1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Yes (to the title question). This also follows from the Jordan-Holder theorem. $\endgroup$ – Qiaochu Yuan Jan 22 '16 at 18:43
  • 1
    $\begingroup$ 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
2
$\begingroup$

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 $N\le K$.

$\endgroup$
  • $\begingroup$ $C_2\times C_2$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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