If $G^{(i)}$ denotes the ith derived subgroup of $G$ then $G^{(i+1)}$ is a proper subgroup of $G^{(i)}$

Question

Is this statement true?

If $G^{(i)}$ denotes the ith derived subgroup of $G$ then $G^{(i+1)}$ is a proper subgroup of $G^{(i)}$ for all $i$.

I think this statement is false by the symmetric group.

Am I correct?

Answer 1

This is false in general. Take for example any perfect group. By definition it satisfies $G=[G,G]=G^{(1)}$ and hence there is always equality for all terms of the derived series.

For the symmetric group $S_n$ it depends on $n$. For example $S_3^{(1)}=A_3\cong C_3$, so that $S_3^{(2)}=1$ is trivial. But of course $S_n^{(2)}=[A_n,A_n]=A_n=S_n^{(1)}$ for all $n\ge 5$, since $A_n$ is perfect for all $n\ge 5$.

Answer 2

As @Dietrich Burde pointed out, this is not true for perfect groups, in fact if one of the (non-trivial) derived series terms coincides with the previous one, then from that point on you hit a perfect group.

If $G$ is finite and the whole derived series is proper indeed, then there must be an $i$, with $G^{(i)}=1$. This is equivalent to $G$ being solvable (e.g. $S_n, n \leq 4$). So, for solvable finite groups your statement is true.