A direct factor $A \le G$ is a subgroup for which there exists some $B \le G$ such that $G = A \times B$. If $G = N_1 \times \ldots \times N_k$ and $K \le G$ is a simple subgroup and a direct factor, then $K \cong N_j$ for some $i = 1,\ldots, k$. This could be seen by the Jordan-Hölder theorem, as the $N_i$ are the composition factors in some series, and the simple direct factors also.

Do you know an example for a simple and normal subgroup $K \le G$, but such that $K$ is not isomorphic to any $N_j$ (and hence $K$ could not be a direct factor)? As shown above the condition that it is a direct factor is sufficent, but is it also necessary?

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    $\begingroup$ Yes (to the title question). This also follows from the Jordan-Holder theorem. $\endgroup$ – Qiaochu Yuan Jan 22 '16 at 18:43
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    $\begingroup$ I don't see where you use the assumption that $K$ is a direct factor in your proof. Doesn't the Jordan-Hölder theorem imply the result anyway? $\endgroup$ – Pierre-Guy Plamondon Jan 22 '16 at 18:43
  • $\begingroup$ Okay, I see, start from $1 \unlhd K \unlhd G$ and refine to a composition series. I somehow thought that I need $K$ to be a direct factor, guess I was just confused. $\endgroup$ – StefanH Jan 22 '16 at 18:50

If all the $N_i$ are simple nonabelian then any simple normal subgroup $K$ must be equal to one of the direct factors: Take a direct factor $N$ such that $K$ projects with nontrivial image on $N$. Then the image of $K$ is a normal subgroup, so $K$ projects onto $N$ and (as elements of a direct product) every nontrivial element of $K$ has a nontrivial $N$-component. If $K\not=N$, there will be an element $n\cdot x\in K$ with $n\in N$, and $x$ in the direct product of the other factors. Pick $m\in N$ which does not commute with $n$. Then $a=nx/(nx)^m=n/n^m\not=1$ (as $x$ commutes with $m$), $a\in N$ and (normality) $a\in K$. But then $N\le K$.

  • $\begingroup$ $C_2\times C_2$ with the diagonal subgroup $D$ isn't a counterexample though, as $D\cong C_2$. $\endgroup$ – Stahl Jan 23 '16 at 1:24
  • $\begingroup$ @Stahl --You're right -- I somehow read it as equality. $\endgroup$ – ahulpke Jan 23 '16 at 1:28
  • $\begingroup$ Thanks for providing this stronger statement. But I guess in the last line it should read "$K \le N$", not "$N \le K$", as this was already established and you use $1 \le N \cap K \unlhd K$ and the simplicity of $K$ here. $\endgroup$ – StefanH Jan 23 '16 at 10:36
  • $\begingroup$ @Stefan I have $N\le K$ because of the normality of $K$. The containment was not yet established. $K$ projects onto $N$, but that does not mean (consider the vector (1,1) projecting on the first component) that $N\le K$. $\endgroup$ – ahulpke Jan 23 '16 at 16:37
  • $\begingroup$ A guess I see, you use $1 \ne N \cap K \unlhd N$, hence $N \cap K = N$, which gives $N \unlhd K$ and then by simplicity $N = K$. $\endgroup$ – StefanH Jan 23 '16 at 16:47

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