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If we have two groups $G,H$ the construction of the direct product is quite natural. If we think about the most natural way to make the Cartesian product $G\times H$ into a group it is certainly by defining the multiplication

$$(g_1,h_1)(g_2,h_2)=(g_1g_2,h_1h_2),$$

with identity $(1,1)$ and inverse $(g,h)^{-1}=(g^{-1},h^{-1})$.

On the other hand we have the construction of the semidirect product which is as follows: consider $G$,$H$ groups and $\varphi : G\to \operatorname{Aut}(H)$ a homomorphism, we define the semidirect product group as the Cartesian product $G\times H$ together with the operation

$$(g_1,h_1)(g_2,h_2)=(g_1g_2,h_1\varphi(g_1)(h_2)),$$

and we denote the resulting group as $G\ltimes H$.

We then show that this is a group and show many properties of it. My point here is the intuition.

This construction doesn't seem quite natural to make. There are many operations to turn the Cartesian product into a group. The one used when defining the direct product is the most natural. Now, why do we give special importance to this one?

What is the intuition behind this construction? What are we achieving here and why this particular way of making the Cartesian product into a group is important?

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    $\begingroup$ Sorry I can't give more details here and now but I advise to look into the Euclidean group as a good motivation for the semi direct product. Translations composed with rotations need for a semi direct product. $\endgroup$ – Learning is a mess Mar 19 '16 at 15:05
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    $\begingroup$ One observation: internal direct product looks for two normal subgroups generating thw whole group with trivial intersection. For the internal semidirect product, one of the groups need not be normal - so, a broader range of groups can be described via this notion. $\endgroup$ – lisyarus Mar 19 '16 at 15:09
  • $\begingroup$ I don't love this order for the semidirect product. For me, I find it more natural to write the normal subgroup on the left because then $(h_1,g_1)(h_2,g_2)=h_1g_1h_2g_2$ and you are using the automorphism to interchange the $g_1$ and the $h_2$, so this equals $h_1\phi_{g_1}(h_2)g_1g_2$. $\endgroup$ – Michael Burr Mar 19 '16 at 15:42
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Forget about the actual construction of the semidirect product for now. I argue that the semidirect product is important because it arises naturally and beautifully in many areas of mathematics. I will list below many examples, and I urge you to find a few that interest you and look at them in detail.

Before doing that let me just give the following extra motivation: say you have a group $G$, and you find two subgroups $H,K$ such that every element of $G$ can be uniquely written as a product $hk$ for $h \in H$, $k \in K$. In other words, you have a set-theoretic bijection between $G$ and $H \times K$. Then certainly you'd want to understand $G$ by studying its smaller components $H$ and $K$. One way to achieve this would be to find a suitable group structure on $H \times K$ intertwining the structures of $H$ and $K$ such that the above bijection becomes a group isomorphism. This can be done; however doing things in this generality becomes rapidly tedious. If instead we restrict our attention to such decompositions with $H$ normal in $G$, the problem becomes much more manageable. In this case we have what we call a split exact sequence $$1 \to H \to G \to K \to 1,$$ and $G$ is called a semidirect product of $H$ and $K$. An existence and uniqueness theorem gives us all the possible semidirect products one can obtain from $H$ and $K$ through the group of homomorphisms from $K$ to $\operatorname{Aut}H$. Note that $K = \mathbb{Z}/2$ appears often in practice, because this guarantees normality of $H$. Now here are some examples:

  • The symmetric group $S_n = A_n \rtimes \mathbb{Z}/2$. The exact sequence is $$1 \to A_n \to S_n \xrightarrow{\mathit{sign}} \mathbb{Z}/2 \to 1.$$

  • The dihedral group $D_n = \mathbb{Z}/n \rtimes \mathbb{Z}/2$. The exact sequence is $$1 \to \mathbb{Z}/n \to D_n \xrightarrow{\mathit{det}} \mathbb{Z}/2 \to 1.$$

  • The infinite dihedral group $D_\infty = \mathbb{Z} \rtimes \mathbb{Z}/2$. The exact sequence depends on your explicit construction. You may take $$1 \to \mathbb{Z} \to \mathbb{Z}/2 * \mathbb{Z}/2 \to \mathbb{Z}/2 \to 1$$ or $$1 \to \mathbb{Z} \to A(1,\mathbb{Z}) \to \mathbb{Z}/2 \to 1,$$ where $A(1,\mathbb{Z})$ is the group of affine transformations of the form $x \mapsto ax + b$, where $a \in \{ \pm 1 \} \cong \mathbb{Z}/2$ and $b \in \mathbb{Z}$.

  • Many matrix groups, thanks to the determinant map. For example, $G = \operatorname{GL}(n,\mathbb{F})$, $O(n,\mathbb{F})$ and $U(n)$ have respective subgroups $H = \operatorname{SL}(n,\mathbb{F}),\operatorname{SO}(n,\mathbb{F}),\operatorname{SU}(n)$ and $K = \mathbb{F}^\times,\mathbb{Z}/2,U(1)$.

  • The fundamental group of the Klein bottle is $G = \langle x,y \mid xyx = y \rangle$. This is just the nontrivial semidirect product of $\mathbb{Z}$ with itself. Interestingly, the other (trivial) semidirect product $\mathbb{Z}^2$ is the fundamental group of the other closed surface of Euler characteristic $0$, namely the torus.

  • The affine group $A(n,\mathbb{F}) = \mathbb{F}^n \rtimes \operatorname{GL}(n,\mathbb{F})$. Its elements are transformations $\mathbb{F}^n \to \mathbb{F}^n$ of the form $x \mapsto Ax + b$, with $A$ an invertible matrix and $b$ a translation vector. The exact sequence is $$1 \to \mathbb{F}^n \to A(n,\mathbb{F}) \xrightarrow{f} \operatorname{GL}(n,\mathbb{F}) \to 1$$ where $f$ forgets the affine structure (the translation part).

  • The hyperoctahedral group $O(n,\mathbb{Z})$ is the group of signed permutation matrices. We have two decompositions $O(n,\mathbb{Z}) \cong \operatorname{SO}(n,\mathbb{Z}) \rtimes \mathbb{Z}/2$ and $O(n,\mathbb{Z}) \cong (\mathbb{Z}/2)^n \rtimes S_n$. In the corresponding exact sequences the surjective map is respectively the determinant homomorphism and the "forget all the signs" homomorphism.

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    $\begingroup$ By the way, I recently wrote some notes on the subject if this perspective interests you. $\endgroup$ – Alex Provost Mar 20 '16 at 17:34
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    $\begingroup$ Updated link $\endgroup$ – Alex Provost Jun 10 '18 at 14:36
  • $\begingroup$ Thanks for your Notes on Semidirect Products. I meet a little problem here: Page 4, Line 3. ..."the left-hand side is precisely $\psi(h)$, and the right-hand side is precisely ($\pi \circ \phi)(h)$." I just can't figure it out. Thanks for your time and help! $\endgroup$ – Andrews Dec 3 '18 at 3:01
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    $\begingroup$ @Andrews You are welcome; thanks for reading! I'm sorry for the confusion, as I will admit upon re-reading that the whole paragraph is quite poorly written. What I should have written is simply this: the map $\psi:H \to \operatorname{Out}K$ always exists abstractly for any (possibly non-split) exact sequence, but depends on the choice of isomorphism $H \cong G/K$ and thus cannot be lifted naturally. The splitting map $s$ gives us a natural choice $\psi(h) = \phi(s(h)) \operatorname{Inn}K$, and now by construction $\phi$ lifts $\psi$. $\endgroup$ – Alex Provost Dec 24 '18 at 16:12
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It is nice to think about $D_4$ as a semidirect product. Namely, $D_4=\langle \sigma,\tau:\sigma^4=\tau^2=1,\tau\sigma=\sigma^{-1}\tau\rangle$. You can see the automorphism because $\sigma$ and $\tau$ do not commute, but the automorphism ($x\mapsto x^{-1}$) tells you how to move the $\tau$ past the $\sigma$.

In general, the direct product is not enough because the operation between elements of the two subgroups is always commutative. On the other hand, if $G$ is a group, $N$ is a normal subgroup, $H$ is a subgroup ($H$ need not be normal like in a direct product), $H\cap N=\{1\}$, and $G=NH$, then $G$ must be a semidirect product. (The operation between elements of $N$ and $H$ need not be commutative.) So, you can argue that the semidirect product classifies all groups constructed in this way.

The big idea in a semidirect product is the following:

  • You have two subgroups $N$ and $H$. You understand the operation when you multiply elements of $N$ and you understand the operation when you multiply elements of $H$.

  • The automorphism is used to compare the operation between elements of $N$ and elements of $H$.

  • You know that $N$ is normal, so for any $n\in N$ and $h\in H$, $hnh^{-1}$ is some element of $N$, and the map $n\mapsto hnh^{-1}$ is an automorphism of $N$. The semidirect product construction describes this conjugation automorphism. Therefore, if the automorphism determined by conjugation is $\phi_h:N\rightarrow N$, then $hn=hnh^{-1}h=\phi_h(n)h$.

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  • $\begingroup$ I have googled "intuition of semidirect product" twice with 1 year apart and both time I enjoyed your answer, but unfortunately it is impossible to upvote twice. :). Do I understand it correctly that $A\rtimes B\simeq \frac{A\times B}{\sim}$ where $\sim=$ relation between $A,B$? $\endgroup$ – C.F.G Nov 14 '20 at 20:50
  • $\begingroup$ Set theoretically, $A\rtimes B$ is $A\times B$ (there's no equivalence relation). The difference is that the product is $A$ and $B$ don't commute with each other, in fact, there is an action of $B$ on $A$ (a function from $B$ to the automorphism group of $A$). $\endgroup$ – Michael Burr Nov 14 '20 at 23:23
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I take a different point of view than most people on this, I think, which is due to the way I first encountered it (along the lines of mathematical physics). To me the direct product is lacking in much structure. You effectively just slap two groups together and call it a day - much like when you combine two subspaces via direct sums.

The semi-direct product is a simple way to really mix two groups together. Let's consider matrices. If we have two groups of square matrices (with the same dimension), say $G$ and $H$, then if we were to multiply elements, we'd have $g_1 h_1 g_2 h_2$. This would be like the product $(g_1,h_1)(g_2,h_2)$ in direct product notation. If $H$ commutes with $G$, then we could rewrite this as $g_1 g_2 h_1 h_2$ which we could realize as being similar to $(g_1g_2,h_1h_2)$ in direct product notation. The two groups don't really see each other in this setting.

We know however that this is not always the case. Instead what you might have is that $H$ acts on $G$ in some way so that if you try to repackage the product $g_1 h_1 g_2 h_2$ in the form $g_1 g_2 h_1 h_2$, $g_2$ gets mixed up a bit by $h_1$. Since we don't want to leave the group $G$, we would need that $h_1$ acting on $g_2$ gives us another element in $G$. Moreover $h_1 I = I h_1$ so $h_1$ would have to permute the elements of $G$ while leaving the identity fixed. Moreover, if we had $g_1 h_1 g_2g_3 h_3$, then acting $h_1$ on each of $g_2$ and $g_3$ independently (and moving over) should give the same result as acting on their product (and moving over) - this is just the homomorphism property.

This does not give rise to the automorphism aspect, but this can be seen by noting that $h^{-1}hg = g$ and so if $h$ mapped $g$ to the identity, you would have a contradiction (unless $g = I$ of course).

In summary: if you had two matrix groups $G$ and $H$ that did not necessarily commute but attempted to reorder the elements in a direct product kind of way, you necessarily need that $H$ acts on $G$ by automorphisms.

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You are looking at this in the wrong way.

The main reason for which we define the direct product of groups is that we like describing/understanding the structure of groups and we noticed that many groups are, well, direct products.

Now not all groups are direct products. For example, the dihedral group is not a direct product. But in this last example, for example, we are able to provide a very useful description of the group in a way that resembles a direct product in a way. As we find this same phenomenon in many contexts, we give it a name and call it semidirect product.

It is wrong, and a source of frustration, to look for the intuition of a definition which is motivated by examples: no one had any intuitive reason to come up with the definition of semidirect products out of thin air.

The definition does not have an intuition to justfy it: it is a useful concept in that it applies to many examples and it encapsulates many useful features which are useful to do things with groups.

You would not as for intution for the definition of the term «tree».


The construction does not seem natural to you simply because you do not know many groups and you have not yet spent much time investigating the structure of groups in any detail — if you do that, then the sheer force of examples will make it natural.

The key point is what it means for a definition to be «natural». And it does almost never mean «one could come up with it out of abstract meditation»: essentally all definitions are made to codify a situation that people encounter often and which, for that reason, is useful to give a name to. Of course, this meaning of «naturality» is relative: what seems unnatural to you would be utterly natural to, say, Burnside.

The punchline of all this is that it is almost never useful or productive to ask for the intuition of definitions when you first encounter them: what will help you is not some etherial intuition but examples, and that is what one should ask for to maximize understanding.


The next question would naturally be «what is the intuition behind the Zappa–Szép product and the answer would be the same: none. But some groups are not direct products not semidirect products but they still have two subgroups which somewhat similar properties as the factors of a direct product, and since this occurs often in practice, we give a name to that situation.

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    $\begingroup$ As evidence, you should notice that most of the time when people ask for intuition they get answered back with examples. $\endgroup$ – Mariano Suárez-Álvarez Mar 19 '16 at 18:44
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    $\begingroup$ While I think you make a good point, I don't think it's quite right. Yes, it's true intuition is relative to context and familiarity, but I think effectively what someone is asking for when asking for intuition is "is there something familiar to me/an average person in my context, that can be related to this notion." Op was quite clear that Direct Products felt familiar and intuitive to them, thus describing their context. So while one could answer solely with examples, it's clear that there is value in answers that try to explain how Direct Products might lead one to Semi-Direct Products. $\endgroup$ – Jonathan Rayner Mar 10 '19 at 23:24
  • $\begingroup$ ... Cameron Williams attempted this sort of explanation, for example (and did not use any examples of groups at all). $\endgroup$ – Jonathan Rayner Mar 10 '19 at 23:26
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Many have given good answers here, so I just want to answer specifically for the intuition behind it.

Semi direct product came into light when we found out that if a group $H$ is a normal subgroup, and another group $K$ is also a subgroup (not necessarily normal) of a bigger group, and $H \cap K = 1$, the multiplication of those two subgroups would yield another group $HK$ with order $\frac{|H||K|}{|H \cap K|} = |H||K|$ (as $H \cap K = 1$).

So we know that the new group $HK$ can be written uniquely in the form of $hk$ where $h \in H$ and $k \in K$. Because of that, the multiplication in $HK$, thus all elements of $HK$, must can always be written in the form of well... $hk$, e.g.:

$(h_1 k_1)(h_2 k_2) = h_1 k_1 h_2 (k_1^{-1} k_1) k_2 = h_1 (k_1 h_2 k_1^{-1}) k_1 k_2$

As $H$ is a normal subgroup, $k_1 h_2 k_1^{-1} \in H$, so it can be re-written as:

$(h_1 k_1)(h_2 k_2) = (h_1 (k_1 h_2 k_1^{-1})) (k_1 k_2) = h_3 k_3$

where $h_3 = h_1 (k_1 h_2 k_1^{-1}) \in H$ and $k_3 = k_1 k_2 \in K$.

But we notice that this left conjugation by $k_1$ which is $k_1 h_2 k_1^{-1}$ is an automorphism of $H$, so of course any automorphism of $H$ would do the job. If we define a homomorphism:

$\varphi: K \rightarrow Aut(H)$,

that homomorphism can be used in the place of left conjugation by $k_1$ and again achieve the same form of $hk$. Rewriting the above derivation with direct product notation and the defined homomorphism, we would get:

$(h_1, k_1)(h_2, k_2) = (h_1 \: \varphi(k_1)(h_2), k_1 k_2) = (h_3, k_3)$

where $h_3 = h_1 \: \varphi(k_1)(h_2) \in H$ and $k_3 = k_1 k_2 \in K$.

which is exactly the definition of semi direct product multiplication you ask.

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