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## An Introduction to Normal and Non-normal Subgroups

February 21, 2015 Leave a comment

We currently are running several reading groups for engineering students and one is on Lie Groups and Representation Theory.  When elementary group theory was introduced yesterday, not surprisingly, there were many questions concerning normal subgroups.  Like many areas of mathematics, the traditional learning paradigm is to “just memorise the definitions and later things will make sense”.  While it is true that later things will make sense — once a critical mass of ideas and definitions has been accumulated, the connections between them can be seen, and the structure becomes self-supporting — we should still strive to make things as intuitive as possible right from the start.  Below is one such attempt.  It tries to introduce group theory based around several examples, one example at a time. Particular emphasis is placed on explaining normal and non-normal subgroups. Key points will be written in bold. (Caveat: The below captures what first came to mind and is likely to be unduly verbose.)

## Free Group with Two Generators

Group theory came from the study of “transformations” of a space, and often, it is useful to think of each element of a group as something which “acts on” or “transforms” something. Later, we will consider permutation groups, which precisely fit this description. (In the reading group, we consider rotation groups, which also fit this description precisely.) Here, we consider a different group first, giving two equivalent definitions of it.

The algebraic definition is that our group $G$ consists of all “words” that can be written using the letters $a$, $b$, $a^{-1}$ and $b^{-1}$, with the cancellation rule that whenever $a$ and $a^{-1}$ appear next to each other, they can be omitted from the word, and the same for $b$ and $b^{-1}$. Besides the “empty word”, denoted $e$, the group $G$ contains elements such as $aaaba$, $ababa$, $ab^{-1}b^{-1}aba^{-1}$ and so forth. The group operation (often written as multiplication) is just concatenation, so that $ab$ times $aabb$ is $abaabb$.

Since $abb^{-1}a$ is the same as $aa$ by definition of our cancellation rule (later, a physical interpretation will be given), note that $ab$ times $b^{-1}a$ equals $aa$.

The axioms of a group require the group operation to be associative, and it is clear that concatenation is associative, i.e., the order of doing the multiplications does not matter: ( (word one) times (word two)) times (word three) gives the same as (word one) times ((word two) times (word three)).  For example, $((ab) \cdot (ab)) \cdot (ab) = ababab = (ab) \cdot ((ab) \cdot (ab))$.

There must also be an identity element. In this case, this is the empty word $e$, and it is clear that adding a “blank” to the start or the end of a word leaves the word unchanged, hence the empty word really is the identity element. (Time precludes me from writing down the group axioms and verifying them in detail.)

Lastly, to verify $G$ is a group, it must be shown that every element has an inverse. This is intuitively clear; for example, the inverse of $ab^{-1}ab$ is $b^{-1}a^{-1}ba^{-1}$ because $ab^{-1}ab \cdot b^{-1}a^{-1}ba^{-1} = e$. This example shows the inverse can be constructed one letter at a time, going from right to left, writing down the inverse of each letter. (Note that the blank word also results if the inverse had been multiplied on the left rather than the right: $b^{-1} a^{-1} b a^{-1} \cdot a b^{-1} a b = e$. If this were not true, then $G$ would not be a group, i.e., the axioms require that each element have a left inverse which is also a right inverse.)

While the above algebraic definition shows that this group, known as the free group with two generators, is easy to work with algebraically, it lacks a physical interpretation. This is now remedied by thinking of the free group with two generators as the fundamental group of the plane with two holes removed; don’t worry if that does not make any sense as I will explain the concept intuitively.

Imagine the floor of a room has an X marked on it, and to the left of the X is a vertical pole sticking out of the ground, and to the right there is another pole.  A spider starts at X and can be commanded to do four things.  Command “a” will make the spider walk from X clockwise around the pole on the left and return to X, all the while leaving a thin web behind him. (If you prefer, think of a robot laying down a rope.) Command “a inverse” will make the spider leave X, walk anti-clockwise around the pole on the left then return to X.  Notice that if the spider does “a” then ”a inverse” the web can be shrunk or contracted back to X, so “a” followed by “a inverse” is the equivalent of “do nothing”, because we want to treat two web configurations as the same if one can be deformed into the other without breaking the web. (Technically, the web is allowed to pass through itself so we do not need to worry about knots forming; we simply care about what happens as the web shrinks to as short a length as possible.)  Note that “a” followed by “a” gives a configuration (denoted $aa$) that is different from just instructing the spider to do “a” (denoted $a$); the former has the web encircling the pole on the left twice while the latter has the web encircling the pole only once, and no amount of jiggling of the web will transform one configuration into the other.

Similarly, commands “b” and “b inverse” can be defined to mean walk clockwise or anti-clockwise around the pole on the right, respectively. The reader should pause to ensure the algebraic definition and this spider-web definition agree intuitively by doing several thought experiments and visualising the resulting web configuration.

A subgroup can be thought of as what results if the spider is no longer allowed to carry out arbitrary commands but can only carry out certain sequences of commands and their inverses.  For example, perhaps the spider can only carry out the command $ab$ and its inverse $b^{-1}a^{-1}$.  Then the resulting subgroup consists of $\cdots, (b^{-1}a^{-1})^3,(b^{-1}a^{-1})^2,b^{-1}a^{-1},e,ab,(ab)^2,(ab)^3,\cdots$. (This subgroup is isomorphic to the group of integers.)  Another subgroup is obtained by simply allowing the spider to carry out command “a” or its inverse.  (This subgroup is also isomorphic to the group of integers.)

Let $H$ denote a subgroup of $G$. Recall the definition of normality: $H$ is a normal subgroup of $G$ if, for every $g \in G$ and $h \in H$, the so-called conjugate $g^{-1} h g$ is in $H$.  This definition will have been motived several times before the end of this article, however, for now, we start with a high level if not somewhat vague explanation: a normal subgroup is one whose elements produce actions that do not get “tangled up” very much with all the other actions, as now discussed.

If $H$ is the subgroup generated by $a$, meaning the elements of $H$ are precisely $\cdots,a^{-3},a^{-2},a^{-1},e,a,a^2,a^3,\cdots$ then $H$ is not normal. For example, $b^{-1} a b$ cannot be written as an element of $H$ because there is no way for the actions of $a$ and $b$ to be exchanged: if first the spider is given the command to wrap a web around the right pole clockwise, then apply a command from $H$, then wrap a web around the right pole anti-clockwise, there is no way for the web around the right pole to “cancel out” and leave a web that is only wrapped around the left pole; the subgroup $H$ is not normal.

Why is normality interesting? Before giving the standard explanation (i.e., the left cosets of a subgroup form a quotient group if and only if the subgroup is normal, and essentially equivalently, a subgroup is the kernel of a group homomorphism if and only if it is normal), a high level pragmatic explanation is proffered: when we are writing down sequences of group operations, we want to be able to simplify things, so we need some way of being allowed to interchange operations. In the simplest of cases, there are so-called Abelian groups which, by definition, are commutative: $gh = hg$ for all $g$ and $h$ in the group. The simplest examples of Abelian groups are vector spaces with the usual vector addition being treated as the group operation (and scalar multiplication is forgotten about, except for multiplication by $-1$ which is the inverse group operator). In an Abelian group, $g^{-1} h g = g^{-1} g h = h$, hence every subgroup is normal. This is the ultimate in “interchangeability”.

A normal subgroup has a property similar to, but a bit weaker than, full commutativity, in the following sense. If $H$ is normal, and $h \in H$, then although $gh$ may not equal $hg$, I can achieve the next best thing: I can find a $\tilde h \in H$ such that $g h = \tilde h g$.  That is, I have been able to interchange the action $g$ with an action from $H$ provided I do not mind using a possibly different action from $H$. (Note that this follows immediately from the definition of normality: $gh = ghg^{-1}g = \tilde h g$ where $\tilde h = g h g^{-1}$ is guaranteed to be in $H$ by normality. Note that $g$ can be replaced by $g^{-1}$ in the definition of normality without changing anything.)

At this point, the reader should be comfortable in believing that being normal is a special property not shared by every subgroup, and if a subgroup is normal, then it is a useful thing, because it allows one to manipulate group operations to achieve simplifications.

## The Symmetric Group on Three Letters

Consider three distinct balls arranged in a row.  A robot can be asked to change their ordering: one action might be to swap the left-most ball with the right-most ball. The set of all distinct actions forms a group $S_3$, called the symmetric group on three letters, with group multiplication $gh$ taken to mean “first perform action $h$ then perform action $g$”, which again is a form of concatenation. Googling will bring up more than enough information about symmetric groups, so here I just wish to emphasise thinking in terms of each element corresponding to asking a robot to perform an action.

Let $H$ denote the subgroup of $S_3$ formed from the action “interchange the left-most and middle balls”.  Denote this action by $(12)$, and note that it is its own inverse. (I am using cycle notation.) This is another example of a non-normal subgroup: consider $(13)^{-1} (12) (13)$, with the notation meaning, first swap the left and right balls, then swap the left and middle balls, then swap the left and right balls.  If the balls (red, green, blue) start off in the order RGB, then applying (13) gives BGR, applying (12) gives GBR and applying $(13)^{-1} = (13)$ gives RBG.  So overall, it transformed RGB to RBG.  Note that the right ball has been changed, yet no command from $H$ can change the right ball, hence $H$ is not normal.  This again is an illustration of the vague intuition that non-normal subgroups perform actions that tangle other actions up.

Besides the empty group and the whole group, it can be shown that the only other normal subgroup of $S_3$ is the so-called alternating group $A_3$ formed from all the even permutations.

Gaining further insight requires looking more closely at applications of normality. Perhaps the most germane to look at is whether a certain way of partitioning a group into equivalence classes leads to a compatible group structure on the equivalence classes. Recall $H$ is the subgroup of $S_3$ generated by (12). It turns out that any subgroup leads to a partition of the original group: $G = \cup_{g \in G} gH$ where $gH = \{gh \mid h \in H\}$ is a left coset. Crucially, for two distinct $g,g' \in G$, either $gH = g'H$, or $gH \cap g'H = \emptyset$. (Intuitively why this is so will be explained later. For now, an algebraic proof is given: Assume there exist $h,h' \in H$ such that $gh = g'h'$. For any $f \in H$, define $f' = (g')^{-1}gf$, so that $gf = g'f'$. The result follows if we can show $f'$ belongs to $H$. Note $f' = (g')^{-1}ghh^{-1}f = h'h^{-1}f$ which is in $H$ because $h'$, $h^{-1}$ and $f$ are all in $H$.)

If we treat a vector space as a group, and take for $H$ a linear subspace, then $gH$ would correspond to the affine subspaces obtained by translations of $H$; perhaps this is the simplest way of visualising left cosets.  (Vector spaces form Abelian groups, hence all their subgroups are normal, and simply considering vector spaces as groups does not lead to insight into non-normal subgroups.)

It can be deduced fairly readily that the elements of $S_3$ are (), (12), (13), (23), (123), (132).  (There are six elements because the left ball can be placed in one of three positions, the middle ball can be placed in one of two positions, and having done so, there is only one spot remaining for the right ball, i.e., there are $1 \times 2 \times 3 = 6$ possible re-arrangements.)

The cosets $()H$ and $(12)H$ are both just $H$ because the elements of $H$ are () and (12).

The coset $(13)H$ consists of the elements (13) and (13)(12). The latter transforms RGB into BRG and hence is equivalent to (123).  So the cosets $(13)H$ and $(123)H$ are the same; they contain the elements (13) and (123) of $S_3$.

The coset $(23)H$ consists of the elements (23) and (23)(12) = (132). Hence the cosets $(23)H$ and $(132)H$ are the same; they contain the elements (23) and (132) of $S_3$.

This $H$ is therefore seen to partition $S_3$ into three sets called equivalence classes: $S_3 = \{ (), (12) \} \cup \{ (13), (123) \} \cup \{ (23), (132) \}$.

Consider multiplying two elements from different equivalence classes of the partition. For example, (13) from the second equivalence class times (23) from the third equivalence class gives (13)(23)=(132), which lies in the third equivalence class. However, (123) from the second equivalence class times (132) from the third equivalence class gives (), which lies in the first equivalence class.  Therefore, there is no consistent way of defining what it means to multiply the second equivalence class with the third equivalence class, since choosing different elements to represent each equivalence class can lead to different answers.  (This should be compared with modulo arithmetic, where things work nicely.) Since $H$ is not normal, it tangles things up too much.

What is required for the equivalence classes to have a group structure?  If $f,f',g,g' \in G$ are such that $fH = f'H$ and $gH = g'H$, that is, $g$ and $g'$ represent the same equivalence class, and the same for $f$ and $f'$, then we want $fgH$ to be the same as $f'g'H$. This must be true in the special case when $f = ()$. If $f = ()$ then $f'$ must be an element of $H$.  (If $H = f'H$ then we can find $h,h' \in H$ such that $h = f'h'$, so that $f' = h(h')^{-1} \in H$.) So a necessary condition is for $gH = f'g'H$ when $f' \in H$. Recall from earlier that if $H$ were normal then we could interchange the order in which $f'$ and $g'$ are multiplied, albeit by having to change $f'$ to a different element of $H$. Thus, if $H$ is normal then $f'g' = g'f''$ for some $f'' \in H$, hence $f'g'H = g'f''H = g'H$, as required. With a little more care, it can be shown that a necessary and sufficient condition for left cosets to have a group structure, is for the subgroup to be normal. One way to see why normality suffices is by writing $(gH)(fH) = (gff^{-1}H)(fH) = (gf)(f^{-1}Hf)H = gfH$ because, by normality, $f^{-1}Hf$ will equal $H$. (This notation requires some justification which I do not elaborate on; treat it as a mnemonic, if you prefer.)

## Why do Cosets form a Partition?

For Abelian groups, such as vector spaces, it is relatively clear why left cosets form equivalence classes. Why does this hold for arbitrary groups, when multiplication can tangle things up?

Let $H$ be a subgroup of $G$.  First note two things: if $h \in H$ then $hH = H$, and if $g \notin H$ then $gH \cap H = \emptyset$. Proving the former is left as an easy exercise. The latter holds because, if $gh$ is in $H$, and if $h$ is in $H$, then $ghh^{-1} = g$ must also be in $H$.

We can now appeal to symmetry. Groups were invented to study symmetry, and in the first instance, this means often we understand what is going on by shifting things back to the “origin”, meaning the identity element of the group. If we want to understand what $gH$ and $g'H$ look like relative to each other, then we can premultiply them by $g^{-1}$. The key point is that this “transformation” is one-to-one and preserves structure, so we are allowed to do this without altering anything important. Thus, comparing $gH$ with $g'H$ is essentially the same as comparing $H$ with $g^{-1}g'H$. But we know from earlier that either $g^{-1}g'$ is an element of $H$, in which case $H = g^{-1}g'H$ and hence $gH = g'H$, or else $g^{-1}g'$ is not an element of $H$, in which case $H$ and $g^{-1}g'H$ are disjoint, and hence $gH$ and $g'H$ are also disjoint.

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