# if $G$ is finite group then polycyclic group is equivalent to super solvable group?

I don't know why this is true? can you help me:

if $G$ is finite group then polycyclic group is equivalent to super solvable group

Definitions-

## Polycyclic group

$G$ is a polycyclic if has a subnormal series s.t. all of factors are cyclic.

## Supersolvable

$G$ is a super solvable if has a normal series s.t. all of factors are cyclic.

• Supersolvable groups are always polycyclic is clear, and note that for a finite group, being solvable is equivalent to being a Polycyclic group . – Bhaskar Vashishth Jun 5 '15 at 18:36
• $A_4$ is polycyclic but nor supersolvable,so what you wrote is not true. – Derek Holt Jun 6 '15 at 9:00

Let $G$ be a finite polycyclic group then $G$ is solvable.

Proof: Let $1\leq N_1 \leq ... \leq N_k=G$ be the subnormal series.

Use induction on $k$,

for $k=1$, it is clear. Assume it is true for $k-1$. Then by assumption $N_{k-1}$ is supersolvable. As $G/N_{k-1}$ is cyclic then $G$ is solvable.

Thus, for finite groups two defination is equivalent.(Being solvable and being polycyclic.) (The othe case is obvious as mentioned in the comments)

• No. Normal subgroup is not a transitive property. The alternating group on 4 symbols is polycyclic but not supersolvable. – ahulpke Jun 5 '15 at 21:43
• Oh sorry, $N$ is supersolvable and $G/N$ is super solvable does not implies $G$ is super solvable. This is the wrong assumtion in my proof. – mesel Jun 5 '15 at 22:22
• @ahulpke : Being subnormal is a transitive , there is no problem in that step. I edited the answer. Thank you. – mesel Jun 5 '15 at 22:25
• @mesel:Is it a counter example?we know that $A_4$is polycyclic but $A_4$ is not super solvable!and $A_4$ is finite – pink floyd Jun 7 '15 at 14:54
• @parisa: After editing, I changed the statement. I claimed that "G is solvable". $A_4$ is solvable hence it is not a counter example. – mesel Jun 7 '15 at 15:56