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 $S_n$.

  • 1
    $\begingroup$ 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. $\endgroup$
    – tkf
    Nov 9 '20 at 20:43
  • 3
    $\begingroup$ Note: $D_6\cong S_3$ $\endgroup$ Nov 9 '20 at 20:45
  • 1
    $\begingroup$ Note that homeomorphisms are not necessarily one to one $\endgroup$ 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.


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