1
$\begingroup$

Let $G$ be a group and $H$ be a simple non-abelian subgroup of $G$ which is ascendant in $G$. Is it true that $H$ is also subnormal in $G$?

Definition Let $G$ be a group and $H$ be a subgroup of $G$. Then we say that $H$ is ascendant in $G$ if we can find an ascendant (also of infinite length) normal series (not necessarily an invariant one) from $H$ to $G$.

$\endgroup$
  • $\begingroup$ What do you exactly mean by "ascendant"? $\endgroup$ – Nicky Hekster Jun 11 '14 at 13:12
  • $\begingroup$ I've specified that in the question now. I hope it is clear. $\endgroup$ – W4cc0 Jun 11 '14 at 13:14
  • $\begingroup$ Yes much better now. $\endgroup$ – Nicky Hekster Jun 11 '14 at 13:15
  • $\begingroup$ I can see no difference between the definitions of subnormal and ascendant. For subnormality the series has to be finite. Is the same true for ascendant? $\endgroup$ – Derek Holt Jun 11 '14 at 14:08
  • $\begingroup$ Yes, for ascendant the series can also be infinite. Is not correct saying "ascendant" to meant also of infinite length? $\endgroup$ – W4cc0 Jun 11 '14 at 14:11
2
$\begingroup$

Not true: take $H=A_5 \lt A_6 \lt G=A_7$. $A_5$ is not subnormal in $A_7$, since $A_7$ is simple.

$\endgroup$
  • $\begingroup$ Oh sorry maybe I used a wrong term. I meant "normal" as "normal in $G$". I know that someone used the term invariant for this. I've corrected the question. I hope now is really clear. $\endgroup$ – W4cc0 Jun 11 '14 at 13:24
  • 2
    $\begingroup$ OK! But now I am totally confused. If there is an ascending normal series then automatically $H$ is subnormal in $G$. $\endgroup$ – Nicky Hekster Jun 11 '14 at 13:27
  • $\begingroup$ Quoting from Wikipedia: The series may be infinite. If the series is finite, then the subgroup is subnormal. I hope this will "deconfues" you. $\endgroup$ – W4cc0 Jun 12 '14 at 12:08

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.