Why are $\langle (1634),(25)\rangle$ and $\langle (16),(34),(25)\rangle$ order-$8$ subgroups of $S_6$?

Question

Let $A=\langle (16342587)\rangle \le S_8$, $B=\langle (1634),(25)\rangle \le S_6$, $C=\langle (16),(34),(25)\rangle \le S_6.$

Show that: $|A|=|B|=|C|=8$

I can understand why would $A$ would be order $8$ but can't understand the rest.

Wouldn't $B$ have an order of $4$ as the $\text{lcm}(4,2) = 4$, and similarly with $C$, wouldn't it be $2$?

Answer 1

The generators of $B$ are disjoint as permutations and thus commute.

Let $a=(1634),b=(25)$. By the above, $ab=ba$. Also $|a|=4$ and $|b|=2$ by inspection. Thus

$$B\le\langle x,y\mid x^4, y^2, xy=yx\rangle\cong\Bbb Z_4\times \Bbb Z_2\tag{1}$$

by this standard result. But $a^mb^n$ are clearly distinct for $0\le m\le 3$ and $0\le n\le 1$, which implies

$$8\le |B|\le |\Bbb Z_4\times \Bbb Z_2|=8.$$

Hence $|B|=8$.

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The argument for $C$ is similar.

Hint:

Consider $$\Bbb Z_2\times \Bbb Z_2\times\Bbb Z_2\cong\langle r,s,t\mid r^2, s^2,t^2, rs=sr, rt=tr, st=ts\rangle.$$

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NB: I abused notation in $(1)$. What I should have said is that $B$ is isomorphic to a subgroup of the group $G$ given by the presentation $$\langle x,y\mid x^4, y^2, xy=yx\rangle,$$ and $G$ is isomorphic to $\Bbb Z_4\times\Bbb Z_2$. With experience, this subtlety gets glossed over with ease, but I couldn't leave it without an explanation, given that you're new to this.