How many homomorphism from $S_3$ to $S_4$? Please find these using fundamental theorem.

I think if $f\colon G \rightarrow G'$ is a group homomorphism then $G/\ker f$ is isomorphic to a subgroup of $G'$. For one choice of $\ker f$, the order of $G/\ker f$ is $k$ and $G'$ has $n$ subgroups of order $k$ . Hence there are $n$ homomorphisms.

But in case of $S_3$ to $S_4$, if $S_3/\ker f$ is a subgroup of $S_4$ then we have three cases:

  1. $\ker f = \{\mathrm{id}\}$: then $S_3$ is isomorphic to a subgroup of $S_4$. There are 4 subgroups in $S_4$ isomorphic to $S_3$ so in this case we have 4 homomorphisms.

  2. $\ker f = A_3$ then $S_3/A_3$ is a subgroup of order 2 in $S_4$. Again, 9 subgroups in $S_4$ so 9 homomorphisms.

  3. $\ker f = S_3$ which gives 0 homomorphism.

So in total 14 homomorphisms. Am I right ?

  • 1
    $\begingroup$ For any two groups $G_1$ and $G_2$, we can always define the so-called trivial homomorphism $x \mapsto e_{G_2}$ for all $x \in G_1$. In particular, the kernel of this homomorphism is the whole group. Point being, case $3$ does add $1$ map to the list. $\endgroup$ – Kaj Hansen May 20 '17 at 1:18
  • $\begingroup$ Please, next time use MathJax to properly write the mathematical terms. It adds much clarity and attracts attention. You can find formatting tips here $\endgroup$ – AspiringMathematician May 20 '17 at 2:18

There are 34 homomorphisms from $S_3$ to $S_4$.

Let's counting homomorphisms by analysis of its kernel.

Case 1. $S_3$ is the kernel: As Kaj Hansen commented, there is the trivial homomorphism and this is the only choice.

Case 2. $A_3$ is the kernel: There are 9 homomorphisms.

Case 3. $1$ is the kernel: Your argument is wrong; as you've already noticed, there are 4 subgroups (the stabilizers of a single letter) that isomorphic to $S_3$. But you also have to take permutations (rename of letters) into account. Hence there are $4 \times 3! = 24$ homomorphisms.

After all, we have $1 + 9 + 24 = 34$ homomorphisms from $S_3$ to $S_4$.

Generally speaking, let $G, G'$ be finite groups and $N$ a normal subgroup of $G$. Suppose $G'$ has $n$ subgroups that isomorphic to (not just the orders are same) $G/N$. What we can say is the number of homomorphisms from $G$ to $G'$ with its kernel $N$ equals $n \times \vert \operatorname{Aut}(G/N) \vert$.

  • $\begingroup$ excellent answer! By the way, Although it's old question, may i ask you simple things?.In the last sentence, How can i construct homomorphism $G$ from $G'$ using $Aut(G/N)$ ? I want to know how to prove #homo = $n \mid Aut(G/N)\mid $. $\endgroup$ – hew Sep 22 '19 at 10:00
  • 1
    $\begingroup$ @hew Let me write the sketch of a proof. Write $H'_1, \dotsc, H'_n$ for subgroups of $G'$ isomorphic to $G/N$. Then there are isomorphisms $i_k \colon G/N \to H'_k$. For each $a \in \operatorname{Aut}(G/N)$, the composition $i_k a p \colon G \to G'$ is a homomorphism with kernel $N$ where $p \colon G \to G/N$ is the canonical map. On the other hand, given a homomorphism $f \colon G \to G'$ with kernel $N$, there is a unique $1 \le k \le n$ with $f(G) = H'_k$. Then you can get $a \in \operatorname{Aut}(G/N)$ by $a(gN) = i_k^{-1}(f(g))$. Please check by yourself that these are mutually inverse. $\endgroup$ – Orat Sep 22 '19 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.