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If $G$ and $H$ are groups with presentations $G=\langle X|R \rangle$ and $H=\langle Y| S \rangle$, then of course $G \times H$ has presentation $\langle X,Y | xy=yx \ \forall x \in X \ \text{and} \ y \in Y, R,S \rangle$. Given two group presentations $G=\langle X|R \rangle$ and $H=\langle Y| S \rangle$ and a homomorphism $\phi: H \rightarrow \operatorname{Aut}(G)$, what is a presentation for $G \rtimes H$? Is there a nice presentation, as in the direct product case? Thanks!

Shaun
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dave
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1 Answers1

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Let $G = \langle X \mid R\rangle$ and $H = \langle Y \mid S\rangle$, and let $\phi\colon H\to\mathrm{Aut}(G)$. Then the semidirect product $G\rtimes_\phi H$ has the following presentation: $$ G\rtimes_\phi H \;=\; \langle X, Y \mid R,\,S,\,yxy^{-1}=\phi(y)(x)\text{ for all }x\in X\text{ and }y\in Y\rangle $$ Note that this specializes to the presentation of the direct product in the case where $\phi$ is trivial.

 

For example, let $G = \langle x \mid x^n = 1\rangle$ be a cyclic group of order $n$, let $H = \langle y \mid y^2=1\rangle$ be a cyclic group of order two, and let $\phi\colon H \to \mathrm{Aut}(G)$ be the homomorphism defined by $\phi(y)(x) = x^{-1}$. Then the semidirect product $G\rtimes_\phi H$ is the dihedral group of order $2n$, with presentation $$ G\rtimes_\phi H \;=\; \langle x,y\mid x^n=1,y^2=1,yxy^{-1}=x^{-1}\rangle. $$

Jim Belk
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  • hi @Jim Belk why is that? which definition of semidirect product works best to derive this presentation? thanks – latelrn Aug 15 '22 at 14:38
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    @l4teLearner I might not be Jim Belk, and I might be a few years late...but to answer your question, it's a little bit complicated. I highly recommend reading through pg 175-176 of Dummit and Foote. But if I were to give a brief explanation, conjugation defines a group action (i.e. $k\cdot h=khk^{-1}$ is a well-defined group action). Since the semidirect product is a group we're looking to construct -- that is, we have yet shown that such a group exists prior to construction -- technically the multiplication of elements in $H$ and $K$ are not defined. Therefore, just so as long... – J.G.131 May 14 '24 at 21:14
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    ...as we define multiplication between such elements in such a way which preserves the fact that $hkh^{-1}$ remains a group action of $K$ on $H$ (that is, just as long as an associated group automorphism on $H$ can be used to define the group action of conjugation), then we reach our well-defined construction of semi-direct product. – J.G.131 May 14 '24 at 21:17
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    So to cut straight-to-the-point, the group automorphism $\phi_{k}\in \text{Aut}(H)$ associated with group action $k\cdot h$ will necessarily satisfy the equation $k\cdot h = \phi_{k}(h)$, or rather $khk^{-1}= \phi_{k}(h).$ – J.G.131 May 14 '24 at 21:20
  • @JAG131 you've been very kind, thanks. I will go through your comments. – latelrn May 14 '24 at 22:55
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    @l4teLearner I appreciate the response. Admittedly, my comments were intended to help those who had similar struggles as you. (I don't suspect you've been pondering on your original question since August 2022...I hope not, at least lol.) But if they are still of any use for you, then hopefully it'll clear things up. :) – J.G.131 May 14 '24 at 23:23
  • How to prove that it's actually the presentation? Kernel of the epimorphism contains this set. But why normal closure of this set is equal to kernel? – jay sri krishna Jun 13 '24 at 09:34