The subgroup generated by all elements of some fixed order is normal

Let G be a group and let a,b \in G.

  1. Let g \in G. Note that g(ab)b^{-1} = gag^{-1} \cdot gbg^{-1}. Prove that |a| = |gag^{-1}|.
  2. Let g \in G. Prove that ga^{-1}g^{-1} = (gag^{-1})^{-1}.
  3. Let N = \langle S \rangle be a subgroup of G. Prove that if gSg^{-1} \subseteq N for all g \in G, then N is normal in G.
  4. Prove that a cyclic subgroup N = \langle x \rangle is normal in G if and only if for each g \in G, gx^{-1}g^{-1} = x^k for some k \in \mathbb{Z}.
  5. Let n be a positive integer. Prove that the subgroup N \leq G generated by all elements of order n is normal in G.

  1. We have g(ab)g^{-1} = ga1bg^{-1} = gag^{-1}gbg^{-1}. Now by a lemma to a previous exercise, (gag^{-1})^n = ga^ng^{-1} for all n \in \mathbb{Z}. Thus |gag^{-1}| \leq |a|. If |gag^{-1}| = k, then ga^kg^{-1} = 1, so that a^k = 1. Thus |a| \leq |gag^{-1}|.
  2. Note that gag^{-1}ga^{-1}g^{-1} = gaa^{-1}g^{-1} = gg^{-1} = 1. By the uniqueness of inverses, (gag^{-1})^{-1} = ga^{-1}g^{-1}.
  3. Let N = \langle S \rangle, and suppose gSg^{-1} for all g \in G. Then for all g \in G, we have gNg^{-1} = g \langle S \rangle g^{-1} = \langle gSg^{-1} \rangle \subseteq \langle N \rangle = N. Thus N is normal in G.
  4. Let N = \langle x \rangle. (\Leftarrow) If N is normal, then for all g \in G we have gxg^{-1} \in \langle x \rangle. Thus gxg^{-1} = x^k for some integer k. (\Rightarrow) If for all g \in G there exists an integer k such that gxg^{-1} = x^k \in N, by the previous point we have N normal in G.
  5. Let S \subseteq G consist of all elements of order n, and let s \in S. For all g \in G, by the first point we have |gsg^{-1}| = |s| = n \in \langle S \rangle. By the third point, \langle S \rangle is normal in G.
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