Every subgroup is contained in its normalizer

Let G be a group and H \leq G.

  1. Show that H \leq N_G(H). Give an example to show that this is not necessarily true of H is not a subgroup.
  2. Show that H \leq C_G(H) if and only if H is abelian.

First we prove a lemma.

Lemma: Let G be a group and let H,K \leq G. If H \subseteq K, then H \leq K. Proof: Let a,b \in H. By the subgroup criterion (in G) we have ab^{-1} \in H. Thus by the subgroup criterion (in K), H \leq K. \square

  1. Let h \in H. If a \in H, then hah^{-1} \in H since H is a subgroup of G. Thus hHh^{-1} \subseteq H. Moreover, note that if a \in H, we have a = h(h^{-1}ah)h^{-1}, so that H \subseteq hHh^{-1}. Thus h \in N_G(H). So H \subseteq N_G(H), and by the lemma we have H \leq N_G(H).

    Now consider G = S_n, n \geq 3, with H = \{ (1\ 2), (1\ 2\ 3) \}. We have (1\ 2)(1\ 2\ 3)(1\ 2) = (1\ 3\ 2), so that (1\ 2)H(1\ 2) \neq H. Thus (1\ 2) \notin N_G(H).

  2. (\Rightarrow) Let a,b \in H. Since H \leq C_G(H), we have aba^{-1} = b, and thus ab = ba. So H is abelian. (\Leftarrow) Let a \in H and b \in H. Since H is abelian, we have ab = ba, so that aba^{-1} = b. Thus a \in C_G(H).
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