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 $G$.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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