Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $G$.

share|improve this question
2  
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
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.