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

### Definition of a Normal Subgroup
A subgroup \( N \) of a group \( G \) is called a **normal subgroup** if it is invariant under conjugation by any element of \( G \). This means that for every element \( g \in G \) and every element \( n \in N \), the element \( gng^{-1} \) is still in \( N \). Formally, \( N \) is normal in \( G \) if:
\[
gNg^{-1} = N \quad \text{for all } g \in G.
\]
This is often denoted as \( N \triangleleft G \).

### Proof that \( HN \) is a Normal Subgroup of \( H \)
Let \( H \) be a subgroup of \( G \), and let \( N \) be a normal subgroup of \( G \). We want to show that the set \( HN \) is a normal subgroup of \( H \).

#### 1. \( HN \) is a Subgroup of \( G \)
First, note that \( HN = \{ hn \mid h \in H, n \in N \} \). To prove that \( HN \) is a subgroup of \( G \), we need to show two things:

- **Closure**: If \( x, y \in HN \), then \( xy \in HN \).
  - Suppose \( x = h_1n_1 \) and \( y = h_2n_2 \) for some \( h_1, h_2 \in H \) and \( n_1, n_2 \in N \).
  - Then, \( xy = (h_1n_1)(h_2n_2) = h_1(n_1h_2)n_2 \).
  - Since \( N \) is normal in \( G \), \( n_1h_2 = h_2n_1' \) for some \( n_1' \in N \). Thus, \( xy = h_1h_2n_1'n_2 \), which is of the form \( h'n' \) where \( h' = h_1h_2 \in H \) and \( n' = n_1'n_2 \in N \).
  - Therefore, \( xy \in HN \), proving closure.

- **Inverses**: If \( x \in HN \), then \( x^{-1} \in HN \).
  - Suppose \( x = hn \) for some \( h \in H \) and \( n \in N \).
  - Then \( x^{-1} = (hn)^{-1} = n^{-1}h^{-1} \).
  - Again, since \( N \) is normal in \( G \), \( n^{-1}h^{-1} = h^{-1}n' \) for some \( n' \in N \).
  - Thus, \( x^{-1} \in HN \), proving that \( HN \) is a subgroup.

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

- Take any \( h_0 \in H \) and any element \( hn \in HN \) where \( h \in H \) and \( n \in N \).
- Consider \( h_0(hn)h_0^{-1} = (h_0h)n(h_0^{-1}) \).
- Since \( N \) is normal in \( G \), \( h_0nh_0^{-1} \in N \).
- Therefore, \( h_0(hn)h_0^{-1} = (h_0h)n' \) for some \( n' \in N \), and \( h_0h \in H \).
- This shows that \( h_0(hn)h_0^{-1} \in HN \).

Since \( h_0 \) and \( hn \) were arbitrary, we have \( h_0(HN)h_0^{-1} \subseteq HN \) for all \( h_0 \in H \), meaning \( HN \) is normal in \( H \).

### Conclusion
The set \( HN \), where \( H \) is a subgroup of \( G \) and \( N \) is a normal subgroup of \( G \), is indeed a normal subgroup of \( H \).

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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 \( \mathbb{Z}/n\mathbb{Z} \) or \( S_n \)?
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.

Learn how to prove that the set HN is a normal subgroup of H if H is a subgroup of G and N is a normal subgroup of G. This proof involves demonstrating subgroup closure and showing that HN is invariant under conjugation within H. Suitable for advanced undergraduate students and anyone studying group theory and normal subgroups.

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