Prove $Q_8$ is not contained as a subgroup in the symmetric group $S_n$, for $n<8$

Question

Prove $Q_8$ is not contained as a subgroup in the symmetric group $S_n$, for $n<8$

I think that the key is considering the elements of order $4$ in $S_n$ but Ido not know how to face up to it.

In fact, it is given the clue that two elements of order $4$ in $S_n$ conmute if they have the same square.

If anyone could help in proving that sentence It would be valuable.

The clue seems unhelpful (in fact wrong) to me! The elements $(1,2,3,4)(5,6)$ and $(1,2,3,4)(5,7)$ of $S_7$ have the same square and do not commute. — Derek Holt, May 11 '21 at 19:17

score 3 · Answer 1 · answered May 11 '21 at 19:28

3

As I said in my comment, the clue appears to be wrong.

But note that if $G \le S_7$ then, since its orbits have size a power of $2$, it must have at least one orbit of size $1$, and so in fact $G \le S_6$, and now we can use the clue.

Now the elements of $S_6$ that have order $4$ have cycle type $(4,2)$ or $(4,1,1)$ and their squares have cycle type $(2,2,1,1)$ in both cases.

The only elements of $S_6$ of order $4$ that square into $(1,2)(3,4)$ are $(1,3,2,4)$, $(1,4,3,2)$, $(1,4,3,2)(5,6)$, and $(1,3,2,4)(5,6)$, but they all commute.

On the other hand, the elements $i$ and $j$ of $Q_8$ do not commute, so we have a contradiction.

answered May 11 '21 at 19:28

Derek Holt

90,008

Could you explain clearer the first paragraph about the orbits please? – Laszlo May 11 '21 at 19:47
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What is unclear about it? Orbits of $Q_8$ have size dividing $8$, so they can only be $1,2,4$ or $8$. But the orbit sizes have to add up to $7$, so there must be at least one size $1$. – Derek Holt May 11 '21 at 22:07
Thank you, i al understanding It better. But why G is a subgroup of S6? – Laszlo May 12 '21 at 07:23
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As I said before, that is because it has an orbit of size $1$. – Derek Holt May 12 '21 at 07:43
And every group which has an orbit of size 1 is contained in S6? I do not know that result. – Laszlo May 12 '21 at 10:34
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Look it's obvious! Think about it. – Derek Holt May 12 '21 at 11:11
I get It. That's because there's a fix point, isn't It? – Laszlo May 12 '21 at 12:14

Prove $Q_8$ is not contained as a subgroup in the symmetric group $S_n$, for $n<8$

1 Answers1