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Problem 5.5.2 ($C_H(K) = N_H(K)$)

Prove that $C_{H} (K) = N_{H} (K)$ .

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Proof. First, $C_{H} (K) \subseteq N_{H} (K)$ is always true. Conversely, let $h \in N_{H} (K)$ . Then for all $k \in K$ , $(h, 1) (1, k) {(h, 1)}^{- 1} \in {1} \times K$ , thus

(h, 1) (1, k) {(h, 1)}^{- 1} = (1, k^{'}) for some k^{'} \in K .

Then $(h, 1) (1, k) = (1, k^{'}) (h, 1)$ , thus

(h, k) = (k^{'} \cdot h, k^{'}) .

Therefore $k^{'} = k$ , and so, for all $k \in K$ ,

(h, 1) (1, k) {(h, 1)}^{- 1} = (1, k) .

This shows that $h \in C_{H} (K)$ (where $h$ is identified with $(h, 1)$ ), so

C_{H} (K) = N_{H} (K) .

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richardganaye

2026-09-12 11:24

Problem 5.5.2 ($C_H(K) = N_H(K)$)

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