A subgroup is normal precisely when it contains its commutator by the whole group

Prove that a subgroup H \leq G is normal if and only if [G,H] \leq H.

(\Rightarrow) Suppose H \leq G is normal. Let g \in G and h \in H; then g^{-1}h^{-1}g \in H, so that g^{-1}h^{-1}gh = [g,h] \in H. Thus [G,H] \leq H.

(\Leftarrow) Suppose [G,H] \leq H. In particular, for all g \in G and h \in H, [g,h] = g^{-1}h^{-1}gh \in H. Thus g^{-1}h^{-1}g \in H. Since h^{-1} is arbitrary in H, H is normal in G.

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  • Steven  On December 8, 2011 at 2:28 pm

    I think the point of this exercise was to use the result in exercise 5.4.1, as well as the second result in proposition 7 (in section 5.4): H \trianglelefteq G if and only if [H,G] \leq H.

    • nbloomf  On December 9, 2011 at 1:48 am

      That works too. 🙂

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