If a group has a unique subgroup of a given order, then that subgroup is normal

Let G be a group, H a subgroup of G, and fix g \in G.

  1. Prove that gHg^{-1} \leq G and that |gHg^{-1}| = |H|.
  2. Deduce that if n \in \mathbb{Z}^+ and H is the unique subgroup of G of order n then H is normal in G.

  1. Note that gHg^{-1} is not empty since g1g^{-1} = 1 \in gHg^{-1}. Now let x = gag^{-1} and y = gbg^{-1} be in gHg^{-1}, where a,b \in H. Then (gag^{-1})(gbg^{-1})^{-1} = g(ab^{-1})g^{-1} \in gHg^{-1}, and by the Subgroup Criterion we have gHg^{-1} \leq G.

    We define a mapping \varphi : H \rightarrow gHg^{-1} by h \mapsto ghg^{-1}. \varphi is clearly surjective. Now suppose \varphi(x) = \varphi(y); then gxg^{-1} = gyg^{-1}, so that x = y. Hence \varphi is injective. Thus |H| = |gHg^{-1}|.

  2. Suppose now that n \in \mathbb{Z}^+ and that G has a unique subgroup H of order n. For all g \in G, gHg^{-1} is a subgroup of order n, hence gHg^{-1} = H. Thus H is normal in G.
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  • me  On May 4, 2012 at 10:40 pm

    You’ve got a typo just before claiming you’ve satisfied the Subgroup Criterion. It should be g(ab^(-1))g^(-1) in gHg^(-1).

    • nbloomf  On August 19, 2012 at 7:38 pm

      Fixed. Thanks!

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