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I'm trying to motivate the definition of a semidirect product, and it seems like it comes from geometric examples. Some classic ones are

$D_{2n} \cong Z_n \rtimes Z_2$

$E(n) \cong R^2 \rtimes O(2)$

$O(n) \cong SO(n) \rtimes Z_2$

It seems like the group that is doing the acting sort of keeps track of the "orientation" of the group that is being acted upon. Can anyone provide a nice reference for understanding these examples, or an explanation?

user94308123
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2 Answers2

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I think the best way to understand the semidirect product is not through geometry, but through pure algebra. Here is my point of view that I prefer (explained in detail in the Dummit & Foote).

Given two subgroups of a group $G$ such that $H$ is normal in $G$, $HK = G$ and $H \cap K = \{1\}$, one can deduce that any element $g \in G$ has a unique expression in the form $g = hk$, where $h$ and $k$ are uniquely determined by $g$. Now if we try to multiply two elements of the group, then $$ g_1 g_2 = (h_1 k_1)(h_2 k_2) = h_1 k_1 h_2 (k_1^{-1} k_1) k_2 = (h_1 (k_1 h_2 k_1^{-1})) (k_1 k_2) = h_3 k_3 $$ In this manner, by unicity we have $k_1 k_2 = k_3$, but $h_3 = h_1 \varphi_{k_1}(h_2)$, where $\varphi_{k_1}$ gives you the way that "$K$ acts on $H$". Now $k_1$ is (except if it is the identity) not an element of $H$, so we cannot say that we obtain elements by conjugation in general ; when one constructs the semidirect product, the conjugation by an element of $K$ is replaced by automorphisms, and to mimic this case we use the abstract definition with the homomorphism $\varphi : K \to \mathrm{Aut}(H)$, but that's essentially what's behind the idea ; you get an "almost direct" product.

I hope that helps,

Patrick Da Silva
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  • Thank you so much. Awesome explanation! – Matha Mota Feb 12 '20 at 16:55
  • This "almost direct product" you mention with $(h,q)\in H\times K$ and a certain action. Does this "induce" a group $(\cdot, G)$ with a certain operation where conjugation satisfies $\Phi_{k_1}(h_2)=k_1h_2k_1^{-1}$ and $H$ is normal in G and $K$ is a subgroup with $H\cap K=1$? – Kadmos May 03 '23 at 18:06
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    @Kadmos : Yes, if you start with groups $H$ and $K$ together with an arbitrary morphism of groups $\varphi : K \to \mathrm{Aut}(H)$, then you can form the semidirect product $H \rtimes_{\varphi} K$ which is basically the set $H \times K$, but with the multiplication rule given as above. You can see that this construction always determines a unique group, and if you call this group $G$ and run the argument in my answer, you get back the exact same thing. In other words, you can speak of "inner" and "outer" semidirect products. – Patrick Da Silva May 13 '23 at 23:51
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    You can also see it this way: multiplication of elements of $H$ and $K$ should already be defined since they should be subgroups of $H \rtimes_{\varphi} K$. So you only have to define multiplications of the form $(h,1)(1,k)$ and $(1,k)(h,1)$. The first one gives $(h,k)$ and the second one gives $(\varphi_k(h), k)$. You can see under this definition that $(1,k)^(-1) (h,1) (1,k) = (\varphi_k(h), 1)$ as expected. – Patrick Da Silva May 13 '23 at 23:54
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Let $G$ be a group and let $S$ be a torsor over $G$; this is a set on which $G$ acts in a way that is (non-canonically) isomorphic to the action of $G$ by left multiplication on itself, and is a natural generalization of an affine space. For example:

  • $n$-dimensional Euclidean space is a torsor over $\mathbb{R}^n$.
  • The set of total orders on a set of size $n$ is a torsor over the symmetric group $S_n$.
  • The set of possible locations of a bug on some vertex of an $n$-gon is a torsor over the cyclic group $\mathbb{Z}/n\mathbb{Z}$.

Then $S$ naturally inherits $G$ as a group of automorphisms (as a set). Now let $H$ be a subgroup of $\text{Aut}(G)$. Then picking an element $s \in S$ also allows us to identify $G$ with $S$ (via the map $g \mapsto gs$), hence defines an action of $H$ on $S$, but this action depends on the choice of $s$.

The subgroup of $\text{Aut}(S)$ (again, as a set) generated by $G$ and $H$ above is the semidirect product $G \rtimes H$ (regardless of the choice of $s$; exercise). Note the geometric significance of the case $G = \mathbb{R}^n, H = \text{O}(n)$.

(Of course in general the homomorphism $H \to \text{Aut}(G)$ is not injective, but I think the geometric significance is clearest when it is. In general perhaps a more algebraic perspective is appropriate; look up "split exact sequence.")

Qiaochu Yuan
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  • I wonder if there is a way to modify this interpretation to somehow interpret that the semidirect product is the direct product if the homomorphism $H\to G$ sends everything to the identity. – alphacapture Apr 14 '18 at 22:46