The set of normal subgroups of a fixed group is closed under arbitrary intersections

Let G be a group.

  1. Let H,K \leq G be normal subgroups. Prove that H \cap N \leq G is normal.
  2. Let \{H_i\}_{i \in I} be a family of subgroups H_i \leq G, all normal. Show that \bigcap H_i \leq G is normal.

  1. Using a lemma to a previous exercise, for all g \in G we have g(H \cap K)g^{-1} = gHg^{-1} \cap gKg^{-1} = H \cap K. Thus H \cap K is normal in G.
  2. Using a lemma to a previous exercise, for all g \in G we have g(\bigcap H_i)g^{-1} = \bigcap gH_ig^{-1} = \bigcap H_i. By Theorem 6 in the text, \bigcap H_i is normal in G.
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Comments

  • Max  On November 15, 2010 at 9:53 pm

    Is there any chance you could direct me to where you proved the lemma that the conjugate of the intersection is the intersection of the conjugates?

    • nbloomf  On November 16, 2010 at 12:05 am

      I can’t remember where it is, so I’ll just re-prove it here.

      Let \varphi be an automorphism of a group G and let H,K \leq G. We want to show that \varphi[H \cap K] = \varphi[H] \cap \varphi[K]. (\subseteq) Suppose x \in \varphi[H \cap K]. Then x = \varphi(y), where y \in H \cap K. In particular, x \in \varphi[H] and x \in \varphi[K], so x \in \varphi[H] \cap \varphi[K]. (\supseteq) Now let x \in \varphi[H] \cap \varphi[K]. Say x = \varphi(h) and x = \varphi(k), with h \in H and k \in K. Since \varphi is an automorphism, it is injective, and we have h = k \in H \cap K. Thus x \in \varphi[H \cap K], and we have \varphi[H \cap K] = \varphi[H] \cap \varphi[K].

      In particular, conjugation by a fixed element g is an automorphism.

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