Let $G$ be a dihedral group of order $2n$ where $n\geq 1$, denoted by $D_n$. We know that $G$ is nilpotent if and only if $n=2^i$ for all $i\geq 1$, a proof of this you can check in the below link 1. Thus, $G$ is nilpotent of fitting length 1 (refer to the link 2 for the definition of fitting length of a group). Now if $n\neq 2^i$ where $i\geq 1$, then the fitting length of $D_n$ is 2 since we have a fitting series $1\subset R \subset G$ where $1$ is the trivial subgroup and $R$ is the rotation group of $G$ and it is clearly of length 2. Is this correct? or there is any mistake?
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I think that's right. – kabenyuk Dec 17 '22 at 14:19
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1The question is exactly Example $3.6$ in Conrad's notes. He gives more details, why the Fitting length is $2$, if $|D_n|$ is not a power of $2$. – Dietrich Burde Dec 17 '22 at 14:32
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Thank you for the reference! It confirms my observation. – Kavita Dec 17 '22 at 15:03