## Group Theory: normal subgroups

I want to show you a simple concept. It has major consequences, but it starts out as a far simpler thing than it usually looks like.

I want to talk about normal subgroups. Clearly, if all subgroups were normal, we wouldn’t need to distinguish normal from non-normal… so I will have to show you a non-normal subgroup. That won’t be a problem.

Before I go into the details, let me give you the answer. Even if you know this material, the answer may surprise you.

A subgroup K of a group G is a normal subgroup of G if and only if K is the kernel of a homomorphism defined on G.

I know… I know: besides “normal”, I’ve used three other terms that I haven’t defined: subgroup, kernel, and homomorphism. If you’ve never seen this material before, you’re lost – but I can fix that.

Again, before I go into the details, let me give you the names of other things I will be talking about: fibers, left cosets and right cosets, and factor groups; I will briefly mention conjugacy classes.

My starting point will be to define a (group) homomorphism. It is just a function between groups which honors the group multiplications. (They do not need to be, and generally are not, the same operation in the 2 groups.)

That is, given 2 groups G and $\Gamma\$, we say that a function $\varphi:\ G \rightarrow \Gamma\$ is a (group) homomorphism if I have written it with 2 different symbols for the operations in order to emphasize that they are, in general, different. Nevertheless, we usually use juxtaposition (as for ordinary multiplication) and write We know that we may be using juxtaposition for two different operations. By the same token, we will denote both identities by 1 in general; they may not be the same, and they may not actually look like “1”. If the function $\varphi\$ is a bijection (one to one and onto; or injective and surjective), then it is called an isomorphism.

I daresay we are most interested in isomorphisms and in surjective homomorphisms (i.e not injective). Let’s look at some examples.

We already know of one isomorphism. We saw in the previous post that the dihedral group D3 of order 6 should be isomorphic to the symmetric group S3 of order 6.

Let me invoke the abstract algebra package and set it to groups: What are the orders of the elements? The left column shows D3… the right is S3; the order is printed under the element. I could wish that the positions of r^2 and r s had been swapped, but I’m not going to work at fixing it. Note that for these two groups, one identity is 1 but the other identity is the permutation {1,2,3}. In addition, each group has two elements of order 3 and three elements of order 2.

Let’s see if the simplest beginning will get us an isomorphism. Suppose I define a function which maps r (of order 3) to {2,3,1} (also of order 3); and which maps s (of order 2) to {1,3,2} (also of order 2). And, of course, define it to map 1 to {1,2,3}.

From that beginning I can use the preservation of the group operations to figure out how to define the rest of the function so that it is an isomorphism.

That is, I need r^2 to map to the product of {2,3,1} with itself, which is Why don’t I simplify things? I have already said that r -> {2,3,1} and s -> {1,3,2}… so let me define those images R and S: Then r s -> RS and r^2 s -> R^2 S… oh, I need to save R^2: I really do have an isomorphism from D3 to S3.

Having shown you just one example, let me define the kernel of a homomorphism as those elements of the domain (G) which are mapped to the identity in $\Gamma\$. In this example, the kernel is just the identity, {1} in G.

Now would be a good time to say that if we were working with vector spaces instead of groups, we could define a vector space homomorphism and its kernel. You’re probably used to them both – they’re called a linear transformation and its null space.

Let’s try another one. This time I want a homomorphism whose kernel is not just the identity. (I’m about to walk a minefield. I’ll come back, I assure you, and show you the mines.)

This time let the codomain $\Gamma\$ be a cyclic group of order 2, say {1, g}: I’m going to let $\varphi\$ map s -> g, but r -> 1. Then $\varphi(r^2) = \varphi(r) \varphi(r) = 1\$.

The kernel of this homomorphism is {1, r, r^2}.

What are the images of r s and r^2 s? if the function is to be a homomorphism, we must have the leftmost equalities on the next two lines, and that let;s us work out the answers: $\varphi(r s) = \varphi(r) \varphi(s) = 1 g = g\$ $\varphi(r^2 s) = \varphi(r^2) \varphi(s) = 1 g = g\$.

The case that gets short shrift in an introductory course – in fact, I think it got short shrift until the 1950s – is the case when $\Gamma\$ is bigger than G. For example, we could define the reverse function from C2 to D3 by $\varphi(g) = s\$.

The image of $\varphi\$, often denoted $\bar{G}\$, is {1, s} in $\Gamma\$; the kernel of $\varphi\$ is {1}; but $\varphi\$ is not an isomorphism because it is not onto D3: nothing in C2 is mapped to the elements r, r^2, r s, or r^2 s of D3.

Now would be a good time to talk about subgroups. Any text has plenty of material, but what it boils down to is that subgroups are perfectly well behaved. A subgroup S of a group G is a subset of the elements of G which is a group in its own right – with the same group operation. It must contain the same identity, and its inverses must be the same as they are in G. basically it’s just a subset which is closed under products and taking inverses.

(In fact, to make this point, Fraleigh talks about a subglob of a glob: a subset which is closed under the operations that make a set into a glob, whatever that may be.)

The easiest way, by far, to generate subgroups of a group G is to start with any one element, and consider all possible positive powers. This gives us a cyclic group which is a subgroup of G.

For example, in D3, each of the elements of order 2 (namely s, r s, r^s) generates a distinct subgroup of order 2: {1, s}, {1, r s}, and {1, r^2 s}. Each of the elements of order 3 (namely r and r^2) generates the same cyclic subgroup of order 3: {1, r, r^2}.

Having shown you all of three examples, let me summarize the properties of a homomorphism $\varphi:G \rightarrow \Gamma\$ with image $\varphi(G) = \bar{G}\$:

• the kernel K of $\varphi\$ is a normal subgroup of G;
• the image $\varphi(G) = \bar{G}\$ is a subgroup of $\Gamma\$;
• the image of the identity is the identity: $\varphi(1) = 1\$;
• the image of the inverse is the inverse of the image: $\varphi(x^{-1}) = \varphi(x)^{-1}\$;
• more generally, the image of any power is that power of the image: $\varphi(x^n) = \varphi(x)^n\$.

What I did in order to construct the isomorphism between D3 and S3, and the homomorphism between D3 and C2, was to define the function on the generators r and s, and use those relationships to extend it to a homomorphism. (But, as I said, I dodged mines in mapping D3 to C2 – when I chose the subgroup of order 3 as the kernel.)

Let me generalize the idea of the kernel. For starters, it is the preimage of the identity in the codomain $\Gamma\$, or in the image $\bar{G}\$. It is the elements of G which are mapped to the identity.

What about the preimages of the other elements of $\bar{G}\$? (If $\varphi\$ is not onto, then there are elements of $\Gamma\$ without preimages; “preimage” only makes sense for subsets of the image $\bar{G}\$. In this case, we are interested only in preimages of singleton sets, the individual elements of $\bar{G}\$.

We call the preimage of any element $\bar{g}\$ in $\bar{G}\$ the fiber over $\bar{G}\$. The kernel is then simply the fiber over the identity. This terminology, “the fiber over something”, shows up in many other places; in particular, it is crucial to grokking a fiber bundle. It’s worth learning.

Well, what are the fibers for our three homomorphims? For the two 1-1 examples – the isomorphism from D3 to S3, and the homomorphism from C2 into D3 – the fibers themselves are singletons.

Here’s the homomorphism from C2 into D3, which mapped 1 -> 1 and g -> s: For the isomorphism from D3 to S3, we have the following fibers. Ho hum. As I said, each fiber is a singleton.

Now here are the fibers for the homomorphism from D3 to C2. We have two fibers, and each contains 3 elements.

Now I want you to notice that one of these fibers is the kernel K – and the other is K s, that is { k s, k in K}, that is each element of K multiplied by s on the right. There’s a name for that: it’s called a right coset of K.

Let me recover the defintion of G, and define K… Then I can ask for the right cosets of K. One o them is K 1, i..e K itself, and the other is K s. Well, what about left cosets? Those should be s K: So the left coset is written as

s K = s {1, r, r^2} = {s, s r, s r^2}.

Time to look at the multiplication table and convert those so that powers of r come first. The multiplication table tells us that s r = r^2 s and s r^2 = r s, so we do in fact get the answer from the package:

s K = s {1, r, r^2} = {s, s r, s r^2} = {s, r^2 s, r s}.

As sets, we have equality:

s K = K s.

Is that a big deal?

Yes. And that’s an understatement.

We can prove that the fibers of any homomorphism are simultaneously the left and the right cosets of G with respect to K. That is, if we have a homomorphism, then the left and right cosets are equal to each other. We could speak of cosets without specifying right or left.

It’s time to see a case where they are not equal. Welcome to the minefield. Instead of taking K = {1, r, r^2} in D3, let’s take K = {1, s}… take a subgroup of order 2 instead of the subgroup of order 3. What are the left cosets? That is, we have

r^2 K = {r^2, r^2 s}

and r K = {r, r s}.

The right cosets? Do you see what I see?

K r^2 = {r^2, r s}

K r = {r, r^2 s}.

They’re scrambled. Each contains 1 element from r K and one from r^2 K. We do not have equality.

r K ≠ K r

r^2 K ≠ K r^2.

The subgroup {1, s} is not a normal subgroup. Otherwise its left and right cosets would be the same. Two ways to state the theorem are:

if $\varphi\$ is a homomorphism on G with kernel K, then the left and right cosets of the kernel K are equal: gK = Kg for all g in G.

Equivalently, if the left and right cosets of K are not equal, then a function with kernel K cannot be a homomorphism.

One of the left cosets, namely K = {1, s}, is the same as a left and as a right coset: 1K = K 1. And that’s the secret to knowing that C3 was a normal subgroup. One of its cosets was 1K = K1 = {1, r, r^2}, and the other cosets, sK and Ks, were the other three elements – so they had to be the same; there weren’t enough elements for making two different cosets.

(All the cosets of a subgroup K are the same size… that common size is called the index of K in G, written [G:K]. I’ve just argued that any subgroup of index 2 in G is normal in G. More importantly, the index of a subgroup in G times the order of the subgroup is equal to the order of G – so we have Lagrange’s theorem:

|G| = |K| [G:K],

where |G| and |K| symbolize the orders of G and K respectively.

We have an immediate corollary that |K| must divide |G|.

How did I know I needed to look at D3? Every subgroup of an abelian (commutative) group is normal. D3 is the smallest nonabelian group, so it’s the smallest possible example of a non-normal subgroup.

Let me come at this from the other side. Never mind the argument that {1,s} cannot be the kernel of a homomorphism… let’s try to build one anyway so we can see where we fail.

Suppose we define a function from D3 to C2 = {1,g} whose kernel is precisely {1, s}. That says both that 1 and s are mapped to 1… and that r, r^2, r s, and r s^2 are not mapped to 1. (Otherwise the kernel would have more elements than just 1 and s.)

If they are not mapped to 1, the only possibility is that they are mapped to g. The fiber over 1 is {1, s} and the fiber over g is { r, r^2, r s, r^2 s}. They’re not even the same size. We could say that the fiber over g is the union of two cosets r K $\cup \$ r^2 K.

Is this a homomorphsm? Consider r s: $\varphi(r s) = g\$ and $\varphi(r) \varphi(s) = g\$.

Good: they’re equal.

What about r^2 s = r (r s) $\varphi(r^2 s) = g\$ but $\varphi(r) \varphi(r s) = g g = g^2 = 1\$.

They’re not equal. No, it’s not a homomorphism.

So. Conceptually, some subgroups can be the kernel of a homomorphism, and some cannot. Those that can, we call normal. In principle, we could prove that a subgroup K was normal in G by finding a homomorphism with kernel K and domain G.

(After all this, it may not surprise you that “normality” is not hereditary. Suppose we have a group G, a normal subgroup K of G, and a normal subgroup H of K. Is H nomal in G? Not necessarily. H normal in K means that there exists a homomorphism defined on K, not on G, with kernel H. We have no reason to believe that that homomorphism can be extended from K to G, i.e. defined on all of G rather than just on K.)

Operationally, if a subgroup is the kernel K of a homomorphism, then its left and right cosets gK and Kg are equal; in practice, this is one way to recognize a normal subgroup.

There are two things I want to close with, but not to go into too much detail on. One is that if instead of writing

gK = Kg,

we write $gKg^{-1} = K\$,

we are actually doing something conceptually different. Two group elements a and b are said to be conjugate if there is some element g such that $g a g^{-1} = b\$.

(It’s not an accident that that looks just like similarity of matrices. In some sense, conjugate group elements are related by a change of coordinates, too.)

Conjugacy is an equivalence relation. For now, just take it that it is sensible to collect elements which are conjugate to each other, into conjugacy classes.

Then the equation $g K g^{-1} = K\$

means that every element of K is conjugate to some element of K: K contains complete conjugacy classes. This is a different characterization of normal subgroups. We also phrase this as: every element g of G normalizes K.

Finally, if the left and right cosets are equal (i.e. if the kernel is normal), then we may “lift” the group operation from the image $\bar{G}\$ to the fibers themselves. This gives us a new group – always remember that its elements are subsets of G, not elements of G ! – and this group is isomorphic to the image $\bar{G}\$. This group is called the factor group (or quotient group) and it is denoted G/K, and it is also called “G mod K”.

Who cares? We already have the homomorphism and its image $\bar{G}\$.

Or do we? True, given a homomorphism, we get its kernel K and image $\bar{G}\$… and the cosets gK (= Kg)….

But if we started with a normal subgroup K, we could compute the cosets gK – which are also the fibers of our as-yet-unknown homomorphism – and construct the factor group G/K… and that would be isomorphic to $\bar{G}\$. So I know what $\bar{G}\$ looks like – and I don’t even know what the homomorphism is. Hell, I don’t even know what the codomain is for the function!

This may be the reason for focusing on the factor group: I don’t need to construct an explicit homomorphism. It’s enough that I know one exists – actually, an infinite number exist, depending on what I choose for the codomain.

We’re done. Time for another deep breath.

A subgroup K is normal in G if and only if any of the following is true:

• K is the kernel of a homomorphism defined on G.
• the left and right cosets of K are equal: gK = Kg.
• every element g of G normalizes K: $K = gKg^{-1}\$.
• we can construct the factor group G/K from the cosets of K.

You might see it differently, but for me the first of those is the primary meaning of “K is normal in G”. All four properties are essential in one way or another. We need to know them all. But it makes sense to me to go from the kernel of a homomorphism to the left and right cosets being equal… then to the relationship between normality and conjugacy… and finally to making a group out of the cosets of a normal subgroup even when we don’t have a relevant homomorphism in hand.

Let me say that again, differently.

If we have a homomorphism with kernel K, then the fibers of the homomorphism are the cosets (left and right) of K. Conversely, if the corresponding left and right cosets of some subgroup K are equal to each other, then we may take them as being the fibers of a function – and that function will be a homomorphism with kernel K. So, K is the kernel of a homomorphism if and only if its left and right cosets are equal. Not all subgroups have this property, and we call those that do, “normal”.

Furthermore, the innocuous-seeming algebra that goes from saying left cosets are equal to right cosets,

gK = Kg

to $K = g^{-1}Kg\$

actually has mathematical content: normal subgroups contain entire conjugacy classes.

Next, we may define a group operation on the fibers of the homomorphism. (This seems natural to me.) Equivalently, since the fibers are the cosets (left and right), we may define a group operation on the cosets. (That came out of left field, for me. Of course it’s the same thing – but my perception was different.) The slick thing is that we don’t need to have the homomorphism in hand in order to work with its cosets, so we can charge ahead and get the factor group G/K, isomorphic to the image of every homomorphism on G whose kernel is K.

Finally, the construction of factor groups is one of the most important tools for decomposing groups into smaller pieces. A group G which has no proper normal subgroups – hence, roughly speaking, no proper factor groups – is called simple. (The identity and the group itself are always normal in G, so we don’t count them. And now you know why I fear I’m punning every time I use the word “simple” in a post about normal subgroups.)

The classification of all the finite simple groups was accomplished by 1980 I think… it was a huge collaboration. The smallest simple group, other than the cyclic groups of prime order, is A5, the alternating group on 5 elements – equivalently, the even permutations in S5, the symmetric group on 5 elements. And, ultimately, the simplicity of An for n >= 5 is why the general polynomial of degree 5 or higher cannot be solved in radicals!

### 2 Responses to “Group Theory: normal subgroups”

1. […] At this point a picture is nice (I got the idea from a post in Rip’s Applied Mathematics Blog, here). […]

2. denciu Says:

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