Understanding the notions of order and index

This survey article studies the notion of order and index of a subgroup; the relation between the orders of different subgroups, and the interplay with number theory.

Other related survey articles at a more advanced level include arithmetic and normal subgroup structure.

Order of a group
The order of a group is the cardinality of its underlying set. A finite group is a group whose order is finite.

The trivial group has order $$1$$. Every group has order at least $$1$$, because it contains the identity element.

The order of an element is the order of the cyclic subgroup generated by that element. Equivalently, it is the smallest positive integer $$n$$ such that the $$n^{th}$$ power of that element is zero; if there is no such positive integer $$n$$, the element is said to have infinite order.

Index of a subgroup
The index of a subgroup is the number of left cosets of the subgroup in the group. Equivalently, it is the number of right cosets of the subgroup in the group.

The index of a subgroup $$H$$ in $$G$$ is denoted $$[G:H]$$.

General setup
Given a subgroup $$H$$ of a group $$G$$, we can consider subsets of $$G$$ of the form $$gH, g \in G$$. Such subsets are termed left cosets of $$H$$. There are two important facts about left cosets of a single subgroup:


 * Left cosets are in bijection via left multiplication: Any two left cosets of $$H$$ have the same size. In fact, if $$xH$$and $$yH$$ are two left cosets, then the left multiplication map by $$yx^{-1}$$ establishes a bijection from $$xH$$ to $$yH$$.
 * Left cosets partition a group: $$G$$ is a disjoint union of the left cosets of $$H$$ in $$G$$.

For finite groups
Together, these facts yield Lagrange's theorem: When $$G$$ is a finite group and $$H$$ is a subgroup, the order of $$G$$ is the product of the order of $$H$$ and the index $$[G:H]$$.

$$|G| = |H| [G:H]$$.

Thus, for a finite group, knowing any two of the three quantities (the order of the group, the order of the subgroup, and the index of the subgroup in the group) allows us to determine the third. Further, we get an important constraint on the orders of possible subgroups: every subgroup has order dividing the order of the group.

For infinite groups
For infinite groups, we again have:

$$|G| = |H|[G:H]$$.

Here, the orders and indices are viewed as infinite cardinals. The order of the subgroup as well as its index in the whole group determine the order of the group. However, knowing the order of the group and the order of the subgroup does not yield the index of the subgroup. Similarly, knowing the order of the group and the index of the subgroup does not yield the order of the subgroup. For instance:


 * In the group of integers $$\mathbb{Z}$$, all the subgroups of the form $$n\mathbb{Z}$$ have the same order (i.e., they are all countable) but they all have different index.
 * In any infinite group, different finite subgroups have the same index, even if their orders differ.

The following deductions can be made for infinite groups:


 * Subgroup of finite index in infinite group has same order
 * Index of finite subgroup in infinite group equals order of group
 * Subgroup of countable index in uncountable group has same order
 * Index of countable subgroup in uncountable group equals order of group

Index is multiplicative
If $$K \le H \le G$$ are groups, then we have a natural surjective map from $$G/K$$ to $$G/H$$, where the inverse image of every element has size equal to the size of $$H/K$$. Thus, in the sense of possibly infinite cardinals, we have:

$$[G:K] = [G:H][H:K]$$.

If any two of these numbers is finite, so is the third, and in that case, the result holds in the sense of multiplication of positive integers.

Lagrange's theorem is a special case of multiplicativity of index, obtained by setting $$K$$ to be trivial.

Product formula
Suppose $$H_1, H_2 \le G$$ are subgroups. Then, we have a natural bijection:

$$H_1H_2/H_2 \leftrightarrow H_1/(H_1 \cap H_2)$$.

Note that $$H_1H_2$$ is not in general a group, but it is still a disjoint union of left cosets of $$H_2$$, so we get, in the finite case:

$$|H_1H_2||H_1 \cap H_2| = |H_1||H_2|$$.

The product formula has a number of important consequences, both in its set-theoretic and in its numerical form.

The transfer inequality for index
If $$H, K \le G$$ are subgroups, then we have:

$$[K:H \cap K] \le [G:H]$$.

This follows directly from the product formula setting $$K = H_1, H = H_2$$, and observing that the number of cosets of $$H$$ in $$G$$ is at least as much as the number of cosets of $$H$$ in $$HK$$.

Note that if $$HK = KH$$, then $$HK$$ is a subgroup, and in that case we get a somewhat stronger result:

$$[K : H \cap K ] = [HK : H ] | [G : H]$$.

In other words, instead of just an inequality, we get a divisor relationship.

The intersection inequality for index
The intersection inequality basically states that subgroups of small index have to have an intersection of small index. Specifically, it states that if $$H, K \le G$$ are subgroups, the index of $$H \cap K$$ is bounded by the product of the index of $$H$$ and the index of $$K$$.

The intersection inequality is a combination of multiplicativity and the transfer inequality.

Index of intersection of permuting subgroups
If $$H, K \le G$$ are permuting subgroups (i.e., $$HK$$ is a subgroup), then we have a divisor relationship:

$$[G:H \cap K] | [G:H][H:K]$$.

This is a stronger statement than the mere inequality we have in general.

The divisor relation holds whenever either of the subgroups is normal, because a normal subgroup permutes with every subgroup.

More on intersections
We just saw two results about intersections for two subgroup $$H$$ and $$K$$:


 * $$[G:H \cap K] \le [G:H][G:K]$$.
 * $$[G:H \cap K] | [G:H][G:K]$$ when $$HK$$ is a subgroup.

The multiplicativity of index also tells us that:


 * $$[G:H] | [G:H \cap K]$$
 * $$[G:K] | [G:H \cap K]$$

Combining these, we get that:

$$\operatorname{lcm}([G:H],[G:K]) | [G:H \cap K]$$.

This gives a lower bound on the index of the intersection.

Intersections of subgroups of relatively prime index
If $$H, K \le G$$ are subgroups of relatively prime finite index in $$G$$, then we have:

$$[G:H \cap K] = [G:H][G:K]$$.

This follows by combining the lcm inequality and the product inequality.

Notice that in this case, we also have $$HK = G$$.

On joins
Given subgroups $$H, K$$ of $$G$$, we have:


 * The order of $$\langle H, K \rangle$$ (the join of subgroups) is a multiple of the lcm of the orders of $$H$$ and $$K$$.
 * The index of $$\langle H, K \rangle$$ is a divisor of the gcd of the indices of $$H$$ and $$K$$.
 * Since $$\langle H, K \rangle \supseteq HK$$, the order of $$\langle H, K \rangle$$ is at least equal to $$|H||K|/(|H \cap K|)$$.
 * If $$HK$$ is a subgroup, it equals $$\langle H, K \rangle$$, and its order is exactly equal to $$|H||K|/(|H \cap K|)$$.

A first step: order of quotient divides order of group
While Lagrange's theorem relates the order of a group and the order of its subgroups, it also does something quite different: it relates the order of a group and the order of its quotient groups. Specifically, if $$\varphi:G \to H$$ is a homomorphism of groups, then the first isomorphism theorem states that if $$N$$ is the kernel of $$\varphi$$, $$\varphi(G) \cong G/N$$, and by Lagrange's theorem, the order of $$G/N$$ divides the order of $$G$$.

Thus, we see that for any homomorphism, the order of the image divides the order of the group.

Homomorphisms between groups of coprime orders are trivial
Suppose $$G$$ and $$H$$ are finite groups whose orders are relatively prime. Then, if $$\varphi:G \to H$$ is a homomorphism, the order of $$\varphi(G)$$ divides the order of $$G$$. Since $$\varphi(G)$$ is a subgroup of $$H$$, the order of $$\varphi(G)$$ divides the order of $$H$$. By our assumption of relatively prime order, we obtain that $$\varphi(G)$$ must be trivial, and thus, the homomorphism is the trivial homomorphism.

Normal Hall subgroups are order-unique
Suppose $$G$$ is a group and $$H$$ is a Hall subgroup of $$G$$. In other words, the order of $$H$$ is relatively prime to its index in $$G$$. Then, if $$H$$ is also normal in $$G$$, it is the only subgroup of its order.

There are two ways of seeing this. One way is to assume there is another Hall subgroup $$K$$ of the same order. In that case, $$HK$$ is a subgroup because $$H$$ is normal, and the order of $$HK$$ is, by the product formula:

$$|HK| = \frac{|H||K|}{|H \cap K|}$$

Thus, the order of $$HK$$ has no prime factors other than those of the order of $$H$$ and the order of $$K$$. The Hall condition then forces that $$HK = H$$, so $$H = K = H$$.

Another approach is to consider another Hall subgroup $$K$$. Now, consider the quotient map $$G \to G/H$$. Consider the restriction of this map to $$K$$. This is a homomorphism between groups $$K$$ and $$G/H$$, whose orders are relatively prime, so it is the trivial homomorphism. Thus, $$K$$ is in the kernel of the quotient map, forcing $$K \le H$$. By order considerations, $$K = H$$.

Both these proofs generalize in slight different ways, giving rise to the notions of pi-core and pi-closure, where pi is a set of prime divisors of the order of the group.