# Number of homomorphisms from $D_6$ to $A_4$

Clarification for notation: $$|D_6|=6$$. Also $$|A_4|=12$$

So I know that there are $$2$$ elements of order $$3$$ and $$3$$ of order $$2$$ in $$D_6$$ (and $$1$$ of order $$1$$) and in $$A_4$$ there are $$3$$ elements of order $$2$$ and $$8$$ elements of order $$3$$.

So my answer is that I must send the $$3$$ elements of order $$2$$ in $$D_6$$ to the $$3$$ elements of order $$2$$ in $$A_4$$ and then I have $$\frac{8!}{2!6!}=28$$ possible homomorphisms (choosing $$3$$ elements of order $$3$$ from the $$8$$ available in $$A_4$$).

I feel $$28$$ is too much compared to other examples so I don't think the answer is right. I have also tried to build an homomorphism to $$S_4$$ since every group is isomorphic to a subgroup of $$S_n$$.

• In a group homomorphism, you cannot map elements independently of each other. For example if $a\mapsto x$ and $b\mapsto y$ then $ab\mapsto xy$ is forced.
– tkf
Nov 9 '20 at 20:43
• Note: $D_6\cong S_3$ Nov 9 '20 at 20:45
• Note that homeomorphisms are not necessarily one to one Nov 9 '20 at 21:27

There's no surjection. Since the image of $$h$$ is a subgroup, it either has order $$6,4,3$$ or $$2$$. But $$4\nmid6$$.
Now, if the image has order $$6$$, then $$h$$ is an isomorphism. But that's out: $$A_4$$ doesn't have a copy of $$S_3$$ as a subgroup. In fact it has no subgroup of order six.
So the image has order $$3$$ or $$2$$. If the image has order $$3$$, the kernel has order two. There are $$3$$ subgroups of order $$2$$, but they are not normal. So the image can't have order $$3$$.
If the image has order $$2$$, the kernel has order $$3$$. There is one subgroup of order $$3$$ in $$D_6$$.
There are $$3$$ copies of $$C_2$$ in $$A_4$$.
Thus we get $$4$$, when we throw in the trivial one.