Recipe

Build a non-normal subgroup by conjugating a point-stabilizer

Use when: you're inside some S_n with n\ge3 and asked to produce a subgroup that isn't normal, or to test whether a 'fixes one point' subgroup is normal

Goal: an explicit point-stabilizer H together with one concrete permutation proving gHg^{-1}\ne H

Taught in Normal subgroups: examples and counterexamples

1. 1

   Choose one specific point k\in\{1,\dots,n\}. Let H=\{\sigma\in S_n:\sigma(k)=k\}, the permutations that leave k fixed.

   Example

   In S_4, fix k=4: H=\{\sigma\in S_4:\sigma(4)=4\}, order 6, one permutation for every way to arrange \{1,2,3\}.

   Why: fixing one point survives multiplying elements together and taking inverses. So H is guaranteed to be a subgroup before any normality question is even asked.

2. 2

   Pick any g with g(k)\ne k, and conjugate: gHg^{-1}.

   Example

   With g=(1\,4), g(4)=1, so gHg^{-1}=\{\tau\in S_4:\tau(1)=1\}.

   • the chosen g moves k — conjugating a "fixes x" subgroup by g produces the "fixes g(x)" subgroup, a genuinely different set.
   • the chosen g also fixes k, so g\in H — pick a different g; conjugating by an element already inside H just gives back H and proves nothing.

   Why: this follows directly from how conjugation renames a permutation's behavior -- the same idea section 2 used for cycles.

3. 3

   Find a specific permutation in H that is not in gHg^{-1}, or the other way around. One clean example is always enough to prove the sets differ.

   Example

   (1\,2\,3)\in H, since it fixes 4, but (1\,2\,3)\notin gHg^{-1}, since it sends 1 to 2, not to 1.

   Why: two subgroups that differ in even one element already prove gHg^{-1}\ne H, which is exactly what "not normal" means.