Question

Prove Proposition 7.2.4(B): If H≤G, Then N(H) Is A Subgroup Of G.

Answer 1

N(H) is a subgroup of G if H is a subgroup of G.

Step-by-step explanation:

When we say "N(H)," we refer to the normalizer of the subgroup H in the group G. The normalizer of H in G, denoted as N(H), is the set of all elements in G that normalize H. In other words, N(H) consists of elements g in G such that gHg⁻¹= H.

To prove Proposition 7.2.4(B), let's consider the definition of a subgroup. If H is a subgroup of G, it means that H is closed under the group operation, contains the identity element, and contains the inverse of each of its elements. Now, for N(H) to be a subgroup of G, it must satisfy the subgroup criteria.

Firstly, note that the identity element e is in N(H) since eHe⁻¹ = H. Secondly, if g and h are in N(H), then (gh)H(gh)⁻¹= g(hHh⁻¹)g⁻¹ = gHg⁻¹, showing closure under multiplication and inverses. Thus, N(H) satisfies the criteria for being a subgroup of G. Therefore, if H is a subgroup of G, then N(H) is indeed a subgroup of G.

In conclusion, the normalizer N(H) is a subgroup of G when H is a subgroup of G. This result is crucial in group theory, providing insights into the structure of groups and their subgroups.

Answer 2

To prove Proposition 7.2.4(B), we need to show that if H is a subgroup of G, then N(H) is also a subgroup of G. The normalizer of H, denoted as N(H), is the set of elements in G that normalize H. To prove that N(H) is a subgroup, we need to show that it satisfies closure, identity, and inverses.

Step-by-step explanation:

To prove Proposition 7.2.4(B), we need to show that if H is a subgroup of G, then N(H) is also a subgroup of G.

First, let's define N(H) as the normalizer of H, which is the set of elements in G that normalize H. An element g in G normalizes H if and only if gHg^(-1) = H, where g^(-1) is the inverse of g.

To prove that N(H) is a subgroup of G, we need to show that it satisfies three conditions: closure, identity, and inverses.

1. Closure: Let x and y be elements in N(H). We need to show that xy is also in N(H). Since x and y normalize H, we have xHx^(-1) = H and yHy^(-1) = H. Now, consider (xy)H(xy)^(-1). By using the associativity of group operation, we can simplify this expression to x(yHy^(-1))x^(-1). Since yHy^(-1) = H, we have (xy)H(xy)^(-1) = xHx^(-1) = H. Therefore, xy also normalizes H, and xy is in N(H).
2. Identity: The identity element of G, denoted by e, is in N(H). This is because eHe^(-1) = H, which means the identity element normalizes H.
3. Inverses: Let x be an element in N(H). We need to show that the inverse of x, denoted by x^(-1), is also in N(H). Since x normalizes H, we have xHx^(-1) = H. Taking the inverse of both sides, we get (xHx^(-1))^(-1) = H^(-1), which simplifies to (x^(-1))^(-1)H^(-1)x^(-1) = H. Therefore, x^(-1) also normalizes H, and x^(-1) is in N(H).

Since N(H) satisfies closure, identity, and inverses, it is a subgroup of G. Hence, Proposition 7.2.4(B) is proven.