During the Twentieth century, mathematicians got in the habit of studying structures by looking at the functions that mapped one structure to another or to itself. Probably the most famous was in topology, where simple objects were made of playdough. Two playdough objects (e.g., a teacup and a donut) were equivalent if one could be deformed into the other without breaking, tearing, or fusing or welding: transform the playdough donut into a teacup by making and expanding a depression that grows into the teacup bowl – making a donut out of the teacup involves reversing the process. But the donut cannot be deformed into a saucer without welding the hole shut, and a saucer cannot be deformed into a donut without tearing open a hole. The notion of *genus*, how many holes a playdough object had, depended on the transformations available to deform one object into another.

We study groups the same way, looking at transformations from one group to another. Suppose we have two groups, G = (*G*, ∘) and H = (*H*, ∗); a *homomorphism* from G to H is a function f : *G* → *H* such that for any g_{1}, g_{2} ∈ *G*, f(g_{1} ∘ g_{2}) = f(g_{1}) ∗ f(g_{2}). Observe that if e is the identity of G, then f(e) is the identity of H, and for any g ∈ *G*, f(g^{-1}) = f(g)^{-1}.

For example, Let G be the symmetry group of the pentagon illustrated in the 30 January 2015 group theory post. Let Z_{2} be the group ({0, 1}, ∗) with the following multiplication table:

For each rotation *r* ∈ *G*, let f(*r*) = 0, and for each reflection m ∈ *G*, let f(*m*) = 1. Thus f outputs 0 if the isometry is direct (i.e. no mirrors) and 1 if it is indirect. This is an example of a homomorphism: recall that while the composition of two direct isometries (or the composition of two indirect isometries: the composition of two reflections across non-parallel mirrors is a rotation) is direct while the composition of a direct isometry and an indirect one is indirect.

One special kind of homomorphism from G to H determines a copy of G inside of H. Say that any kind of function f from *G* to *H* is *one-to-one* (or *injective*) if, for any g_{1}, g_{2} ∈ *G*, if g_{1} ≠ g_{2} then f(g_{1}) ≠ f(g_{2}). For example, the function f_{1}(*x*) = *x*^{3} from the real numbers to the real numbers is one-to-one, while the function f_{2}(*x*) = *x*^{2} is not (as -1 ≠ 1 while f_{2}(-1) = f_{2}(1)). A one-to-one homomorphism is called an *embedding* (or *monomorphism*). For example, take the group of integers mod 5, call it Z_{5}, and for each i ∈ {0, 1, 2, 3, 4}, let f(i) be the rotation of the pentagon by 72i degrees counterclockwise. Then f is an embedding from Z_{5} to the *rotation group* of the pentagon, and we can say that f *embeds* Z_{5} in the symmetry group of the pentagon.

Another special kind of homomorphism from G to H maps G onto all of H. Say that any kind of function f from *G* to *H* is *onto* (or *surjective*) if, for any h ∈ *H*, there is a g ∈ *G* such that f(g) = h. For example, the function f_{1}(x) = tan x maps the reals onto the reals, but the function f_{2}(x) = sin x does not. An onto homomorphism is called an *epimorphism*. Looking at the previous paragraphs, our mapping of the symmetry group of the pentagon onto Z_{2} was an epimorphism, while our embedding of Z_{5} into that symmetry group was not.

If we put these notions together we get: a function f from *G* to *H* is *bijective* if it is one-to-one and onto. A bijective homomorphism is called an *isomorphism*. For example, the map from Z_{5} onto the rotation group of the pentagon is an isomorphism defined by f(i) = the rotation by 72i degrees is an isomorphism. An isomorphism is often regarded as a relabeling of the names of group elements. For example, Z_{5} is a group of integers and the (noncrystallographic) point group **5** of rotations by multiples of 72 ° are not the same thing – one consists of integers and the other of rotations – but there is an isomorphism from one to the other, so we call them *isomorphism* and often treat them as if they were the same.

One special kind of isomorphism is an isomorphism from a group onto itself: such an isomorphism is called an *automorphism*. For example, if we mapped the rotation group of the pentagon onto itself by mapping a rotation by 72i degrees to a rotation by 144i degrees, that is an automorphism of that group.

Now that we have some functions from groups to groups, let’s briefly look at groups that are part of bigger groups, much as the rotation group of the pentagon is part of the symmetry group of the pentagon. If all the elements of a group G are elements of a group H, and if they have the same binary operator, we say that G is a *subgroup* of H, and that H is a *supergroup* of G. There are two popular ways of specifying a subgroup. In both, we have a group H and a set X of elements of H, and we want the subgroup of H that is “generated” by X. For example, given the symmetry group of the pentagon, the subgroup generated by the rotation by 72 ° is the rotation group of the pentagon, while the subgroup generated by one of the reflections is the group consisting of that reflection and the identity. (And the subgroup generated by any two reflections is the entire group.) What do we mean by “generated by”?

This is the kind of question a mathematician *would* ask, and here are two answers. First, the answer in all the mathematics texts. Let H be a group. It is a fact (see if you can verify it) that any intersection of subgroups of H is also a subgroup of H. Then one answer is to say that the *subgroup generated by* X is the intersection of all subgroups of H that contain all the elements of X; this subgroup is often denoted 〈 X 〉. This answer is nice and neat, which is why the texts use it, but it doesn’t tell us how to compute 〈 X 〉.

An alternative is the following recipe for constructing the subgroup generated by X. Let *G*^{0} = X. For any nonnegative integer n, given *G*^{n}, let *G*^{n+1} = {g, g ∘ h^{-1} : g, h ∈ *G*^{n}}. We get *G*^{0}, *G*^{1}, *G*^{2}, *G*^{3}, … . If H is finite, this sequence will eventually stop at some n where *G*^{n} = *G*^{n+1}, and this is (the set of elements of) our desired subgroup generated by X. If H is infinite, the sequence of sets *G*^{n} can keep growing forever, and it is their union *G*^{∞} that is the subgroup generated by X.

For example, given the symmetry group of the pentagon, and letting X be the set containing the rotation by 72 °, we have *G*^{0} = {Rot_{72 °}}, *G*^{1} = {Rot_{0 °}, Rot_{72 °}, Rot_{144 °}}, and *G*^{2} is the rotation group of the pentagon.

As an example, let’s look at a kind of subgroup that shows up a lot in crystallography. Let G be a group of bijective functions on a set or structure X. For any x ∈ X, let G(x) = {g(x) : g ∈ *G*}: this is the set of points that x is mapped to by the functions in *G*. For example, consider the *infinite dihedral group* (also known as **pm**) which acts on the number line and consists of two kinds of functions:

*Reflections across integers.*For each integer*i*, let Ref_{i}be the function Ref_{i}(*x*) = 2*i*–*x*. This reflects the number line across the integer*i*.*Translations by integers.*For each integer*i*, let T_{i}be the function T_{i}(*x*) =*x*+*i*.

Suppose that we had a real number *x* and we used the infinite dihedral group to move it around. If *x* was on an orange dot between *i* and *i* + 1/2 for some integer *i*, *x* would be translated to the left and right by increments of 1 onto something like the orange dots in this picture (where the blue mirrors stand at the integers).

Meanwhile, a mirror would reflect *x* to a real number between *j* – 1/2 and *j* for some integer *j*, and hence onto a green dot. In this picture, the orange dots are the images of *x* via translations while the green dots are images of *x* via compositions of reflection and translation.

If we started with *x* was between *i* – 1/2 and *i*, we would get the same picture, only now the green dots would be the images of *x* under translations while the orange dots would be the images under reflections + translations. Either way, the green and orange dots together make up the orbit of *x*.

But if *x* was an integer, or if *x* = *i* + 1/2 for some integer *i*, then things are a little different. If *x* was an integer, the translations and reflections map it to other integers, and the orbit of *x* is just the integers. Or if *x* = *i* + 1/2, then the translations and reflections map *x* to other half-way points, and the orbit of *x* is {*j* + 1/2 : j an integer}.

Mirrors and other landmarks appear in crystallography, so one note about these. Let G be a group acting on X, and suppose that the only function in *G* that maps *x* to itself is the identity. We then say that G *acts freely* on *x*. For example, for all *x* except the integers and the half-way points *i* + 1/2, *i* integer, the infinite dihedral group acts freely on *x*. On the other hand, suppose that several functions in *G* map *x* to itself. For example, in the infinite dihedral group, the reflection across the integer *i* maps *i* to itself. The *stabilizer* of *x* (which we can denote Stab(G, *x*)) is the subgroup of G that maps *x* to itself.

One important kind of subgroup comes from homomorphisms. Let f : *G* → *H* be a homomorphism from G to H. The *kernel* of f is the subgroup ker f = {g ∈ *G* : f(g) = e_{H}}, where e_{H} is the identity of H. Ker f is a subgroup of G, and it has an important property. Call a subgroup N of G *normal* if, for every g ∈ *G* and every n ∈ *N*, g ∘ n ∘ g^{-1} ∈ *N*. This is a peculiar condition, but as we will see in a moment, a useful one.

Suppose that N is a subgroup of G. For any g ∈ *G*, the *left coset* of N by g is the set g ∘ *N* = { g ∘ n : n ∈ *N* } while the *right coset* of N by g is the set *N* ∘ g = { n ∘ g : n ∈ *N* }. The left cosets form a partition of *G*, as do the right cosets, and N is a normal subgroup of G if and only if the left cosets form the same partition of *G* that the right cosets do. The reason is that N is normal in G if and only if g ∘ *N* ∘ g^{-1} = { g ∘ n g^{-1} : n ∈ *N* } = *N*, so that multiplying both sides of g ∘ *N* ∘ g^{-1} = *N* by g – on the right hand side of each – produces g ∘ *N* = *N* ∘ g.

For example, consider the point group **4m**, which consists of four mirrors passing through the origin. The four rotations of **4m** (counting the identity as a rotation by zero degrees) form a subgroup **4**. And **4** is a normal subgroup of **4m**: if *r* was a rotation (by a multiple of 90°), then for any *R* in **4m**:

- If
*R*was a rotation, then*R*∘*r*∘*R*=*r*is a rotation, and - If
*R*was a reflection, then*R*∘*r*∘*R*=*r*is a rotation.

On the other hand, consider the wallpaper group **p4**, with the unit square being a unit cell, and hence 90° rotation centers at all points (*i*, *j*), *i*, *j* integers. The stabilizer Stab(**p4**, (0, 0)) is not normal. To see this, let *r* be the 90° rotation about (1, 0), and we claim that *r* ° Stab(**p4**, (0, 0)) ° *r*^{-1} ≠ Stab(**p4**, (0, 0)). Let *R* be the rotation by 90° rotation about (0, 0), and it suffices to show that *r* ° *R* ° *r*^{-1} ∉ Stab(**p4**, (*0*, *0*)). If it was, then *r* ° *R* ° *r*^{-1}(0, 0) would be (0, 0); in fact, *r* ° *R* ° *r*^{-1}(0, 0) = *r* ° *R* (1, 1) = *r* (-1, 1) = (0, -2).

Returning to homomorphisms, a subgroup is normal if and only if it is a kernel of a homomorphism. First of all, given a kernel of a homomorphism f : *G* → *H*, call it ker f, for any g ∈ *G*, g ° ker f ° g^{-1} = ker f. To see this, take any k ∈ ker f, and observe that f(g ° ker f ° g^{-1}) = f(g) * f(k) * f(g^{-1}) = f(g) * f(g^{-1}) (as f(k) is the identity in H) = f(g) * f(g)^{-1}, which is the identity in H.

Secondly, given a normal subgroup N of G, let G/N be the set of left cosets of N in G: G/N = { g ∘ *N* : g ∈ G }. If N is normal, G/N is a group with binary operator ° as follows: (g ° *N*) ° (h ° *N*) = ((g ° (*N* ° h)) ° *N*) = ((g ° (h ° *N*)) ° *N*) = (g ° h) ° (*N* ° *N*) = (g ° h) ° *N* as *N* ° *N* = { n_{1} ° n_{2} : n_{1}, n_{2} ∈ *N* } = *N*. As *N* is the identity of G/N, and *N* is the set of elements of *G* that f maps to *N*, *N* = ker f.

G/N is called the *quotient group* of G and N. As we shall see, the point groups are quotient groups of the space groups.