Math Problem Statement

Define normal subgroup. If H is a subgroup of G and N is a normal subgroup of G, then prove that HON is a normal subgroup of H

Solution

Definition of a Normal Subgroup

A subgroup NN of a group GG is called a normal subgroup if it is invariant under conjugation by any element of GG. This means that for every element gGg \in G and every element nNn \in N, the element gng1gng^{-1} is still in NN. Formally, NN is normal in GG if: gNg1=Nfor all gG.gNg^{-1} = N \quad \text{for all } g \in G. This is often denoted as NGN \triangleleft G.

Proof that HNHN is a Normal Subgroup of HH

Let HH be a subgroup of GG, and let NN be a normal subgroup of GG. We want to show that the set HNHN is a normal subgroup of HH.

1. HNHN is a Subgroup of GG

First, note that HN={hnhH,nN}HN = \{ hn \mid h \in H, n \in N \}. To prove that HNHN is a subgroup of GG, we need to show two things:

  • Closure: If x,yHNx, y \in HN, then xyHNxy \in HN.

    • Suppose x=h1n1x = h_1n_1 and y=h2n2y = h_2n_2 for some h1,h2Hh_1, h_2 \in H and n1,n2Nn_1, n_2 \in N.
    • Then, xy=(h1n1)(h2n2)=h1(n1h2)n2xy = (h_1n_1)(h_2n_2) = h_1(n_1h_2)n_2.
    • Since NN is normal in GG, n1h2=h2n1n_1h_2 = h_2n_1' for some n1Nn_1' \in N. Thus, xy=h1h2n1n2xy = h_1h_2n_1'n_2, which is of the form hnh'n' where h=h1h2Hh' = h_1h_2 \in H and n=n1n2Nn' = n_1'n_2 \in N.
    • Therefore, xyHNxy \in HN, proving closure.
  • Inverses: If xHNx \in HN, then x1HNx^{-1} \in HN.

    • Suppose x=hnx = hn for some hHh \in H and nNn \in N.
    • Then x1=(hn)1=n1h1x^{-1} = (hn)^{-1} = n^{-1}h^{-1}.
    • Again, since NN is normal in GG, n1h1=h1nn^{-1}h^{-1} = h^{-1}n' for some nNn' \in N.
    • Thus, x1HNx^{-1} \in HN, proving that HNHN is a subgroup.

2. HNHN is Normal in HH

Now, we must show that HNHN is a normal subgroup of HH, meaning h(HN)h1HNh(HN)h^{-1} \subseteq HN for all hHh \in H.

  • Take any h0Hh_0 \in H and any element hnHNhn \in HN where hHh \in H and nNn \in N.
  • Consider h0(hn)h01=(h0h)n(h01)h_0(hn)h_0^{-1} = (h_0h)n(h_0^{-1}).
  • Since NN is normal in GG, h0nh01Nh_0nh_0^{-1} \in N.
  • Therefore, h0(hn)h01=(h0h)nh_0(hn)h_0^{-1} = (h_0h)n' for some nNn' \in N, and h0hHh_0h \in H.
  • This shows that h0(hn)h01HNh_0(hn)h_0^{-1} \in HN.

Since h0h_0 and hnhn were arbitrary, we have h0(HN)h01HNh_0(HN)h_0^{-1} \subseteq HN for all h0Hh_0 \in H, meaning HNHN is normal in HH.

Conclusion

The set HNHN, where HH is a subgroup of GG and NN is a normal subgroup of GG, is indeed a normal subgroup of HH.


Would you like more details or have any further questions? Here are five related questions to explore further:

  1. What is an example of a normal subgroup in a familiar group like Z/nZ\mathbb{Z}/n\mathbb{Z} or SnS_n?
  2. How does the concept of a normal subgroup relate to the construction of quotient groups?
  3. Can the intersection of two normal subgroups be non-normal? Why or why not?
  4. What are the properties of the cosets of a normal subgroup?
  5. How does the isomorphism theorem depend on the normality of subgroups?

Tip: When working with subgroups, remember that normality is essential for defining quotient groups, which are central to many areas of group theory.

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Math Problem Analysis

Mathematical Concepts

Group Theory
Normal Subgroups
Subgroup Closure
Conjugation

Formulas

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Theorems

Normal Subgroup Criterion

Suitable Grade Level

Advanced Undergraduate