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Question How many group homomorphisms are there from $D_4$ to $S_4$?

My solution. Here, $$D_4:=\langle r,s| r^4=s^2=e, ~ sr=r^3s\rangle$$ Let $f:D_4 \to S_4$ be a homomorphism then $|f(r)|=1,2 ~\text{or} ~4$ and $|f(s)|=1 ~\text{or} ~2$. Now there are 9 elements of order 2 in $S_4$ and 6 elements of order 4 in $S_4$. Therefore total choice of $f(r)$ is 1.9.6=54. And that of $f(s)$ is 1.9=9.

Hence the total number of such $f$ is 54.9=486.

Is this correct? Please let me know where I made mistake. Thank you.

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  • $\begingroup$ No, you can't just randomly send elements to ones of appropriate orders and get a homomorphism. $\endgroup$ – Tobias Kildetoft Aug 10 '18 at 7:13
  • $\begingroup$ Then how to proceed..? $\endgroup$ – Indrajit Ghosh Aug 10 '18 at 7:14
  • $\begingroup$ But I didn't maps randomly...I first take a homomorphism $f$ and then this homomorphism can be determined if we can choose what value it can take at the generators....finally applying $f(ab)=f(a)f(b)$ we get really a homomorphism.....where is the mistake? $\endgroup$ – Indrajit Ghosh Aug 10 '18 at 7:18
  • $\begingroup$ @IndrajitGhosh you're treating $D_4$ as if it were the free group in two generators, but it is not: you must take into account the relations between these. $\endgroup$ – Guido A. Aug 10 '18 at 7:20
  • $\begingroup$ You have to make sure that the images of $r$ and $s$ satisfy the same relations as they do in $D_{4}$, or else your homomorphism will not be well-defined. You can imagine that a random choice of an element $f(s)$ of order $2$ and an element $f(r)$ of order (say) $4$ need not satisfy $f(sr) = f(r^{3}s)$. $\endgroup$ – Alex Wertheim Aug 10 '18 at 7:22
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Normal subgroups of $D_4$ are $N_1:=\text{Id}$, $N_2:=\langle r^2\rangle$, $N_3:=\langle r\rangle$, $N_4:=\langle r^2,s\rangle$, and $N_5:=D_2$. Let $n_k$, where $k\in\{1,2,3,4,5\}$, denote the number of group homomorphisms $\psi:D_4\to S_4$ such that $$\ker(\psi)=N_k\,.$$

Obviously, $n_5=1$ as the only group homomorphism $D_4\to S_4$ with kernel $N_5=D_4$ is the trivial homomorphism $x\mapsto 1_{S_4}$ for all $x\in D_4$. Now, $D_4/N_4\cong C_2$, where $C_r$ denote the cyclic group of order $2$. There are in total $$\binom{4}{2}+\frac{1}{2}\,\binom{4}{2}=9\text{ subgroups of }S_4\text{ isomorphic to }C_2\,.$$ Since $\big|\text{Aut}(C_2)\big|=\varphi(2)=1$, where $\varphi$ is Euler's totient function, we conclude that $n_4=1\cdot 9=9$.

Next, we again have $D_4/N_3\cong C_2$. As before, we obtain $n_3=9$. Now, $D_4/N_2\cong C_2\times C_2$. There are $$1+\frac{1}{2}\,\binom{4}{2}=4\text{ subgroups of }S_4\text{ isomorphic to }C_2\times C_2\,.$$ Since $\big|\text{Aut}(C_2\times C_2)\big|=(2^2-1)(2^2-2)=6$, we get $n_2=6\cdot 4=24$.

To calculate $n_1$, we must first count the number of subgroups of $S_4$ isomorphic to $D_4$. Since $|D_4|=8=2^3$ and $2^3$ is the largest power of $2$ that divides $|S_4|=4!=24$, we conclude that $D_4$ is a $2$-Sylow subgroup of $S_4$. The $2$-Sylow subgroups of $S_4$ are all conjugates. Since the only normal subgroups of $S_4$ are $\text{Id}$, $V$, $A_4$, and $S_4$ (where $V$ is isomorphic to the Klein $4$-group $C_2\times C_2$ and $A_4$ is the alternating group of order $\dfrac{4!}{2}=12$), and none of which is of order $8$, we see that $D_4$ is not a normal subgroup of $S_4$. Note that the intersection of two distinct $2$-Sylow subgroups of $S_4$ is exactly $V$. Since the number of $2$-Sylow subgroup is an odd number and we already know that this number is not $1$, and since $$|V|+j\,\big(|D_4|-|V|\big)=4+j\,(8-4)> 24-4=|S_4|-\Big|\big\{x\in S_4\,\big|\,x\text{ is of order 3}\big\}\Big|$$ for $j\geq 5$, we conclude that the number of $2$-Sylow subgroups in $S_4$ is precisely $3$. (Alternatively, the number $s$ of $2$-Sylow subgroups of $S_4$ is an odd number dividing $\dfrac{|S_4|}{2^3}=3$, and since $s>1$, we get $s=3$.) From this link, we know that $\big|\text{Aut}(D_4)\big|=4\cdot\varphi(4)=8$. Hence, $n_1=8\cdot 3=24$. For a list of all subgroups of $S_4$ isomorphic to $D_4$, see here.

In summary, the number of group homomorphisms from $D_4$ to $S_4$ is $$n_1+n_2+n_3+n_4+n_5=24+24+9+9+1=67\,.$$ In other words, $$\big|\text{Hom}(D_4,S_4)\big|=67\,.$$

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