How many homomorphisms are there from $S_6$ to $D_8$?

Question

I tried to look at the generators of $S_6$, however there are multiple generators for $S_6$, so I was not sure to look at which one. I chose $(1 \ 2)$ and $(1 \ 2 \ 3 \ 4 \ 5 \ 6)$, because it is simple. So element $\sigma$ of order 2 and element $\tau$ of order 6.

Then we know that, $f(\sigma)$ has to have an order that divides 2, so either an order of 1 or an order of 2. There are 10 elements in $D_8$ that have such an order.

We also know that $f(\tau)$ has to have an order that divides 6, so an order of 1,2,3 or 6 is sufficient. There are once again 10 elements in $D_8$ that have such an order.

So in total there would be 100 homomorphisms, according to this reasoning. I feel, however, that I am doing something wrong. My questions are:

• Is there some sort of extra structure in $D_8$ that I should account for? If so, how can I do that?
• When there are multiple generators for a group, does it matter which one you take for the amount of homomorphisms?
• In $D_8$, do all the reflections have an order of 2?

Thanks for reading,

K.

Answer 1

It's not quite that simple. You need to also account for all the relations among your generators, and ensure that they are satisfied by any candidate mapping into the target group.

However, in this case, your job is simplified a bit because the symmetric group $S_6$ has very few normal subgroups: $1$, $A_6$ and $S_6$ itself. Therefore, the image of any homomorphism must be isomorphic to $S_6 \cong S_6/1$, $C_2 \cong S_6/A_6$ or $1 \cong S_6/S_6$. The first one cannot occur, and the last one is unique. Now you can just look for suitable subgroups of $D_8$, and you'll end up with a much shorter list of candidates.

Answer 2

Let $\phi: S_6 \rightarrow D_8$ be a group homomorphism. Recall that $\operatorname{Ker}(\phi)$ is a normal subgroup of $S_6$ and therefore $\operatorname{Ker}(\phi)\in \{ \{ \operatorname{id} \}, A_6, S_6 \}$. Thus, we have

$$ \vert \operatorname{Ker}(\phi) \vert \in \{ 1 , 360, 720 \} .$$

Furthermore we have

$$ \vert \operatorname{Im}(\phi) \vert = \frac{\vert S_6\vert}{\vert Ker (\phi) \vert} =\frac{720}{\vert \operatorname{Ker} (\phi) \vert}. $$

This implies that $\vert \operatorname{Im}(\phi) \vert \in \{ 1 , 2 \}$. Set

$$ s_1=(12), \ s_2=(23), \ s_3=(34), \ s_4=(45), \ s_5=(56).$$

These elements generate $S_6$ and all relations come from the relations

$$ s_j s_i = s_i s_j \quad j\neq i \pm 1 $$

and

$$ (s_i s_{i+1})^3 = \operatorname{id} \quad i\in \{1, \dots, 4\}.$$

The order of $\phi(s_j)$ divides $2$. Thus, either $\phi(s_j)=e$ or is of order two. Assume that $\phi$ is not the trivial group homomorphism. Then there exists a smallest $j_0\in \{1, \dots, 5 \}$ such that $\phi(s_{j_0})=g\in D_8\setminus \{e\}$. As the image has order 2, we have $\phi(s_{j_0+1})\in \{ e, g \}$ By the second relation we must have

$$ (g \phi(s_{j_0+1}))^3 = e. $$

As $g$ has order 2, this implies $\phi(s_{j_0+1})=g$. By the same argument, we get $\phi(s_j)=g$ for all $j\in \{ 1, \dots, 5\}$.

One easily sees that both relations are satisfied. Thus,

$$ \phi(s_j)=g $$

really defines a group homomorphism. We finally get that the number of group homomorphisms from $S_6$ to $D_8$ is

$$ 1 + \vert \{ g\in D_8 : \operatorname{ord}(g)=2 \} \vert =10.$$

Added: Here is a much easier proof without generators and relations. Let $\phi: S_6 \rightarrow D_8$ be a non-trivial group homomorphism. As we have $\operatorname{Ker}(\phi)=A_6$ and $\vert \operatorname{Im}(\phi) \vert=2$, we have that $\phi$ is of the form

$$ \phi(\sigma) = \begin{cases} e,& \sigma \in A_6; \\ g,& \sigma\notin A_6 \end{cases} $$

where $g$ has order two. One checks that this really is a group homomorphism and thus one gets that the number of group homomorphisms is the number of elements of order 2 in $D_8$ plus 1 (for the trivial homomorphism).