By Arthur Cayley

This quantity is made from electronic pictures from the Cornell college Library historic arithmetic Monographs assortment.

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**Additional info for An elementary treatise on elliptic functions**

**Sample text**

If Gb and Gr are finite groups of order m and n, respectively, then order of the group Gb Gr or Gr Gb is mn. Thus, taking the direct product of two finite groups, each having more than one element, gives us the simplest method of forming a larger group. If a group G is isomorphic to the group Gb Gr, then it is said to be the direct product group. With this definition, elements of the direct product group need not be in the form of pairs. The direct product of more than two groups can be defined in the same manner.

Let a be an arbitrary element of G. Consider the set K a–1Ha a–1 {h1, h2, h3, …}a {a–1h1a, a–1h2a, a–1h3a, …} Since a, a–1, and hi all belong to G, the elements a–1hia of K, being the products of these elements, also belong to G. Thus, K is a subset of G. We will prove that K is a subgroup of G. 18 Group Theory for High Energy Physicists Let {k1, k2, k3, …} be the elements of K. To prove that K is a subgroup of G, we have to show that if ki and kj belong to K, then the element kikj–1 also belongs to K.

We will show that the set D(G) of matrices forms a group. We have G {R, S, …} K: D(G) {D(R), D(S), …} The correspondence K is indicated by arrows. Since the correspondence between the elements of the group G and the set D(G) is preserved under combination, we have (RS) n D(RS) 46 Group Theory for High Energy Physicists Now (RS) RS Therefore, D(RS) D(R)D(S) As shown next, this equation for the closure property leads to the conclusion that D(G) forms a group. Let us denote the identity element of G by E.