Q35E - Organic Chemistry (Mcmurry)

Question

Let   be a group and assume that for each positive integer i , N_i  is a normal subgroup of G. If every element of   can be written uniquely in the form n_i1 .n_i2 ...n_ik with i₁ < i₂ < ... < i_k and n_ij$\in$N_ij , prove that G$\cong$$\sum _{\mathrm{i}=1}$ N_i   (see Exercise 34). [Hint: Adapt the proof of Theorem 9.1 by defining f(a₁, a₂...) to be the product of those a_i that are not the identity element.]

Step-by-Step Solution

Answer

It has been proved that, G$\cong \sum _{\mathrm{i}=1}$N_i .

Given that

Given that is a group and for each positive integer i , N_i  is a normal subgroup of G.

 Every element of G can be written uniquely in the form n_i1 .n_i2 ...n_ik  with i₁ < i₂ < ... < i_k   and n_ij$\in$N_ij .

Define a map which is surjective

As per the hint, define a map f: N₁ x N₂ x .... $\to$G by f (n₁, n₂, ...) = n₁ n_2.... .

Since every element of G n₁ n₂ .... can be written in the form and by the hypothesis, f is surjective.

Prove injective

Let f(n₁, n₂, ....) = f (m₁ , m₂, .... ) .

Then, n₁ n₂ = m₁ m₂ .

By uniqueness hypothesis, n_i = m_i for each i.

Therefore, (n₁ n₂...) = (m₁ m₂ ...) in N₁ x N₂ x ...

Thus, f is injective.

Prove homomorphism

It is clear that, N's are mutually disjoint subgroups and Lemma 9.2 implies that n_i m_j = m_j n_i for n_i $\in$$\in$ and .

Find as:

 

Thus, .

Hence, is homomorphism and isomorphism.

Since is an isomorphism, it can be concluded that .