29
$\begingroup$

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!

$\endgroup$

1 Answer 1

48
$\begingroup$

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 $$ G\rtimes_\phi H \;=\; \langle x,y\mid x^n=1,y^2=1,yxy^{-1}=x^{-1}\rangle. $$

$\endgroup$
6
  • $\begingroup$ hi @Jim Belk why is that? which definition of semidirect product works best to derive this presentation? thanks $\endgroup$ Aug 15, 2022 at 14:38
  • 1
    $\begingroup$ @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... $\endgroup$
    – JAG131
    May 14 at 21:14
  • 1
    $\begingroup$ ...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. $\endgroup$
    – JAG131
    May 14 at 21:17
  • 1
    $\begingroup$ 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 $khk^{-1}= \phi_{k}(h).$ $\endgroup$
    – JAG131
    May 14 at 21:20
  • 1
    $\begingroup$ @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. :) $\endgroup$
    – JAG131
    May 14 at 23:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .