An Illustrated Proof: Normality and Quotients

On this page we write a proof that a normal subgroup can be used to form a factor group.  We illustrate the proof with images from Group Explorer.  The intent is to show that Group Explorer can be used to make some nice intuition-strengthening images, and such images can be useful when seeking to appreciate a proof.

Our website also contains another illustrated proof; there we show that Z_m x Z_n is isomorphic to Z_mn if and only if m and n are relatively prime.  For more information about normal subgroups in Cayley diagrams, see the example on the Editing Cayley Diagrams page, or the example on the Editing the Multiplication Table page.

Theorem:  If a subgroup H of a group G is normal in G then the set of right cosets ^G/_H forms a group under the operation HaHb = Hab.

Proof:  Assume that H is a normal subgroup of G.  We will show that ^G/_H forms a group.

When we write ^G/_H, we mean the set of all right cosets Ha of H, for any a in G.  The group operation defined on this is HaHb = Hab.  But this definition is a bit worrying for the following reason.  It is defined in terms of a and b, which are certain elements chosen from G to form the cosets Ha and Hb respectively.  Whenever a definition begins with a choice, it is important to demonstrate that whatever choice is made, the same result is achieved.  This is called ensuring that the operation is "well-defined."  So one important thing for us to show is that defining Ha times Hb to be Hab is independent of the choices a and b.

The fact that we have to work with is that H is normal in G, which means that its left and right cosets coincide.  Let us take a moment to appreciate this visually.

What do left and right cosets look like?  In the Cayley diagrams that Group Explorer creates, arrows are read as left multiplication by a generator, and so the left coset aH can be seen by looking at the subgroup H and then following the a-arrows from it.  This is illustrated in the images to the right. Begin with the polar Cayley diagram for the group S3, with subgroup < f > highlighted in yellow. Following the red arrows, which correspond to the generator r, from < f >, leads us to the left coset, highlighted here in yellow. On the other hand, if we wish to find a right coset Ha, we must first follow the arrow from the identity to a, and then generate the subgroup H at that new location, because the arrows of the diagram represent left multiplication, and the H is on the left.  Using the same diagram again, but this time taking a right coset, we find the illustrations to the right. Begin with the same Cayley diagram for S3, with subgroup < f > highlighted in yellow. Follow the red arrow for the generator r, from the identity at the top, then consider the two-element segment shaped like H; it the right coset Ha, highlighted in yellow. As these examples show, left and right cosets do not always coincide.  The left cosets, for example, aren't even always shaped like H.  That is, the a-arrows from H don't all agree on which H-shaped piece of G to point to.  If they did, the left coset aH would also be a right coset Ha.  This is the defining characteristic for normal subgroups.  For a visual example of it, refer to the Example on Exhibiting Normal Subgroups on the Editing Cayley Diagrams page.

We return to the question of whether, given Ha times Hb, whether the coset Hab is different based on which a and b are chosen from the cosets Ha and Hb.  First we answer the question the standard way, by doing a careful proof.  Then we'll consider what the steps in our proof look like visually.

Let's take an arbitrary element from Ha and another from Hb, and prove that when we multiply them we always get something in Hab.  An element from Ha will look like h₁a for some h₁ in H, and one from Hb will look like h₂b for some h₂ in H.  So is h₁ah₂b in Hab?  That is, does h₁ah₂b = hab for some h in H?  Well, recall that ah₂ is in aH, which is the same as Ha, so there is some h₃ in H such that ah₂ = h₃a.  So if we let h = h₁h₃ then h₁ah₂b = h₁h₃ab = hab, which is in Hab.

Okay, but what does all this look like?  Let's take a look.

The following Cayley diagram shows the semidirect product of Z3 x Z3 with Z2.  It is organized so that each coset of the subgroup in yellow has its own color, and is chunked together by a translucent gray box.  Let us say that the generator a is shown with red arrows, and b with green lines.  Let us consider whether we can see that h1ah2b is in Hab just by looking at the diagram. Consider how we would find the element h1ah2b by following arrows from the identity element, which is the topmost yellow node.  Begin by following the b arrow to the right.  This puts us in the coset bH, or Hb.  Next, multiply by the element h2, which is an element of H, and so simply moves us within the purple coset; though we do not know which purple element we're at, we know that it's purple.  Next, multiply by the element a, which means following a red arrow.  But since red arrows take purple nodes to light blue ones, we must now be at a light blue node (having moved counterclockwise).  Lastly, apply the element h1, which is an element of H and therefore just moves us within the light blue coset. Thus we can see that the element h1ah2b must be in the light blue coset, so we ask, "Is the light blue coset Hab?"  Let us find the element ab by by following first a green arrow and then a red arrow from the identity.  We end up at a light blue node, and if we then proceed to generate the H-shaped portion of the diagram from that light blue node, we get the entire light blue coset mentioned earlier. So we can not only prove mathematically that h1ah2b is in Hab, we can also prove it visually.

We have shown that the operation HaHb = Hab is well-defined, and it is easy to check that it has all the properties of a group, because it inherits them from G.

	Associativity: (HaHb)Hc = (Hab)Hc = H(ab)c = Ha(bc) = Ha(Hbc) = Ha(HbHc)
	Identity: HeHa = Hea = Ha = Hae = HaHe
	Inverses: HaHa-1 = Haa-1 = He = Ha-1a = Ha-1Ha

Q.E.D.