Exercise 2.4.7 ($D_8 \simeq \langle(1\ 2), (1\ 3)(2\ 4) \rangle$)

Proof. Put $\sigma =(12)$, $\tau =(13)(24)$ (and $e=()$ the identity element of $S_4$).

Let $H=⟨\sigma ,\tau ⟩$. Then

Then

and

Therefore

that is

Since these eight permutations are distinct, This shows that $|H|\ge 8$.

(To avoid the cumbersome proof that these eight permutations form a group, we use the presentation of $D_8$.)

Note that

Moreover, since $\mathit{\sigma \tau }\in ⟨\sigma ,\tau ⟩$, then $⟨\sigma ,\mathit{\sigma \tau }⟩\subseteq ⟨\sigma ,\tau ⟩$, and since $\tau =\sigma (\mathit{\sigma \tau })\in ⟨\sigma ,\tau ⟩$, then $⟨\sigma ,\tau ⟩\subseteq ⟨\sigma ,\mathit{\sigma \tau }⟩$, so $⟨\sigma ,\tau ⟩=⟨\sigma ,\mathit{\sigma \tau }⟩$. This gives

Since a presentation of $D_8$ is given by

and $H=⟨\rho ,\sigma ⟩$, where ${\rho }^4={\sigma }^2=e,\mathit{\sigma \rho }={\rho }^3\sigma$, there exists a surjective homomorphism $\phi :D_8\to H$ such that $\phi (r)=\rho ,\phi (s)=\sigma$. (*)

Hence $8=|D_8|\ge |H|$, and by (1), $|H|\ge 8$. Therefore $|H|=|D_8|$ and $\phi$ is bijective, so $\phi$ is an isomorphism. This shows that

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