Final answer:
If H be a subgroup of G, and g be an element in G, then NG(Hg)= NG(H)g.
Step-by-step explanation:
In group theory, let H be a subgroup of a group G, and let g be an element in G. The notation NG(H) represents the normalizer of H in G, defined as the set of elements in G that normalize H. The left coset of H by g, denoted as Hg, is the set h ∈ H. Now, we want to show that the normalizer of Hg is equal to the normalizer of H multiplied by g, i.e., NG(Hg) = NG(H)g.
To prove this, we first observe that if x is in NG(Hg), then xHgx^(-1) = Hg, meaning that conjugating Hg by x leaves it unchanged. We aim to show that x is in NG(H)g. Consider the left coset Hg, and note that xHgx^(-1) = Hg implies xH = Hgx^(-1). This shows that x normalizes H, so x is in NG(H). Consequently, x can be written as gk for some k in G. Therefore, NG(Hg) is a subset of NG(H)g.
Now, let x be in NG(H)g. Then, x can be expressed as gk for some k in G. Conjugating Hg by x, we get xHgx^(-1) = gkg^(-1)Hg = Hg. This implies that x is in NG(Hg). Thus, NG(H)g is a subset of NG(Hg). Combining both inclusions, we conclude that NG(Hg) = NG(H)g.