A normal subgroup $C_2$ | Generated by rotation 180° |
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Four conjugate subgroups $C_2$ | Each generated by a reflection |
A normal subgroup $C_4$ | Rotations |
Two normal subgroups $C_2^2$ | One is the stabilizer of a diagonal: |
One is stabilizer of a midline: |
Pick finitely many elements $α _ i ∈ K$, $i = 1, … , n$ such that $σ (α _ i) = α _ i$ for $i = 1, … , n$ implies $σ $ is the neutral element of $G$. Set \[ L = K^ G(\{ σ (α _ i); 1 ≤ i ≤ n, σ ∈ G\} ) ⊂ K \] and observe that the action of $G$ on $K$ induces an action of $G$ on $L$. We will show that $L$ has degree $|G|$ over $K^ G$. This will finish the proof, since if $L ⊂ K$ is proper, then we can add an element $α ∈ K$, $α ∉ L$ to our list of elements $α _1, … , α _ n$ without increasing $L$ which is absurd. This reduces us to the case that $K/K^ G$ is finite which is treated in the next paragraph.
Assume $K/K^G$ is finite. By Lemma 9.19.1 we can find $α ∈ K$ such that $K = K^G(α)$. By the construction in the first paragraph of this proof we see that $α $ has degree at most $|G|$ over $K$. However, the degree cannot be less than $|G|$ as $G$ acts faithfully on $K^G(α) = L$ by construction and the inequality of Lemma 9.15.9.