I have seen it claimed online that:

Given two groups $G = \langle X \mid R \rangle$ and $H = \langle Y \mid S \rangle$ with some action $\phi\colon H \to \text{Aut}(G)$, then $$ G\rtimes_\phi H \;=\; \langle X, Y \mid R,\,S,\,yxy^{-1}=\phi(y)(x)\text{ for all }x\in X\text{ and }y\in Y\rangle. $$

Specifically I have seen it on stack exchange, and on group props, however neither of these sources have a proof, or any references.

I have searched all over for a references to a book or paper, however I have yet to find any.

To be more clear: I am looking for proof that the group formed by the semidirect product $G\rtimes_\phi H$ is isomorphic to the group given by the presentation $$\langle X, Y \mid R,\,S,\,yxy^{-1}=\phi(y)(x)\text{ for all }x\in X\text{ and }y\in Y\rangle.$$

Does anyone have any ideas? Thanks!

  • $\begingroup$ Hmm, not sure of a reference, as it is just immediate from the definition. $\endgroup$ – Tobias Kildetoft Feb 5 '16 at 22:03
  • $\begingroup$ What do you mean by a proof? A proof that it is a group? That is a straightforward exercise. $\endgroup$ – David Hill Feb 5 '16 at 22:03
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    $\begingroup$ I understand that the presentation defines a group upto isomorphism, however how can one show that group is isomorphic to the semi direct product? $\endgroup$ – Chris Feb 5 '16 at 22:10
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    $\begingroup$ It is proved in Section 10.2 of the book "Presentations of Groups" by D.L. Johnson, The result proved there is more general and covers all extensions, not just semidirect products. But the proof is not exactly difficult! $\endgroup$ – Derek Holt Feb 5 '16 at 22:42
  • $\begingroup$ @DerekHolt: I can find neither Section 10.2 nor that presentation in Section 10 of that book. $\endgroup$ – Shaun Jan 12 '18 at 17:22

Derek Holt commented exactly what I was looking for:

Section 10.2 of the book "Presentations of Groups" by D.L. Johnson


  • $\begingroup$ There is no Section 10.2, at least not in my copy of the book. $\endgroup$ – Shaun Jan 12 '18 at 16:52
  • $\begingroup$ My copy is from the 1970s, that's why; it's "Presentation of Groups" instead of "Presentations of Groups". $\endgroup$ – Shaun Jan 17 '18 at 13:17
  • $\begingroup$ More specifically, it is the corollary 1 in the page 140 of the book $\endgroup$ – No One Oct 4 '18 at 19:53

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