Normal Subgroups

Let us now return to the study of the properties of the subgroups of a group. Our major task will be to introduce the extremely important concept of a normal subgroup.

Throughout this section, let G be a group. If A and B are arbitrary subsets of G, let us set

AB = {a · b | a A, b B}.

Since multiplication in G is associative, we see that for any subsets A, B, and C of G, we have

We will denote each of these sets by ABC. If A = {a}, we will write aB (respectively Ba) instead of {a}B (respectively, B{a}). Note that this notation coincides with our previous use of aB and Ba for left and right cosets in the case where B is a subgroup of G.

Proposition 1: Let H be a subgroup of G, a G. Then aHa^-1 is a subgroup of G.

Proof: From the definition of our notation, we see that

aHa^-1 = {aha^-1 | h H}.

Let , aHa^-1. By Proposition 2 of the section on subgroups, it suffices to show that ·^-1 aHa^-1. Since , aHa^-1, there exist x,y H such that

= axa^-1    = aya^-1.

Now ^-1 = (aya^-1)^-1 = ay^-1a^-1. Therefore,

·^-1=(axa^-1)(ay^-1a^-1) = a(xy^-1)a^-1 aHa^-1.

We say that the subgroup aHa^-1 is conjugate to H. Usually, aHa^-1H. That is, most subgroups H have conjugates other than themselves. However, the subgroups which coincide with all of their conjugates are a very important class of subgroups. They play a very special role in group theory. Let us single them out in the following:

Definition 2: Let H be a subgroup of G. If for every a G, we have aHa^-1 = H, then H is called a normal subgroup of G. (The terms invariant subgroup and self-conjugate subgroup are also used.) A normal subgroup H of G is denoted HG.

Before giving examples of subgroups which are and are not normal, it is convenient to establish a simple criterion which can be used to determine whether or not a subgroup is normal.

Proposition 3: Let H be a subgroup of G. Then H is normal if and only if for every a G, we have aHa^-1 H.

Proof: Assume first that H is normal. Then for every a G, we have aHa^-1 = H, so that obviously we have aHa^-1 H. Conversely, assume that aHa^-1 H for all a G. Let us prove that H is normal. Let b G. Then, by our assumption,

bHb^-1 H.

However, since (b^-1)^-1 = b,

b^-1Hb = b^-1H(b^-1)^-1 H.

Therefore, if h H, h' = b^-1hb H. so that

h = bh'b^-1 bHb^-1.

Since h is any element H, we see that

(2)

H bHb^-1.

Therefore, by (1) and (2) we see that

H = bHb^-1.

Since our choice for b G was arbitrary, we have proved that H is normal.

Example 1: If G is any group, {1} and G are both normal subgroups. These normal subgroups are said to be trivial.

Example 2: If G is abelian, every subgroup H is normal. For if a G, h H, we have aha^-1 = aa^-1h = h.

Example 3: Let G = D_n, H = [R]. We assert that H is a normal subgroup of G. For certainly we have

R^aHR^a H    (a = 0, 1, 2,..., n-1)

Further since FRF^-1 = R^-1 = R^n-1, we have

FHF^-1 H.

Therefore,

(FR^a)H(FR^a)^-1 = F(R^aHR^a)F^-1 FHF^-1 H

for a = 0, 1,...,n-1. But every element of G is of the form R^a or FR^a (a = 0, 1,...,n-1). Therefore, by Proposition 3, H is normal.

Example 4: Let G = D_n, H = [F]. Since FRF = R^-1, we have

and since F = F^-1, R^-1 = R^n-1, Rⁿ = I,

Since n > 3, we see that 0 < n - 2 < n, so that R^n-2 I. Therefore, FR^n-2 H, so that RFR^-1 H so RHR^-1 H. Thus, H is not a normal subgroup of G.

Let H be a subgroup of G and let us consider the sets H\G and G/H of right and left cosets respectively. A typical left coset is a set of the form aH (a G); a typical right coset is a set of the form Ha (a G). It is not generally true that a left coset is also a right coset. The important thing to note, however, is that if H is normal, then for every a G we have aHa^-1 = H. Therefore,

Thus we have proved

Proposition 4: If H is a normal subgroup, then every right coset Ha of G with respect to H is also a left coset - namely aH; and conversely, every left coset is a right coset. In other words if H is normal, then G/H = H\G.

We will now go through a construction which will appear in various guises a number of times in our study of algebra. So let us try to get some sort of broad insight into the general philosophy which underlies the general philosophy which underlies the construction. The primary mathematical objects we have been discussing so far have been sets. Not just arbitrary sets, however. Our sets in this chapter have been endowed with some extra structure - they are groups. We have already considered the problem of translating this extra structure from the original set to one of its subsets. What we arrived at were subgroups. Now we will consider a similar problem. Suppose that we have an equivalence relation on our original set. Can we translate the structure on the original set to the set of equivalence classes? To be more specific, let us consider a group G and a subgroup H. In or proof of Lagrange's theorem, we considered the equivalence relation on G defined by

a ~ b if and only if ba^-1 H.

The equivalence classes turned out to be left cosets of G with respect to H. The set of equivalence classes is therefore G/H. In reference to our general question raised above, we can now ask: Can we define a group structure of G/H? Of course we aren't interested in just any group structure on G/H. We are interested in a group structure which is naturally connected with the group structures on G and H. The first law of multiplication which comes to mind is the one defined by

The very simplicity of this definition leads us to suspect that it should be the "natural" multiplication of G/H. However, we run into problems. It is not in general true that (4) defines a binary operation on G/H. What can go wrong is that the "product" defined by (4) may depend on the choice of a and b and not merely on the cosets aH and bH. However, this difficulty goes away if H is a normal subgroup of G. Let us prove this non-trivial statement.

What we claim is that (4) defines a binary operation on G/H in the case where H is a normal subgroup of G. We remarked above that G/H is the set of equivalence classes of G with respect to the equivalence relation ~ defined by (*). Provided that ~ is compatible with multiplication in G, The rules of an equivalence relation imply that (4) defines a binary operation on G/H. This a, b, c, d G, and assume that a ~ b and c ~ d. Let us show that ac ~ bd. This will imply the compatibility of ~. But ac ~ bd if and only if (ac)(bd)^-1 H. However, since a ~ b and c ~ d, we see that b = ah₁, d = ch₂ (h₁,h₂ H). Therefore, (ac)(bd)^-1=acd^-1b^-1=ah₂^-1b^-1. However, since H is normal in G, Hb^-1 = b^-1H, so that h₂^-1b^-1 = b^-1h₃ for some h₃ H. Thus, (ac)(bd)^-1 = ab^-1h₃ = h₁^-1h₃, and thus (ac)(bd)^-1 H, which implies that ac ~ bd, which is what we desired to prove.

Theorem 5: Let H be a normal subgroup of G. Then with respect to the rule of multiplication (4), G/H is a group. The identity element is the coset H = 1H, and the inverse of aH is a^-1H.

Proof: It suffices to verify the group axioms (G1-G3).

Associativity:aH ·(bH · cH) = aH · bcH = (a · bc)H = (ab · c)H = abH · cH = (aH · bH) · cH, where we have made essential use of the associativity of multiplication in G.

Identity: 1H · aH = (1 · a)H = aH. Similarly, aH · 1H = aH.

Inverse: aH · a^-1H = aa^-1H = 1H. Similarly, a^-1H · aH = 1H.

One further comment about the definition (4) of multiplication of G/H is in order. It is possible to interpret the product aH · bH as the product of sets - aH and bH - which we defined as ah · bH = {m · n | m aH, n bH} = {ahbh' | h,h' H}. If H is normal, then Hg = gH for all g G, so that aH · bH, when interpreted as a product of two sets, equals a · Hb · H = a · bH · H = ab · H · H = abH. Therefore, the product of aH and bH as subsets of G coincides with the product defined by (4), and out notation should not lead to any confusion.

Definition 6: Let H be a normal subgroup of the group G. Then the set of cosets G/H with the group structure defined above, is called the quotient group of G modulo H. The term quotient group is used interchangeably with the term factor group.

Example 5: Let G denote the group Z with respect to addition, and let H = nZ. It is clear that H is a subgroup of G, and H is normal since G is abelian. A typical element of Z/nZ is a coset a + nZ, which consists of all integers of the form a + nz (z Z). Thus, we see that the element of Z/nZ is a residue class modulo n. The law of addition in the quotient group Z/nZ is just the addition of residue classes which we defined earlier. Thus Z/nZ is the group Z_n of residue classes modulo n.

Example 6: Let G be the group of all functions f : RR, where the group operation is addition of the functions; let

H = {f | f(0) = 0}.

Then H is a normal subgroup of G. The elements of G/H are cosets f + H (f ). Of course, not all of these cosets are distinct. Therefore, let us now describe the elements of G/H more explicitly. For each real number a, let us choose a function f_a such that f(0) = a. Then the cosets f_a + H belong to G/H. Let us prove that every element of G/H is a coset of this form and that all these cosets are distinct. If f + H G/H, let us choose a R such that f(0) = a. Then

Therefore, f - f_a H f + H = f_a + H, so that every function of G/H is of the form f_a + H for some a R. If f_a + H = f_b + H, then f_a = f_b + h for some h H. But since h(0) = 0, f_a(0) = f_b(0), and thus a = b. Therefore, all the cosets f_a + H are distinct and

G/H = {f_a + H | a R).

It is easy to see that (f_a + H) + (f_b + H) = f_a+b + H. Therefore, the mapping

k:R G/H

defined by k(a) = f_a + H is a surjective isomorphism, so that G/H R.