Let N be a normal subgroup of a group G.
Prove that the cosets of N
in G
form a group G/N
of order [G:N]
First, we need to show that the cosets of N in G are well-defined and partition G.
To show that they are well-defined, we need to show that if a and b are elements of G such that aN = bN, then b^-1a is an element of N.
This can be shown as follows:
aN = bN
=> b^-1aN = N
=> (b^-1a)N = N (using the property that (xN)(yN) = xyN for any x, y in G)
Hence, b^-1a is an element of N.
To show that the cosets partition G, we need to show that every element of G is in exactly one coset of N in G. This follows immediately from the definition of a coset: for any g in G, we have gN = { gn : n in N }. Since every element of G can be written as gn for some g in G and n in N, it follows that every element of G is in exactly one coset of N in G.
Now, we can define the group operation on the cosets of N in G. Let aN, bN be two cosets of N in G. Then we define their product (or multiplication) as follows:
(aN)(bN) = abN
Note that abN is a coset of N in G since N is normal.
We need to show that this operation is well-defined (i.e., it does not depend on the choice of representatives for the cosets) and that it satisfies the group axioms.
To show that the operation is well-defined, we need to show that if aN = a'N and bN = b'N, then abN = a'b'N. This can be shown as follows:
aN = a'N => a' = an for some n in N
bN = b'N => b' = bm for some m in N
=> a'b' = (an)(bm) = ab(nbm) (using the associativity of G)
=> a'b'N = abN (since nbm is in N and thus ab(nbm)N = abN)
Now, we check that the group axioms are satisfied:
- Closure: We just showed that the product of two cosets is a coset.
- Associativity: This follows from the associativity of G.
- Identity: The identity of G/N is the coset containing the identity element e of G, i.e., N itself. It is easy to check that for any aN in G/N, we have (aN)(N) = aN = N(aN), so N acts as the identity element.
- Inverse: The inverse of aN in G/N is the coset containing a^-1, i.e., (aN)^-1 = a^-1N. This is easily verified by checking that (aN)(a^-1N) = N = (a^-1N)(aN).
Hence, we have shown that G/N is a group of order [G:N].
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To prove that the cosets of N in G form a group G/N of order [G:N], you need to show that these cosets satisfy all the group axioms, namely closure, associativity, identity, and inverse.
1. Closure: Let aN and bN be two cosets of N in G. We need to show that (aN) * (bN) is also a coset of N in G. Since aN and bN are cosets, there exist elements a and b in G such that aN = Na and bN = Nb. Then, (aN) * (bN) = (Na) * (Nb) = N(ab) by the group multiplication property. Thus, (aN) * (bN) is indeed a coset of N in G.
2. Associativity: This property follows directly from the associativity of the group multiplication in G. For any three cosets aN, bN, and cN, we have (aN * bN) * cN = (aNbN) * cN = ((ab)N) * cN = (ab)NcN = (ab)(NcN) = aN(bcN) = aN * (bN * cN). Therefore, the cosets of N in G are associative.
3. Identity: The identity element of the group G is denoted as e. We claim that Ne is the identity element of G/N. For any coset aN, we have (aN) * (Ne) = (aN) * (N) = N(ae) = aN. Similarly, (Ne) * (aN) = N(ea) = aN. Therefore, Ne is the identity element in G/N.
4. Inverse: For any coset aN, we claim that the inverse of aN is (aN)^(-1) = (a^(-1))N. First, note that a^(-1) exists in G since a is an element of G and G is a group. Then we have (aN) * ((a^(-1))N) = aN(a^(-1))N = (aa^(-1))N = Ne = (a^(-1)a)N = a^(-1)(aN) = (a^(-1))N * (aN). Therefore, (aN)^(-1) = (a^(-1))N.
By showing that the cosets of N in G satisfy the group axioms, we have proven that G/N is a group. The order of G/N is the number of distinct cosets of N in G, denoted as [G:N].