Sage

Sage is able to create homomorphisms (and by extension, isomorphisms and automorphisms) between finite permutation groups. There is a limited supply of commands then available to manipulate these functions, but we can still illustrate many of the ideas in this chapter.

The principal device for creating a homomorphism is to specify the specific images of the set of generators for the domain. Consider cyclic groups of order \(12\) and \(20\): \[\begin{aligned}G &= \{a^i\vert a^{12}=e\} & H &= \{x^i\vert x^{20}=e\}\end{aligned}\] and define a homomorphism by just defining the image of the generator of \(G\), and define the rest of the mapping by extending the mapping via the operation-preserving property of a homomorphism. \[\begin{aligned}\phi: G\rightarrow H, &\quad\phi(a)=x^5 \\ \Rightarrow &\quad\phi(a^i) = \phi(a)^i = (x^5)^i = x ^{5i}\end{aligned}\] The constructor PermutationGroupMorphism requires the two groups, then a list of images for each generator (in order!), and then will create the homomorphism. Note that we can then use the result as a function. In the example below, we first verify that C12 has a single generator (no surprise there), which we then send to a particular element of order \(4\) in the codomain. Sage then constructs the unique homomorphism that is consistent with this requirement.

Note that the element c must therefore be in the kernel of phi.

We can then compute the subgroup of the domain that is the kernel, in this case a cyclic group of order \(3\) inside the cyclic group of order \(12\). We can compute the image of any subgroup, but here we will build the whole homomorphic image by supplying the whole domain to the .image() method. Here the image is a cyclic subgroup of order \(4\) inside the cyclic group of order \(20\). Then we can verify the First Isomorphism Theorem.

Here is a slightly more complicated example. The dihedral group \(D_{20}\) is the symmetry group of a \(20\)-gon. Inside this group is a subgroup that is isomorphic to the symmetry group of a \(5\)-gon (pentagon). Is this a surprise, or is this obvious? Here is a way to make precise the statement \(D_{20}\) contains a copy of \(D_{5}\).

We build the domain and find its generators, so we know how many images to supply in the definition of the homomorphism. Then we construct the codomain, from which we will construct images. Our choice here is to send a reflection to a reflection, and a rotation to a rotation. But the rotations will both have order \(5\), and both are a rotation by \(72\) degrees.

Condensed: the full section is in Judson, Abstract Algebra: Theory and Applications.