Showing There is No Group Epimorphism from $S_n \longrightarrow \mathbb{Z}/2\times\mathbb{Z}/2$, $n\geq 1$

Question

I know that the function $\text{sgn}: S_n\longrightarrow \mathbb{Z}/2$ is a unique group epimorphism. I am having trouble proving that there does not exist such an epimorphism bewteen $S_n\longrightarrow \mathbb{Z}/2\times\mathbb{Z}/2$. My approach is as follows

As a suggestion by my instructor, I consider a homomorphism $\phi: \mathbb{Z}/2\times\mathbb{Z}/2\longrightarrow\mathbb{Z}/2$, defined by $(a,b)\longmapsto a+b$, and suppose there exists an epimorphism $\psi: S_n\longrightarrow \mathbb{Z}/2\times\mathbb{Z}/2$. Then $\phi\circ\psi: S_n\longrightarrow \mathbb{Z}/2$. We compare this to the $\text{sgn}$ function since we know it has the same domain and codomain as this composite function. So I assume we are supposed to identify some choice (or rather, conclude there is no such choice of) of $\psi$ such that this composite is equivalent to $\text{sgn}$. I tried defining a $\psi$ like $(0,\cdots,n-1)\mapsto(n_i,n_j)$, where $n_i$ and $n_j$ are some natural numbers in that $n$-cycle, and they are modded by 2. I have stated that this is a surjection since the image will be equal to $\{(0,0),(0,1),(1,0),(1,1) \}=\mathbb{Z}/2\times\mathbb{Z}/2$ depending on $n$. Then, $\phi(n_i,n_j)=n_i+n_j=\{0,1 \}$. I think that this is equal to $\text{sgn}$ since in class we were told that the signum of an $n$-cycle 0 if $n$ is even, and 1 if $n$ is odd. However, I suspect my choice of $\psi$ is not actually a homomorphism, and I am having trouble showing whether or not it is. Furthermore, I feel not confident about just choosing a $\psi$, as this doesn't really show this is the case for all $\psi$. Another thing to note is that my professor suggested testing the homomorphisms for $\phi$ defined by $(a,b)\mapsto a$ and $(a,b)\mapsto b$. I tried them with the same $\psi$, but I am failing to see how this helps exactly.

Any guidance will be much appreciated. Thanks!

Answer 1

Here is a simpler approach (I think). It avoids things like maps out of $\mathbf Z/(2) \times \mathbf Z/(2)$ to a smaller group, and all it relies on is that $\mathbf Z/(2) \times \mathbf Z/(2)$ is abelian with more than $2$ elements.

The case $n = 1$ is trivial, so let $n \geq 2$.

When $n \geq 2$, the group $S_n$ is generated by transpositions, which all have order $2$ and are all conjugate to each other. So if we have a homomorphism $f : S_n \to A$ where $A$ is an abelian group written additively, then $f$ has the same value on all transpositions, which implies $f$ is completely determined by the single value $f((12))$. When a permutation $\sigma \in S_n$ is a product of $r$ transpositions, we have (i) $2f((12)) = 0$ in $A$ and (ii) $f(\sigma) = rf((12))$. Thus the image of $f$ is the subgroup generated by $f((12))$, which has order $1$ or $2$. So when $|A| > 2$, a group homomorphism $S_n \to A$ can't be surjective.

Answer 2

Let $p_1(a,b)=a$ and $p_2(a,b)=b$. Note that these are epimorphisms. Assume for the sake of contradiction that there exists an epimorphism $f$ from $S_n$ to $\mathbb Z/2 \times \mathbb Z/2$.

Then, $f_1=p_1 f$ and $f_2=p_2 f$ are epimorphisms. Since $\text{sgn}$ is the unique epimorphism from $S_n$ to $\mathbb Z/2$, we have that $f_1=\text{sgn}=f_2$. This means that since $0\not \equiv 1 \mod 2$, we have that $(0,1)$ is not in the image of $f$, so $f$ is not surjective and so is not an epimorphism.

Answer 3

By contradiction, if it existed, then $S_n$ would have a normal subgroup of order $\frac{n!}{4}$ (first homomorphism theorem). But, for $n\ge5$, the only proper normal subgroup of $S_n$ is $A_n$, of order $\frac{n!}{2}$. On the other hand, $S_4$ hasn't got any normal subgroup of order $6$. The cases $n=3,2$ are trivial. This, jointly with the fact that $S_4$ has a normal subgroup of order $4$, shows more, indeed: that the only possible epimorphism from $S_n$, for every $n\ge2$, is onto $C_2$, and the case $n=4$ yields as additional possible epimorphisms those onto either $C_6$ or $S_3$.