Welcome back, STEEMathematicians! This is the fourth and final addition to your introductory series on group theory - in case you missed them, here's the first, second and third posts. We will bring this series to a close by considering Lagranges theorem, and its consequences before discussing cosets, conjugates and classes. By then end of this post, we should be ready to begin tackling representation theory, which will be the subject of the next series. Lets get stuck in.
Really big groups, with prime order, constructed in the regime of elliptic curve cryptography are extremely useful for public-key cryptography! Indeed, one of the very earliest encryption protocols, Caesar's cipher can be interpreted as a group operation (see, this isn't all for nothing!).
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Theorem: If G is a finite group of order g, and H is a subgroup of G, of order h, then g is an integer multiple of h.
OK, great. Now lets dig a bit deeper and try to demonstrate that this is true!
We will use an equivalence relation ~ that, given some X and Y belonging to G, it is the case that X ~ Y if X-1Y belongs to H. This is exactly the same as saying that Y ∈ XH. In words, we can say that X and Y are left-congruent with respect to H.
Now lets justify that it is indeed an equivalence relation. If this is unfamiliar to you, check out post three. We have:
The classes of this equivalence relation divide S into the left cosets of H.
Each of these cosets XH contains h elements, which is the order of H. Now, since it is the case that different cosets have no common elements, and each element of G is in one of the cosets, the number of elements in S is equal to the sum of the number of elements of the individual cosets. The number of cosets is integer, therefore we can confidently state that G is an integer multiple of h.
Definition: The number of left cosets of H in G is known as the index of H in G and is written [G : H] = g/h.
For some subgroup J of subgroup H, we have [G : H] [H : J] = [G : J].
Corollary: Two cosets are either disjoint or identical
This follows from the property of the equivalence relation we introduced, above. If we suppose that cosets X1H and X2H have an element in common, i.e. X1Y1 = X2Y2, for some Y1, Y2 in H, then, X1 = X2Y2Y1-1. Now, since Y1 and Y2 both are a part of H, so too is Y2Y1-1. Thus, X1 is part of the left coset of X2H. We can make a similar argument to justify that X2 is part of the left coset X1H. As a consequence of this, either the two cosets are identical, or they do not have an element in common.
Corollary: Two cosets X1H and X2H are identical exactly if X2-1X1 is part of H
If X2-1X1 is a part of H, then X1 = X2Yi for some Yi, and X1H = X2YiH. By the permutation law (see post two), we are able to write YiH = H. Therefore, the two cosets are identical. Now lets think about this from the converse perspective, by supposing that X1H = X2H. Then, what naturally follows is the statement, X2-1X1H = H. However, it is the case that one element of H is Y; thus, X2-1X1 must also be an element of H.
Lets introduce an equivalence relation between elements X and Y of a group G, by denoting X ~ Y if there exists an element Gi ∈ G for which Y = Gi-1XGi. If X ~ Y, they are called conjugates. We justify that this is indeed an equivalence relation, since we have:
These neat results establish conjugacy as an equivalence relation, and thus show that it divides G into classes.
Corollary: If Z is in the class containing I, then Z = Gi-1IGi = Gi-1Gi = I. Thus, since any conjugate of I is equal to I, the identity is in a class on its own.
I'll leave this one for you to show!
Corollary: If X is in a class on its own, then Y = Gi-1XGi must imply that Y = X. Then, since X = GiGi-1XGiGi-1 for any Gi, we have X = Gi(Gi-1XGi)Gi-1 = GiYGi-1 = GiXGi-1, i.e. XGi = GiX ∀ Gi. In words, commutation with all elements of the group is a necessary and sufficient condition for a group element to be in a class on its own.
This one too (aren't I kind)!
Corollary: In any group G, the set S of elements in classes by themselves is an Abelian subgroup. This is known as the centre of G.
To show this, we take note that we already know that I belongs to S. Then, for X, Y ∈ S, i.e. XGi = GiX and YGi = GiY∀ Gi ∈ G, we have, (XY)Gi = XGiY = Gi(XY). This demonstrates the closure of S. It is also the case that XGi = GiX, which implies that X-1Gi = GiX-1. That is, the inverse of X belongs to S. We can therefore state that S is an Abelian group.
And there we go folks, we now know a respectable amount of group theory! I do hope you enjoyed reading these posts as much as I enjoyed putting them together, as this is a topic very close to my heart! If you found these posts interesting, the next (closely related) topic I'll be writing about is representation theory, which is quite a powerful tool as it allows us to study groups (and other algebraic objects) by representing their elements as linear transformations of vector spaces, which allows us to do some pretty interesting maths. One application of representation theory in the physical sciences is in determining the symmetry adapted molecular orbitals either side of the fermi energy, which is useful in elucidating the band structure of molecules. As always, if there are any questions, post away in the comments and I'll do my best to get back to you. Cheers, D.
