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What exactly does $\text{Aff}(2) = \mathbb{R}^2 \rtimes SL_{2}(\mathbb{R})$ mean? I know that it is the group of area preserving affine transformations of (oriented) $\mathbb{R}^2$. But how would you interpret the $\rtimes$ symbol?

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It is semidirect product. Look up up on wikipedia. (Basically, it is a generalisation of direct product. In a direct product you have two subgroups $H, K\lhd G$, $G=HK$ and $H\cap K=1$. In a semidirect product, only one of $H$ and $K$ is necessarily normal. It is helpful to know what a group action is, as the non-normal group acts on the normal one via a non-trivial action.) – user1729 Dec 21 '12 at 14:52
@user1729: What is both subgroups are normal? Are groups that can be "factored" into two normal subgroups better than groups that cannot? – ekrnkjenrkjenrek Dec 21 '12 at 14:55
I've expanded a bit on this in my answer, below. – user1729 Dec 21 '12 at 15:01
It is a symbol used on maps of highways to show that a rest stop does have picnic tables, but indicating that some of the tables have been pushed over by vandals. – Will Jagy Dec 21 '12 at 21:15
@amWhy: There's no need to add [abstract-algebra] to posts tagged group theory, unless they fundamentally speak about objects other than groups. See this meta thread. – Asaf Karagila Nov 7 '13 at 15:38

The symbol $\rtimes$ means "semidirect product". Look up up on wikipedia.

Basically, it is a generalisation of direct product. In a direct product you have two subgroups $H, K\lhd G$, $G=HK$ and $H\cap K=1$. In a semidirect product, only one of $H$ and $K$ is necessarily normal. It is helpful to know what a group action is, as the non-normal group acts on the normal one via a non-trivial action.

In a direct product, the two subgroups acts on each other trivially - this means that they commute with one another, that $k^{-1}hk=h$ for all $h\in H$ and $K\in k$. In a semidirect product, assuming $H$ is normal then $k^{-1}hk=h^{\prime}\in H$ and it is not necessary that $h=h^{\prime}$.

In the case you give here, $G=\operatorname{Aff}(2)$ contains two subgroups $H$ and $K$ with $H\cong \mathbb{R}^2$ and $K\cong SL_2(\mathbb{R})$, $H\cap K=1$, $G=HK$ and $H\lhd G$. They symbol $\ltimes$ would mean $K$ was normal rather than $H$.

For example, there are two groups of order $6$. One is a direct product, $C_6=C_2\times C_3$, whilst the other is a semidirect-product of $C_3=\{1, a, a^{-1}\}$ with $C_2=\{1, b\}$ where $b^{-1}ab=a^{-1}$ (note that $b=b^{-1}$). This is $D_3$.

Helpful Tip: When you do not know what a symbol means then pick up your favourite text on the subject and find the "Notation" section. This gives you all the notation used in the book, and if the book is general enough, used in the subject. For example, my favourite general group theory book is "A Course in the Theory of Groups" by Derek J. S. Robinson. The notation section here is at the beginning, starting on page xv. Half-way down page xvi I see the symbol we are after - "Semidirect products" it says. I go to the index and find that semidirect products are discussed on page 27. Score!

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The symbol $\;$ "$\rtimes$" $\; $ is used to denote the semi-direct product.

  • "In mathematics, specifically in group theory, a semidirect product is a particular way in which a group can be constructed from two subgroups, one of which is a normal subgroup. A semidirect product is a generalization of a direct product."

Let $G$ be a group with identity element $e$, $N$ a normal subgroup of $G$ and $H$ a subgroup of $G$. The following statements are equivalent:

  • $G = NH$ and $N \cap H = \{e\}$.
  • $G = HN$ and $N \cap H = \{e\}$.
  • Every $g\in G$ can be written as a unique product of an element of $N$ and an element of $H$.
  • Every $g\in G$ can be written as a unique product of an element of $H$ and an element of $N$.
  • The natural embedding $H \to G$, composed with the natural projection $G \to G / N,$ yields an isomorphism between $H$ and the quotient group $G / N.$
  • There exists a homomorphism $G \to H$ which is the identity on $H$ and whose kernel is $N.$

If any (and hence every) statement above is true, then $G$ is a semidirect product of $N \lhd G$ and $H$, it is usually written: $G = N \rtimes H$, and one can say: "$G$ splits over $N$". Or one can say: "$G$ is a semidirect product of $H$ acting on $N$."

Relation to direct product: Suppose $G$ is a semidirect product of the normal subgroup $N$ and the subgroup $H$. If $H$ is also normal in $G$, or equivalently, if there exists a homomorphism $G \to N$ which is the identity on $N$, then $G$ is the direct product of $N$ and $H$.

The direct product of two groups $N$ and $H$ can be thought of as the outer semidirect product of $N$ and $H$ with respect to $\phi(h) = id_N \;\;\forall h \in H$.

  • Note that in a direct product, the order of the factors is not important, since $N \times H$ is isomorphic to $H \times N.$
  • This is not the case for semidirect products, as the two factors play different roles.

So, for example, in your specific example, $G=\text{Aff}(2)$ is the semidirect product of two of its subgroups: $N$ and $H$ where we have $N\cong \mathbb{R}^2$ and $H\cong SL_2(\mathbb{R})$, and such that $N\cap H=1$, $G =\text{Aff}(2)=NH$ and $N = \mathbb{R^2} \lhd G$.

TIPS for notation-related questions:

You might find Wikipedia's List of Mathematical Symbols helpful, when the occasion requires it. But user1729's suggestion to look at a text's "Notation" section, if it has one, is probably the first course of action to take, since some symbols mean different things, in different contexts, and occasionally may vary from author to author.

Also see Detexify, which allows you to "draw" a symbol, then churns out various possible matches, their "names", AND the $TeX$ code you can use for formatting.

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The particular semidirect product that motivated your question is actually a very good example to see what's "really" involved in general semidirect products. The group Aff(2) of affine transformations of the plane contains the subgroup $T$ of rigid translations and the subgroup $F$ of affine transformations that fix the origin; the former is isomorphic to $\mathbb R^2$ and the latter to $SL_2(\mathbb R)$. Furthermore, any affine transformation can be uniquely expressed as the composition of a member of $F$ followed by a member of $T$. The idea here is that, given an affine transformation $\alpha$, you can first find a member $\phi$ of $F$ that has the same effect as $\alpha$ on directions and on lengths in any direction; then $\alpha=\tau\circ\phi$, where $\tau\in T$ translates the origin to the same place where $\alpha$ moves the origin. The fact that any $\alpha\in\text{Aff}(2)$ is given by a member of $F$ and a member of $T$, suggests that we're dealing with a direct product, $T\times F$, but that's only correct in terms of the sets, not in terms of the group structure. To see why the group structure is more complicated, consider what happens when we follow the $\alpha$ above with another affine transformation $\alpha'=\tau'\circ\phi'$ (where $\tau'\in T$ and $\phi'\in F$ as before). The effect of $\alpha'\circ\alpha=\tau'\circ\phi'\circ\tau\circ\phi$ on directions and on lengths is given nicely by $\phi'\circ\phi$, but where $\alpha'\circ\alpha$ moves the origin is not given simply by $\tau'\circ\tau$. The reason is that, once the origin has been moved to a new position by $\tau$, it will be moved again by $\phi'$ before $\tau'$ acts on it. Specifically, if $\tau$ was translation by a vector $\vec v$ and $\tau'$ by $\vec v'$, then $\alpha'\circ\alpha=\tau'\circ\phi'\circ\tau\circ\phi$ moves the origin not to $\vec v+\vec v'$ but to $\phi'(\vec v)+\vec v'$. So the multiplication differs from what you'd have in a direct product; the difference is that the $F$ component of the left factor modifies the $T$ component of the right factor before the $T$ components are combined.

In general, then, a semidirect product of groups $A$ and $B$ consists of pairs $(a,b)$ but the product $(a',b')\cdot(a,b)$ is not $(a'\cdot a,b'\cdot b)$ but $(a'\cdot\xi(b')(a),b'\cdot b)$. Here $\xi(b')$ tells how the presence of $b'$ modifies the $a$ factor before it gets multiplied by $a'$. In the example, $\xi$ of an element of $F$ modifies translations by acting on the associated vector $\vec v$. In general, $\xi(b')$ needs to be an automorphism of the group $A$, and the map $\xi$ from $B$ into the group of automorphisms of $A$ needs to be a homomorphism, in order for the semidirect product to be a group. To specify a semidirect product, one needs to specify not only the two groups $A$ and $B$ but also the homomorphism $\xi$.

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Most everything is covered in the previous answers, but I want to emphasize something that confused me for a long time about the semidirect product notation.

Every semidirect product $N\rtimes H$ is formed using a specified homomorphism $\varphi:H\rightarrow \operatorname{Aut}(N)$. Different $\varphi$ can produce radically different groups, so strictly speaking the correct way to write a semidirect product is $N\rtimes_\varphi H$ to indicate which homomorphism is being used. When you see $N\rtimes H$ with the $\varphi$ omitted, the writer is implicitly assuming that the homomorphism being used is obvious or that context should make it unambiguous.

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The hardest thing about the notation $N \rtimes H$ is remembering which of the groups is the normal subgroup and which is the group acting. There are two good mnemonics for this. The first is that $\rtimes$ contains the symbol $\vartriangleleft$ for a normal subgroup and the normal subgroup is on the correct side of the symbol. The other mnemonic is to hold your arms in front of you and move them up and down. In profile you will make a $\rtimes$ and then you just remember that you are acting on something so the hands go on the side with $N$ which is being acted on while $H$ goes on the side that you're on doing the acting.

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A problem with the first mnemonic rule is that it breaks with the notation for bicrossproducts, in which neither subgroup is normal :-) – Mariano Suárez-Alvarez Nov 7 '13 at 15:41

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