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Recipe

Find the normalizer of a subgroup

Use when: you're given a subgroup H\le G with a known generating set and asked for every g\in G satisfying gHg^{-1}=H

Goal: the complete set N_G(H), and whether it equals H, G, or something strictly between

Taught in The normalizer of a subgroup

Quick recipe

  1. Include H itself for free
  2. Test each remaining candidate by conjugating H's own generator(s)
  3. Collect the survivors and compare to H and G
  1. 1

    List every element of H as an automatic member of N_G(H). No conjugation needs to be checked for them.

    Example

    For H=\{e,(1\,2)\} in S_3, both e and (1\,2) are automatically in N_{S_3}(H).

    Why: every h\in H satisfies hHh^{-1}=H by closure, so testing elements already inside H would just repeat a fact already proved.

  2. 2

    For g\notin H, conjugate a generator of H by g, not every element of H, and check whether the result lands back inside H.

    Example

    For H=\{e,(1\,2)\}, generated by (1\,2) alone, test g=(2\,3): (2\,3)(1\,2)(2\,3)^{-1}=(1\,3), which is not in H, so (2\,3)\notin N_{S_3}(H).

    • every generator's conjugate lands back inside H — g belongs to N_G(H), since conjugation sends products to products, so the rest of H follows automatically.
    • even one generator's conjugate lands outside H — g is excluded immediately; no need to test the rest of H.

    Why: conjugation by a fixed g sends products to products, so one generator landing inside H forces every product of that generator, meaning all of H, to land inside H too.

  3. 3

    Repeat Step 2 for the rest of G's elements, then gather every survivor together with all of H from Step 1. The result always sits somewhere between H and G.

    Example

    Repeating Step 2 for the remaining elements of S_3 rules out all four elements outside H, so N_{S_3}(H)=\{e,(1\,2)\}=H exactly.

    • the result equals H exactly — the normalizer is as small as it can possibly be.
    • the result equals G — H is normal, since N_G(H)=G is exactly the definition of normal restated.
    • the result sits strictly between H and G — H is not normal, but more elements protect it under conjugation than H alone provides.

    Why: H\le N_G(H)\le G always, so the finished set has to land somewhere in that sandwich, and which end (or middle) it lands at is the whole answer.

Common mistakes

  • Testing whether g commutes with every element of H (the centralizer condition) instead of whether g sends H's elements back into H.
  • Checking only one candidate element of H against g, when H needs more than one generator to be pinned down completely.
  • Stopping the search early after finding a few elements matching H, without checking every remaining element of G.
  • Concluding H is normal from a normalizer that's merely bigger than H, instead of checking it equals all of G.