number of homomorphism from $K_4\to S_4$

Question

I want to calculate number of homomorphism from $K_4\to S_4$. let $\phi:K_4\to S_4$ be homomorphism then for each element $x\in S_4$ WHERE $|x|=2$ we have 3 cases , $ord(\phi(a))=1$ and $ord(\phi(b))=2$, or $ord(\phi(b))=1$ $ord(\phi(a))=2$ or both $ord(\phi(a))=2~,~ord(\phi(b))=2 $ we have 9 elements of order 2 , for each case we have 3 homomrphism, then total $9*3=27$ homorphism and $1$ for identity so we have 28 homomorphism with images having order 2 and 1,

now we need homomorphism having images with order four, we have 4 subgroups of order 4 which are not cyclic and since $Aut(K_4)=6$ we got $6*4$ homomorphisms and hence total homomorphisms are $24+28=52$

is it correct?

$K_4$ is Klein-4 group.

"for each element except identity we have 3 homomrphism" - Can you explain why you think so? (As best I can tell, you're assuming the order of $\phi(a)$ equals the order of $\phi(b)$) — Brian Moehring, Jun 18 '24 at 05:42
i have 3 case let $c\in S_4$ of order 2, then $\phi(a)=e~~and~~ \phi(b)=c$, or $\phi(b)=e~~and~~ \phi(a)=c$, or $\phi(a)=c~~and~~ \phi(b)=c$ — Ricci Ten, Jun 18 '24 at 05:53
Okay, you weren't counting how I expected you were. Where does $\phi(a)=(12), \phi(b)=(34)$ appear in your count? — Brian Moehring, Jun 18 '24 at 06:13
Again, you need to explain how you're counting. As best I can tell, you're including functions like $a \mapsto (12), b\mapsto (23)$, which don't generate a homomorphism. — Brian Moehring, Jun 18 '24 at 06:31
From the [tag:solution-verification] tag wiki: "A question with this tag should include an explanation for why the argument presented is not convincing enough." — Shaun, Jun 18 '24 at 11:31
You should say, in the first part, for each of the $3$ cases we have $9$ homomorphisms. And point out how it works in the third case. Namely that $\phi(a)=\phi(b).$ — suckling pig, Jun 19 '24 at 03:24

suckling pig · Accepted Answer · 2024-06-18T09:04:38.757

First all the images of elements of $K_4$ need to have orders dividing $2.$ (I'll let you supply the reason.)

Now I'll take a stab.

There are four $K_4$'s in $S_4.$ Then use that $\rm Aut(K_4)\cong S_3.$ Composing, that gives $6\cdot4=24.$

Indeed I count $9$ elements of order two in $S_4.$

Now suppose that the image is $\Bbb Z_2.$ There's $9$ choices for the image of $(1,0).$ $(0,1)$ goes to the identity. And we can reverse the generators. That gives $18.$

Both generators can also go to the same element of order two, for $9$ more.

Finally we have the trivial homomorphism.

We get $52.$

number of homomorphism from $K_4\to S_4$

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