Semidirect Products, Split Exact Sequences, and all that

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Ken's Summary of Semidirect Products and Split Exact Sequences

One of the things I've butted heads with in studying Lie Groups is the semidirect product and its relationship to split exact sequences. It quickly became apparent that this was a pretty sizeable hole in my basic knowledge, so I decided to clarify this stuff once and for all.

Normal Subgroups and Quotient Groups

First, a brief refresher on Normal subgroups and Quotient groups. We are given a group $G$ and subgroup $H\subseteq G$.

  • Left cosets are written $gH$ and right cosets are written $Hg$. Each is a set of elements in $G$. Not all left cosets are distinct, but any two are either equal or disjoint. Ditto for right cosets.
  • The left (right) cosets form a partition of $G$, but they do not in general form a group. We can try to imbue them with a suitable product, but there are obstructions to the group axioms. For example $g^{-1}H$ is not a useful inverse since $(gh)^{-1}= h^{-1}g^{-1}$, so neither left cosets nor right cosets multiply as desired. More generally $(gg')H$ does not consist of a product of an element of $gH$ and an element of $g'H$.
  • We define the Quotient Set $G/H$ to be the set of left cosets. As mentioned, it is not a group in general. There is an equivalent definition for right cosets, written $H\setminus{}G$, but it doesn't appear often. In most cases we care about the two are the same.
  • It is easy to see that the condition which removes the obstruction is that $gH=Hg$ for all $g$. Equivalently, $gHg^{-1}=H$ for all $g$. If this holds, the cosets form a group. Often the stated condition is that the sets of left and right cosets are the same. But $g\in gH,Hg$ so this is the same exact condition.
  • $H$ is a Normal Subgroup if it obeys the conditions which make the cosets into a group.
  • Usually a Normal Subgroup is denoted $N$, and we write $N\triangleleft G$ (or $N\trianglelefteq G$).
  • For a Normal subgroup $N$, the Quotient Set $Q=G/N$ has (by definition) the natural structure of a group. It is called the Quotient Group.
  • We have two natural maps associated with a Normal Subgroup:
    • $N\xrightarrow{i} G$ is an inclusion (i.e. injective), defined by $h\rightarrow h$ (where the righthand $h$ is viewed in $G$). This is a homomorphism defined for any subgroup, not just normal ones
    • $G\xrightarrow{q} Q$ is the quotient map (surjective), defined by $g\rightarrow gN$ (with the righthand viewed as a coset, i.e. an element of $G/N$). This map is defined for any subgroup, with $Q$ the Quotient Set. For Normal Subgroups, it is a group homomorphism.
  • We know there is a copy of $N$ in $G$. Though $Q$ is derived from $G$ and $N$, and possesses no new info, there may or may not be a copy of it in $G$. Two natural questions are when that is the case, and how $G$, $N$, and $Q$ are related in general.
Let's also recall the First Isomorphism Theorem for groups. Given any two groups $G$ and $H$ and a homomorphism $\phi:G\rightarrow H$, the following hold:
  • $\ker \phi$ is a Normal Subgroup of $G$
  • $\im \phi$ is a subgroup of $H$
  • $\im \phi$ is isomorphic to the Quotient Group $G/\ker\phi$.
Again, we have to ask: since $\ker\phi$ is a Normal Subgroup of $G$, and $\im\phi$ is isomorphic to the Quotient Group $G/\ker\phi$ which "sort of" may have an image in $G$, is it meaningful to write something like (playing fast and loose with notation) $G\stackrel{?}{=} \ker\phi \oplus \im \phi$ (being very loose with notation)? The answer is no, it's more complicated.

Exact Sequences

Next, a very brief review of exact sequences. We'll use $1$ for the trivial group. The usual convention is to use $1$ for general groups and $0$ for Abelian groups. An exact sequence is a sequence of homomorphisms between groups $\cdots \rightarrow G_n \xrightarrow{f_n} G_{n-1}\xrightarrow{f_{n-1}} \cdots$ where $\im f_n= \ker f_{n-1}$ for every pair. Here are some basic properties:
  • $1\rightarrow A \xrightarrow{f} B\cdots$ means that $f$ is injective.
  • $\cdots A\xrightarrow{f} B\rightarrow 1$ means that $f$ is surjective.
  • $1\rightarrow A\rightarrow B\rightarrow 1$ means $A=B$.
  • Short Exact Sequence (SES): This is defined as an exact sequence of the form: $1\rightarrow A\xrightarrow{f} B\xrightarrow{g} C\rightarrow 1$.
  • For an SES, $f$ is injective, $g$ is surjective, and $C=B/\im f$
  • SES's arise all the time when dealing with groups, and the critical question is whether they "split".
We're now ready to define Split SES's.
  • Right Split SES: There exists a homomorphism $h:C\rightarrow B$ such that $g\circ h=Id_C$. Basically, we can move to $B$ and back from $C$ without losing info --- which means $C$ is in some sense a subgroup of $B$.
  • Left Split SES: There exists a homomorphism $h:B\rightarrow A$ such that $h\circ g=Id_A$. Basically, we can move to $B$ and back from $A$ without losing info --- which means $A$ is in some sense a subgroup of $B$.
  • These two conditions are not in general equal, or even equivalently restrictive. The Left Split condition is far more constraining than the Right Split one in general. The direction of the homomorphisms in the SES introduce an asymmetry. [My note: it seems likely that the two are dual in some sense.]

External vs Internal View

We're going to described 3 types of group operations: the direct product, semi-direct product, and group extension. Each has a particular relationship to Normality and SES's. There are two equivalent ways to approach this, depending whether we prefer to define a binary operation between two distinct groups or to consider the relationship amongst subgroups of a given group.
  • External view: We define a binary operation on two distinct, unrelated groups. Two groups go in, and another group comes out.
  • Internal view: We define a relationship between a group and various groups derived from it (ex. Normal or Quotient).
  • These approaches are equivalent. The Internal view describes the relationship amongst the two groups involved in the External view and their issue. Conversely, the derived groups in the Internal view may be recombined via the External view operation.
We must be a little careful with notation and terminology. When we use the symbol $HK$, it can mean one of two things.
  • Case 1: $H$ and $K$ are distinct groups. $HK$ is just the set of all pairs of elements $(h,k)$. I.e. it is the direct product set (but not group).
  • Case 2: $H$ and $K$ are subgroups of a common group $G$ (or have some natural implicit isomorphisms to such subgroups). In this case, $HK$ is the set of all elements in $G$ obtained as a product of an element of $H$ and an element of $K$ under the group multiplication.
  • Note that we may prefer cases where two subgroups cover $G$, but there are plenty of other possibilities. For example, consider $Z_{30}$ (the integers mod 30). This has several obvious subgroups ($Z_2$, $Z_3$, $Z_5$, $Z_6$, $Z_{10}$, $Z_{15}$). $Z_2$ and $Z_3$ only intersect on $0$ (the additive identity). However, the two do not cover (or even generate) the group! Similarly, $Z_2$ and $Z_{10}$ do not cover the group (or even generate it) but intersect on a nontrivial subset!
  • Going the other way, we'll say that $G=HK$ if $H$ and $K$ are subgroups and every element $g$ can be written as $hk$ for some $h\in H$ and $k\in K$. Note that $H$ and $K$ need not be disjoint (or even cover $G$ set-wise).
Another potentially confusing point should be touched on. When we speak of "disjoint" subgroups $H$ and $K$ we mean that $H\cap K=\{e\}$, NOT that it is the null set. I.e., $H\cap K= 1$, the trivial group.

Semidirect Product

The semidirect product may seem a bit arbitrary at first but, as we will see, it is a natural part of a progression which begins with the Direct Product. Here are the two ways of defining it.
  • External view (aka Outer Semidirect Product): Given two groups $H$ and $K$ and a map $\phi:K\rightarrow Aut(H)$, we define a new group $H\rtimes K$. We'll denote by $\phi_k(h)$ the effect of the automorphism $\phi(k)$ on $h$ (and thus an element of $H$). Set-wise, $H\rtimes K$ is just $H\times K$ (i.e. all pairs $(h,k)$). The identity is $(e,e)$. Multiplication on $H\rtimes K$ is defined as $(h,k)(h',k')= (h\phi_k(h'),kk')$. The inverse is $(h,k)^{-1}= (\phi_{k^{-1}}(h^{-1}),k^{-1})$.
  • Internal view (aka Inner Semidirect Product): Given a group $G$ and two disjoint subgroups $N$ and $K$, such that $G=NK$ and $N$ is a Normal Subgroup, $G$ is called the Semidirect product $N\rtimes K$. The normality of $H$ constrains $K$ to be isomorphic to the Quotient Group $G/N$.
There are a few important things to note about this.
  • There are (potentially) many Semidirect products of two given groups, obtained via different choices of $\phi$. The notation is deceptive because it hides our choice of $\phi$. Given any $H,K,\phi$ there exists a Semidirect product $H\rtimes K$. The various Semidirect products may be isomorphic to one another, but in general need not be. I.e., a given $H$ and $K$ may have multiple distinct semidirect products. This actually happens. Wikipedia mentions that there are 4 non-isomorphic semidirect products of $C_8$ and $C_2$ (the former being the Normal Subgroup in each case). One is a Direct Product, and the other 3 are not.
  • It also is possible for a given group $G$ to arise from several distinct Semidirect products (of different pairs of groups). Again from Wikipedia, there is a group of order 24 which can be written as 4 distinct semiproducts of groups.
  • Yet another oddity is that a seemingly nontrivial $H\rtimes K$ can be isomorphic to $H\oplus K$.
  • If $\phi= Id$ (i.e. every $k$ maps to the identify map on $H$), then $G=H\oplus K$.
  • To go from the External view to the Internal one, we note that, by construction, $H$ is a Normal Subgroup of $H\rtimes K$ and $K$ is the Quotient Group $G/H$. To be precise, the Normal Subgroup is $(N,e)$, which is isomorphic to $N$, and the Quotient Group $G/(N,e)$ is isomorphic to $K$.
  • To go from the Internal view to the External one, we choose $\phi_k(h)= khk^{-1}$ as our function. I.e., $\phi$ is just conjugation by the relevant element.
  • It may seem like there is an imbalance here. For a specific choice of Normal Subgroup $N$, the External view offers complete freedom of $\phi$, while the Internal view has a fixed $\phi$. Surely the latter is a special case of the former. The fallacy in this is that we must consider the pair $(G,N)$. We very well could have non-isomorphic $G,G'$ with Normal Subgroups $N,N'$ where $N\approx N'$. I.e. they are the same Normal Subgroup, but with different parent groups. We then would have different $\phi$'s via our Internal view procedure. The correspondence is between $(H,K,\phi)$ and $(G,N,K)$ choices. Put differently, the freedom in $\phi$ loosely corresponds to a freedom in $G$.
  • Note that, given $G$ and a Normal Subgroup $N$ --- with the automatic Quotient Group $G/N$ --- we do NOT necessarily have a Semidirect product relationship. The condition of the Semidirect product is stricter than this. As we will see it requires not just isomorphism, but a specific isomorphism, between $H$ and $G/N$. Equivalently, it requires a Right-Split SES (as we will discuss).
  • The multiplication defined in the External view may seem very strange and unintuitive. In essence, here is what's happening. For a direct product, $H$ and $K$ are independent of one another. Each half of the pair acts only on its own elements. For a semidirect product, the non-normal half $K$ can twist the normal half $H$. Each element of $K$ can alter $H$ in some prescribed fashion, embodied in $\phi(k)$. So $K$ is unaffected by $H$ but $H$ can be twisted by $K$.
  • It is interesting to compare the basic idea to that of a Fiber bundle. There, the fiber can twist (via a group of homeomorphisms) as we move around the base space. Here, the normal subgroup can twist as we move around the non-normal part. Each generalizes a direct product and measures our need to depart from it.
  • The semidirect product of two groups is Abelian iff it's just a direct product of abelian groups.

Group Extensions

As with Semidirect products, there are 2 ways to view these. To make matters confusing, the notation speaks to an Internal view, while the term "extension" speaks to an External view.
  • External view: Given groups $A$ and $C$, we say that $B$ is an extension of $C$ by $A$ if there is a SES $1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1$.
  • Internal view: Given a group $G$ and Normal Subgroup $N\triangleleft G$, we say that $G$ is an extension of $Q$ by $N$, where $Q=G/N$ is the Quotient Group.
  • Note that the two are equivalent. If $B$ is an extension of $A$ by $C$, then $A$ is Normal in $B$ and $C$ is isomorphic to the Quotient Group $B/A$.
  • Put simply, the most general form of the Group, Normal Subgroup, induced Quotient Group trio is the Group Extension.

Direct Products, Semidirect Products, and Group Extensions

In the External view, we've mentioned three means of getting a group $B$ from two groups $A$ and $C$:
  • Direct Product: $B=A\oplus C$. This is unique.
  • Semidirect Product: $B=A\rtimes C$. There may multiple of these, corresponding to different $\phi$'s.
  • Group Extension: A group $B$ for which there are 2 homomorphisms forming a SES $1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1$. There may be many of these, corresponding to different choices of the two homomorphisms.
Equivalently, we have several ways of describing the relationship between two subgroups $H,K\subseteq G$ which are disjoint (i.e. $H\cap K=\{e\}$).
  • Direct Product: $G=H\oplus K$ requires that both be Normal Subgroups.
  • Semidirect Product: $G=H\rtimes K$ requires that $H$ be normal (in which case, $Q=G/H$, and $\phi$ is determined by it). For a given $H$ there may be multiple, corresponding to different $G$'s.
  • Group Extension: Both $H$ and $K$ sit in $G$ to some extent. $H$ must be Normal.
Note that not every possible relationship amongst groups is captured by these. For example, we could have two non-normal subgroups or two homomorphisms which don't form an SES, or no relationship at all. An excellent hierarchy of conditions was provided by Arturo Magidin in answer to someone's question on Stackoverflow. I roughly replicate it here. Unlike him, I'll be sloppy and not distinguish between subgroups and groups isomorphic to subgroups.
  • Direct Product ($G=H\oplus K$): $H,K$ both Normal Subgroups. $H,K$ disjoint. $G=HK$
  • Semidirect Products ($G=H\rtimes K$): $H$ Normal Subgroup, $K$ Subgroup. $H,K$ disjoint. $G=HK$. I.e., we lose Normality of $K$.
  • Group Extension ($G$ is extension of $H$ by $K$): $H$ Normal Subgroup, $G/H\approx K$. I.e. $K$ remains the Quotient Group (as before), but the Quotient Group may no longer be a subgroup of $G$ at all!

Now is a good time to mention the relationship between the various SES Splitting conditions:

  • For all groups: Left Split is equivalent to $B=A\oplus C$, and they imply Right Split. (LS=DP) => RS always.
  • For abelian groups, the converse holds and Right split implies Left Split and Direct Sum. I.e. the conditions are equivalent. LS=DP=RS for Abelian.
  • For nonabelian groups: Right Split implies $B=A\rtimes C$ (with $\phi$ depending on the SES map). We'll discuss this shortly.
Back to the hierarchy, now from a SES standpoint:
  • Most general case: There is no SES at all. Given groups $A,B,C$, there may be no homomorphisms between them. If there are homomorphisms, there may be none which form an SES. Consider a general pair of homomorphisms $f:A\rightarrow B$ and $g:B\rightarrow C$, with no assumptions. We may turn to the first isomorphism theorem for help, but that does us no good. The first isomorphism theorem says that $\ker f \triangleleft B$ and $\im f\approx A/\ker f$, and $\ker g \triangleleft C$ and $\im g\approx B/\ker g$. This places no constraints on $A$ or $C$.
  • Group Extension: Any SES defines a group extension. They are the same thing.
  • Semidirect Product: Any SES which right-splits corresponds to a Semidirect Product (with the right-split map determining $\phi$)
  • Direct Product: Any SES which left-splits (and thus right-splits too) corresponds to a direct product.
So, when we see the standard SES: $1\rightarrow N\rightarrow G\rightarrow G/N\rightarrow 1$, this is a group extension. Only if it right splits can we write $G= N\rtimes G/N$, and only if it left splits can we write $G= N\oplus G/N$.

Some Notes

  • Group Extensions are said to be equivalent if their $B$'s are isomorphic and there exists an isomorphism between them which makes a diamond diagram commute. It is perfectly possible for the $B$'s to be isomorphic but for two SES's not to be equivalent extensions.
  • Subtlety referred to above. A quotient group need not be isomorphic to a subgroup of $G$. It only is defined when $N$ is normal, and there automatically is a surjective homomorphism $G\rightarrow Q$. But we don't have an injective homomorphism $Q\rightarrow G$, which is what would be need for it to be isomorphic to a subgroup of $G$. This is precisely what the right-split furnishes. In that case, it is indeed a subgroup of $G$. The semidirect product may be thought of as the statement that $Q$ is a subgroup of $G$.
  • In the definition of right split and left split, the crucial aspect of the "inverse" maps is that they be homomorphisms. A simple injective (for right-split, or surjective for left-split) map is not enough!
  • It is sometimes said that the concept of subgroup is dual to the concept of quotient group. This is intuitive in the following sense. A subgroup can be thought of as an injective homomorphism. By the SES for normal/quotient groups, we can think of a quotient group as a surjective homomorphism. Since injections and surjections are categorically dual, it makes sense to think of quotient groups and subgroups as similarly dual. Whether the more useful duality is subgroup <-> quotient group or normal subgroup <-> quotient group is unclear to me.

Some Resources

Wikipedia Article on Semidirect Products
Discussion of Subgroup/Quotient-Group duality
An excellent comparison of Extensions, Semidirect Products, and Direct Products (scroll down to Arturo Magidin's answer)
Notes on Semidirect Products (from an Algebra course by Dave Bayer)
K.M. Halpern