of $N_1$, $N_2$ and $N_3$ (although it is the direct product of any two of these). – Derek Holt Dec 10 '12 at 20:08
@DerekHolt Ah, I was being careless and not paying attention to the fact that it was more than two subgroups--of course you need that this is true, you need each of them to intersect the product of the rest trivially. – Alex Youcis Dec 10 '12 at 20:12
up vote 3 down vote accepted

Yes, $G$ is a direct product of the subgroups $N_i<G$ if and only if the following conditions are satisfied:

  1. $\prod_i N_i = G$,
  2. for any $i$, $N_i\cap \prod_{j\neq i} N_j = \{1\}$,
  3. $N_i\unlhd G$ for all $i$.

Note that the second condition is actually stronger than the requirement for the pairwise intersections to be trivial.

All this is a good exercise, and you should prove all of this.

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