# the smallest number for faithful operation

For each of the following groups, find the smallest integer $$n$$ such that the group has a faithful operation on a set of order $$n$$: (a) $$D_4$$, (b) $$D_6$$, (c) the quaternion group $$H$$.

For $$D_4$$ the $$n$$ is $$4$$: As $$|D_4| = 8$$ and we can the the square symmetry group to show this.

For $$D_6$$ the $$n$$ is $$6$$: As $$D_6$$ has a cycle group of order is $$6$$, so it can not be less than $$6$$.

For $$H$$ the $$n$$ is $$5$$: As $$H$$ has three cycle group of order is $$4$$, so it can not be $$4$$ and we can find three cycle of group in $$S_5$$.

I wonder is my thought is correct and there is a easy way to prove it?

Your thoughts are, for the most part, correct, but you still get the wrong answer. That said your mistakes are perfectly understandable, so let's take a second to talk this through.

Recall that a group $$G$$ acts faithfully on a set of size $$n$$ if and only if $$G$$ is a subgroup of $$S_n$$, the symmetric group on $$n$$ letters. So secretly we're interested in the subgroups of $$S_n$$.

For $$D_4$$, you're completely correct that we get a faithful action on a set of size $$4$$ by permuting the vertices of a square. You're also correct that this is best possible. Indeed, $$|S_3| = 6 < 8 = |D_4|$$, so it's not possible for $$D_4$$ to be a subgroup of any symmetric group smaller than $$S_4$$. Thus here, $$n=4$$ works.

For $$D_6$$, you want to say that it has an element of order $$6$$, so can't embed in $$S_n$$ for $$n < 6$$. Since there's an obvious faithful action of $$D_6$$ on six points (permute the vertices of a regular hexagon), this should give the answer. But notice $$|S_5| = 5! = 120$$ is easily big enough to contain a $$D_6$$ (of size $$12$$) as a subgroup, and $$S_5$$ does have elements of order $$6$$ (indeed, $$(1 \ 2 \ 3)(4 \ 5)$$ is an example). So there's no obvious reason $$S_5$$ couldn't contain a $$D_6$$, and in fact it does! See here for a proof. The next obvious question is whether $$S_4$$ contains a copy of $$D_6$$. Since $$|S_4| = 24$$ this might still be possible, but now your idea comes to the rescue: $$S_4$$ has no elements of order $$6$$. See here for a proof. So now $$n=5$$ works.

Lastly, for $$H$$ the quaternion group, looking at the cycle structure is a good idea, but in fact here $$n=8$$ is the best we can do. Obviously $$n=8$$ works, since $$H$$ acts on itself by left multiplication. But it's kind of subtle to see why $$n=7$$ fails. See here for a proof (in the question) as well as a slick version of the same proof (in the answer).

As an aside, this question is quite hard in general, so you shouldn't feel bad for struggling with it. This $$n$$ is usually denoted $$\mu(G)$$ for a group $$G$$, and computing $$\mu(G)$$ for every finite group is still super unsolved! We know the answer when $$G$$ is abelian, and in other special cases too, but there's lots of work to be done. See here and here for more information.

As another aside, $$\mu(D_n)$$ is known too, and it can be much smaller than $$n$$. We already see this with $$D_6$$ embedding into $$S_5$$, but astonishingly $$D_{360}$$ is a subgroup of $$S_{22}$$! Even more spectacularly, $$D_{9240}$$ is a subgroup of $$S_{34}$$! In fact, as long as $$n$$ has many prime divisors, none of which is "too small", then $$D_n$$ embeds into an $$S_m$$ with $$m \ll n$$. See an old blog post of mine here for more details.

I hope this helps ^_^

• Thanks very much for such a great comprehensive answer. This help me a lot. I think this is a exercise for expand reading. Jul 12 at 3:26