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Exhibit Dihedral group as a subgroup of Symmetric group via regular representation


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.2 Exercise 4.2.3

Solution: To save effort we will perform this computation in $S_{D_8}$ and then use the labeling bijections $D_8 \rightarrow \{1,2,\ldots,8\}$ to produce subgroups of $S_8$. Let $\lambda : D_8 \rightarrow S_{D_8}$ be the left regular representation.

Clearly $\lambda(1) = 1$.

We have $$\lambda(r)(1) = r, \lambda(r)(r) = r^2,$$ $$\lambda(r)(r^2) = r^3, \lambda(r)(r^3) = 1,$$ $$\lambda(r)(s) = rs = sr^3,$$ $$\lambda(r)(sr^3) = rsr^3 = sr^6 = sr^2,$$ and $$\lambda(r)(sr^2) = rsr^2 = sr^5 = sr.$$ Thus $$\lambda(r) = (1\ r\ r^2\ r^3)(s\ sr^3\ sr^2\ sr).$$Note that $\lambda(r^2) = \lambda(r)^2$. Thus $$\lambda(r^2) = (1\ r^2)(r\ r^3)(s\ sr^2)(sr\ sr^3).$$ Note that $\lambda(r^3) = \lambda(r)^3$. Thus $$\lambda(r^3) = (1\ r^3\ r^2\ r)(s\ sr\ sr^2\ sr^3).$$We have $$\lambda(s)(1) = s, \lambda(s)(s) = 1,$$ $$\lambda(s)(r) = sr, \lambda(s)(sr) = r,$$ $$\lambda(s)(r^2) = sr^2, \lambda(s)(sr^2) = r^2,$$ $$\lambda(s)(r^3) = sr^3,  \lambda(s)(sr^3) = r^3.$$ Thus $$\lambda(s) = (1\ s)(r\ sr)(r^2\ sr^2)(r^3\ sr^3).$$Note that $\lambda(sr) = \lambda(s)\lambda(r)$. Thus $$\lambda(sr) = (1\ sr)(r\ sr^2)(r^2\ sr^3)(r^3\ s).$$Note that $\lambda(sr^2) = \lambda(s)\lambda(r^2)$. Thus $$\lambda(sr^2) = (1\ sr^2)(r\ sr^3)(r^2\ s)(r^3\ sr).$$Note that $\lambda(sr^3) = \lambda(s)\lambda(r^3)$. Thus $$\lambda(sr^3) = (1\ sr^3)(r\ s)(r^2\ sr)(r^3\ sr^2).$$
Now via the labelling $$1 \mapsto 1 r \mapsto 2 r^2 \mapsto 3 r^3 \mapsto 4$$ $$s \mapsto 5 sr \mapsto 6 sr^2 \mapsto 7 sr^3 \mapsto 8$$ we have the isomorphic image$$1 \mapsto 1$$ $$r \mapsto (1\ 2\ 3\ 4)(5\ 8\ 7\ 6)$$ $$r^2 \mapsto (1\ 3)(2\ 4)(5\ 7)(6\ 8)$$ $$r^3 \mapsto (1\ 4\ 3\ 2)(5\ 6\ 7\ 8)$$ $$s \mapsto (1\ 5)(2\ 6)(3\ 7)(4\ 8)$$ $$sr \mapsto (1\ 6)(2\ 7)(3\ 8)(4\ 5)$$ $$sr^2 \mapsto (1\ 7)(2\ 8)(3\ 5)(4\ 6)$$ $$sr^3 \mapsto (1\ 8)(2\ 5)(3\ 6)(4\ 7).$$ Similarly, via the labeling $$1 \mapsto 1 r \mapsto 3 r^2 \mapsto 5 r^3 \mapsto 7$$ $$s \mapsto 2 sr \mapsto 4 sr^2 \mapsto 6 sr^3 \mapsto 8$$ we have the isomorphic image$$1 \mapsto 1$$ $$r \mapsto (1\ 3\ 5\ 7)(2\ 8\ 6\ 4)$$ $$r^2 \mapsto (1\ 5)(3\ 7)(2\ 6)(4\ 8)$$ $$r^3 \mapsto (1\ 7\ 5\ 3)(2\ 4\ 6\ 8)$$ $$s \mapsto (1\ 2)(3\ 4)(5\ 6)(7\ 8)$$ $$sr \mapsto (1\ 4)(3\ 6)(5\ 8)(7\ 2)$$ $$sr^2 \mapsto (1\ 6)(3\ 8)(5\ 2)(7\ 4)$$ $$sr^3 \mapsto (1\ 8)(3\ 2)(5\ 4)(7\ 6)$$.
It is easy to see that these two subgroups are different.


Linearity

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