Can someone please help to understand and identify the following groups?

  1. $G_1 = \langle a, b \mid a^{p^2} = b^q = 1, b a b^{-1} = a^i, \operatorname{ord}_{p^2}(i) = q \rangle$.
  2. A family of groups given by, $$ \langle a, b, c \mid a^p = b^p = c^q = 1, c a c^{-1} = a^i, c b c^{-1} = b^{i^t}, ab = ba, \operatorname{ord}_p(i) = q \rangle $$ where $p, q$—distinct primes, $p > q$ and $q \mid p - 1$.

(Original image here.)

Can the first group be written as, $$ \mathbb{Z}_{p^2} \rtimes_\varphi \mathbb{Z}_q $$ and the second group as, $$ (\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes \mathbb{Z}_q. $$ What does $\operatorname{ord}_{p^2}(i) = q$, $\operatorname{ord}_p(i) = q$ mean?

  • $\begingroup$ $\text{ord}_{p}(i)=q$ means that the multiplicative order of $i$ modulo $p$ is $q$ $\endgroup$ – A. Goodier Aug 26 '18 at 10:05
  • $\begingroup$ The answers to your first two questions are both yes. $\endgroup$ – Derek Holt Aug 26 '18 at 10:09
  • $\begingroup$ Thanks. Can I regard them as an extension of the semidirect product between $Z_p$ and $Z_q$? $\endgroup$ – Buddhini Angelika Aug 26 '18 at 10:35
  • $\begingroup$ Can I say that the above two semidirect products are the direct products? $\endgroup$ – Buddhini Angelika Aug 26 '18 at 17:14
  • $\begingroup$ No they are not direct products. If they were direct products then they would be abelian groups, which they are not. $\endgroup$ – Derek Holt Aug 26 '18 at 18:59

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