When considering finite groups $G$ of order, $|G|=p^2qr$, where $p,q,r$ are distinct primes, let $F$ be a Fitting subgroup of $G$. Then $F$ and $G/F$ are both non-trivial and $G/F$ acts faithfully on $\bar{F}:=F/ \phi(F)$ so that no non-trivial normal subgroup of $G/F$ stabilizes a series through $\bar{F}$.

a) Case $|F|=p^2$ and $F \cong C_{p}^2$. Then $G=C_{p}^2 \rtimes H$ with $H$ of order $qr$ and $H$ embeds into $Aut(C_{p}^2)=GL(2,p)$.

b) Case $|F|=pr$. In this case $\phi(F)=1$ and $Aut(F)=C_{p-1} \times C_{r-1}$. Thus $G/F$ is abelian and $G/F \cong C_{p} \times C_{q}$.

Can some one please help clarify how to represent the above two groups using notation form? i.e. is it correct if I represent them using notations as follows?

a) $G \cong C_{p}^2 \rtimes (C_{q} \times C_{r})$ or $G \cong C_{p}^2 \rtimes (C_{q} \rtimes C_{r})$ or $G \cong C_{p}^2 \rtimes (C_{r} \rtimes C_{q})$

b) $G \cong (C_{p} \times C_{r}) \rtimes (C_{p} \times C_{q})$ or $G \cong (C_{p} \rtimes C_{r}) \rtimes (C_{p} \times C_{q})$ or $G \cong (C_{r} \rtimes C_{p}) \rtimes (C_{p} \times C_{q})$

Moreover there exists two nilpotent groups of order $p^2qr$. What are the representations for the two nilpotent groups of order $p^2qr$?

Please help me with these questions.

Thanks a lot in advance.


Semidirect products include direct products, but if you know for certain that something is a direct product then it is more helpful and informative to denote it as a direct product. So it is not wrong to denote it as a semidirect product but it is unhelpful and potentially misleading.

Since $C_{pq} \cong C_p \times C_q$ for coprime $p$ and $q$, you could use wither of those notations - they are both correct. So, there are multiple correct answers to your final question.

  • $\begingroup$ Thanks @DerekHolt But the meaning is not very clear to me. Is it possible to mention some of the possible groups? $\endgroup$ – Buddhini Angelika Mar 1 '19 at 17:01
  • $\begingroup$ And are the answers suggested for a) and b) by myself, correct? $\endgroup$ – Buddhini Angelika Mar 1 '19 at 17:02

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