# How many homomorphism from $S_3$ to $S_4$?

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 ?

• 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. – Kaj Hansen May 20 '17 at 1:18
• Please, next time use MathJax to properly write the mathematical terms. It adds much clarity and attracts attention. You can find formatting tips here – 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 $n \times \vert \operatorname{Aut}(G/N) \vert$.

• 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 = $n \mid Aut(G/N)\mid$. – hew Sep 22 '19 at 10:00
• @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. – Orat Sep 22 '19 at 12:47