Find three subgroups of $QD16$ of order $8$, and find the isomorphism type of each group.

Question

Consider the group of order $16$ with the following presentation: $QD_{16} = ⟨σ, τ \mid σ^8 = τ^2 = 1, στ = τσ^3⟩$ Called the quasidihedral group of order $16$.

Find three subgroups of $QD16$ of order $8$, and find the isomorphism type of each group.

I found three subgroups of order $8$ which are $\langle σ \rangle,\langle σ^2,τ \rangle$ and $ \langle τσ,σ^2 \rangle$ .

Since $\langleσ \rangle$ is a cycle group with order 8 clearly $\langle σ \rangle \cong \mathbb{Z}_8$ . For $\langle σ^2,τ \rangle$ and $\langle τσ,σ^2 \rangle$ I know these groups are isomorphic to $D_8$ and $Q_8$ respectively but I can't state the reasons clearly.

Here is what I tried,

Since $D_8=\langle s,r\mid r^4=s^2=1,rs=sr^{-1}\rangle$, I showed $(σ^2)^4=τ^2=1$ and $σ^2τ=τ(σ^2)^{-1}$ therefore $\langle σ^2,τ \rangle=\langleσ^2,τ\mid (σ^2)^4=τ^2=1, σ^2τ=τ(σ^2)^{-1}\rangle$. Is this enough to show that $D_8 \cong \langle σ^2,τ \rangle$?

I used a similar argument for $\langle τσ,σ^2 \rangle$.

Thank you.

Answer 1

It is a good idea to realize the given group as a group of permutations. Consider inside the symmetric group $S_8$ with $8!$ the subgroup generated by $s=(12345678)$ and $(15)(28)(46)$. The subgroup generated by $s$ is the cyclic group $C$ with eight elements $\{s^k\ :\ 0\le 7<8\}$. It contains in a picture the rotation symmetries of the regular octogon,

               7                    Order 8: (12345678) k-> k+1 mod 8
       8       *       6                     (14725836) k-> k+3 mod 8
        *             *                      (16385274) k-> k+5 mod 8
                                             (18765432) k-> k+7=k-1 mod 8
                      
   1 *                   * 5        Order 4: (1357)(2468)
                                             (1753)(2864)
                     
        *             *             Order 2: (15)(26)(37)(48)
       2       *       4
               3                    Order 1: (1)(2)(3)(4)(5)(6)(7)(8)

There are other four permutations of order two, they exchange the places joined by a line in the picture:

               7
       8       *       6
        *             *
        |             |
        |             |
   1 *--|-------------|--* 5        (15) . (28)(46)
        |             |
        |             |             and also the "rotated" versions
        *             *             (26) . (13)(57)
       2       *       4            (37) . (24)(68)
               3                    (48) . (35)(17)

Let $t$ be any one of them, to fix ideas, we pick $t=(15) \;(28)(46)$.

And there are four other permutations of order four, obtained by cyclicly permuting in the same time, but as displayed with different orientations the o-vertices, and the *-vertices in the picture:

               7
       8       o       6
        o             *
                       
                       
   1 *                   * 5        (1256)(3874) and the inverse (1652)(3478)
                       
                                    and also the rotated versions 
        *             o
       2       o       4            (1854)(2367) and its inverse (1458)(2763)
               3

Then the conjugation of $s$ by $t$ is $${}^ts=s^t=tst=(1^t2^t3^t4^t5^t6^t7^t8^t)=(58361472)=(14725836)=s^3\ .$$

The elements $s,t$ satisfy the presentation of the OP group $QD_{16}$, and has the expected order, (the order does not drop,) so the group $G=\langle s,t\rangle$ is (isomorphic to) the group $QD_{16}=\langle \sigma,\tau\rangle$ (via $\sigma\to s$, $\tau\to t$).

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We have now an explicit construction of the involved group, it is easy to check the structure of the three subgroups of index two, order eight from the question.

$(1)$ The cyclic group $C=\langle s\rangle$ has clear structure, one generator of order eight.

$(2)$ The subgroup generated by $s^2$ and $t$ contains order four elements $s^{\pm2}$ which have the cycle structure $s^2=(1357)(2468)$, $s^6=(1753)(2864)$, and it turns out they are $t$-conjugated, $$ ts^2t=(1^t3^t5^t7^t)(2^t4^t6^t8^t) =(5317)(8642)=(1753)(2864)=s^{-2}\ . $$ We thus have a cyclic group of order four, $\langle s^2\rangle$, extendend/twisted/wedged by $t$ via the inverse automorphism. This semiproduct has the dihedral group structure $D_4$. The same argument is present in the question.

$(3)$ The subgroup generated by $s^2$ and $ts=(15)(28)(46)\ (12345678) =(1854)(2367)$ (an element of order four) contains again the cyclic group $\langle s^2\rangle$ with four elements. But the twist is $$ (s^2)^{ts}=(1^{ts}3^{ts}5^{ts}7^{ts})(2^{ts}4^{ts}6^{ts}8^{ts}) =(8642)(3175)=(1753)(2864)=s^{-2}\ . $$ Here, the twist is implemented by an element of order four. The two cyclic subgroups $\langle s^2\rangle$ and $\langle ts\rangle$ share the same element of order two, $s^4=(ts)^2$, which is the only element of order two in the group. So we have the $Q_8$-structure.

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For short, we distinguish the cases by listing the orders of the elements of the three groups. The lists are respectively:

$(1)$ $C_8$ has the orders $1, 2, 4, 4, 8, 8, 8, 8$.

$(2)$ $D_4$ has the orders $1, 2, 2, 2, 2, 2, 4, 4$.

$(3)$ $Q_8$ has the orders $1, 2, 4, 4, 4, 4, 4, 4$.