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

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 $$2$$?

• Jun 2, 2023 at 22:29
• $C = \{ (), (1 6), (3 4), (2 5), (1 6)(3 4), (1 6)(2 5), (3 4)(2 5), (1 6)(3 4)(2 5)\}$. You are confusing it with $\langle (1 6)(3 4) (2 5)\rangle = \{(),(1 6)(3 4) (2 5)\}$. Jun 2, 2023 at 22:37
• $B\cong \mathbb Z_4\times \mathbb Z_2$ and $C\cong \mathbb Z_2\times \mathbb Z_2\times \mathbb Z_2$ Jun 2, 2023 at 23:01
• Just use, for example, Size(Group([ (1,6), (3,4), (2,5) ])); in GAP. Jun 2, 2023 at 23:09
• One could argue that your question is in fact two. Jun 2, 2023 at 23:11

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 $$|B|=8$$.

The argument for $$C$$ is similar.

Hint:

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

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.