Lagange's theorem
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Cosets and a Proof of Lagrange's Theorem
In order to prove Lagrange's theorem we need to define an object called a coset of H in G. We do this as follows Let G be a group and H be a subgroup of G. We write [Equation goes here - download the original to see it.] to signify that H is a subgroup of G. For each element [Equation goes here - download the original to see it.] Example (3) [Equation goes here - download the original to see it.] Solution The elements of [Equation goes here - download the original to see it.] [Equation goes here - download the original to see it.] The group table for this group is [Diagram goes here - download the original to see it.] To illustrate the construction of this table, consider the element pb. This means b followed by p. Under b [Equation goes here - download the original to see it.] Under p [Diagram goes here - download the original to see it.] Therefore, for pb we have [Diagram goes here - download the original to see it.] That is [Diagram goes here - download the original to see it.] which is q. That is [Equation goes here - download the original to see it.] The other entries are found similarly. Now we find the cosets of H in S3. We have [Equation goes here - download the original to see it.] and [Equation goes here - download the original to see it.] [Equation goes here - download the original to see it.] [Equation goes here - download the original to see it.] This illustrates that if x, y are distinct elements of G and H in a subgroup of G then the cosets xH and yH are not necessarily distinct. We need a criterion for demonstrating when xH = yH. This is given by [Equation goes here - download the original to see it.] An equivalent to this is given by [Equation goes here - download the original to see it.] We will firstly show that [Equation goes here - download the original to see it.] We will also illustrate the meaning of this criterion for deciding when two cosets of a group are identical by looking again at our example. Here, [Equation goes here - download the original to see it.] We will now prove in general that [Equation goes here - download the original to see it.] [Equation goes here - download the original to see it.] The cosets of H in G form a partition of G. What this means is that if two cosets of H in G are not identical then they do not share any element in common. The proof of this is by contradiction. Thus, suppose [Equation goes here - download the original to see it.] are two cosets of H in G, but that they share at least one element in common. Let this common element be t. That is [Equation goes here - download the original to see it.] Therefore, [Equation goes here - download the original to see it.] [Equation goes here - download the original to see it.] Thus, if two cosets of H in G share an element in common, then they must be completely identical. Hence, the cosets of H in G partition G. This means that every element of G is in one, and only one, coset of H in G. The number of elements of each coset of H in G is the same. That is, [Equation goes here - download the original to see it.] This is because each coset is formed by taking an element g of G and combining it with each distinct element h of H. For each distinct h in H we get a different element gh in G. Indeed, if [Equation goes here - download the original to see it.] then [Equation goes here - download the original to see it.] and thus [Equation goes here - download the original to see it.] follows. Hence, there is a one-one correspondence between elements of H and elements of any coset xH of H in G. nFurther, G is divided into a finite number of cosets xH of H in G. Thus, [Equation goes here - download the original to see it.] (The order of G is equal to the product of the number of cosets of H in G, and the order of H.) That is, [Equation goes here - download the original to see it.] The order of H divides the order of G, which proves the theorem. In summary, the outline of the proof is as follows Let H be a subgroup of G. That is [Equation goes here - download the original to see it.] Then the cosets of H in G partition (divide up) G in such a way that (1) each coset has exactly the same number of distinct elements as H; (2) every element of G is in one and only one coset of H. Hence, [Equation goes here - download the original to see it.] (The order of G is equal to the product of the number of cosets of H in G, and the order of H.) which means that the order of any subgroup of G must divide the order of G.
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Contents of
Lagange's theorem
1 Lagrange's theorem
2 Cosets and a Proof of Lagrange's Theorem
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