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Here is the question from I.N. Herstein book.

enter image description here

Now point (a) is pretty straight forward. For part (b) I took inspiration from group $Q_8$ and defined the group as $<r,s|r^p=1,s^q=1, rqr^{-1}=q^2 >$. Is this correct?

Now question says there is only one such group but I can change $rqr^{-1}=q^2$ to $rqr^{-1}=q^3$ or any other power of $q$ except $1$. Am I right or I am missing something?

Arturo Magidin
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Arun
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    Do not use images, especially for something like this text. Why? (1) Images are not searchable, so people searching for, for example, $p\nmid (q-1)$ would not be able to find this post. (2) Images are not accessible: people who use screenreaders cannot read your post. (3) Images do not adapt to different displays; your post may appear with images that are too large, too small, or illegible, depending on the display. This image, in particular, is just text and very light math formulas. You can certainly type its content. Do so. – Arturo Magidin Jul 04 '21 at 23:28
  • It makes no sense to say $rqr^{-1} = q^2$ when $q$ is a prime number. You probably meant $rsr^{-1} = s^2$. In any case, please follow Arturo's advice and write up the problem completely in your post instead of relying on an image scan from Herstein's book. In Herstein's problem, "unique" means "unique up to isomorphism" and it's correct. Consider $S_3$, $D_3$, and ${\rm GL}_2(\mathbf Z/(3))$. These are non-abelian of order $2 \cdot 3$ and $2 \mid (3-1)$ but they are isomorphic, so this doesn't contradict the intended meaning: a non-abelian group of order $6$ is unique up to isomorphism. – KCd Jul 05 '21 at 02:09

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