Group presentation for semidirect products

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!

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 $$G\rtimes_\phi H \;=\; \langle x,y\mid x^n=1,y^2=1,yxy^{-1}=x^{-1}\rangle.$$
• @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... May 14 at 21:14
• ...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. May 14 at 21:17
• 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 $khk^{-1}= \phi_{k}(h).$ May 14 at 21:20