Determine all elements of group $\operatorname{Hom}(S_{3}, \mathbb{Z}_{2})$.

Question

Determine all elements of group $\operatorname{Hom}(S_{3}, \mathbb{Z}_{2})$.

I was thinking of mapping generators of group $D_{3}$ (which is isomorphic to the group $S_{3})$ to elements of group $\mathbb{Z}_{2}$. So, if $r,f$ are generators of $D_{3}$, respecting the the rule that homomorphism maps neutral element to neutral element, I believe that there are $2$ homomorphisms, $$f_{1} (r) = f_{1}(f) = 0 $$ and $$f_{2} (r) = 0,\quad f_{2}(f) = 1.$$ Is that good thinking?

Answer 1

In the dihedral group $D_3$ (or $D_6$ depending on notation) you have the relations $r^3=f^2=rfrf=e$ (and those relations generate all relations). Hence, if you define a map to another group $G$ $$ \varphi: D_3 \rightarrow G, \ \varphi(r)=g, \ \varphi(f)=h $$ then this will define a group homomorphism iff (now I use multiplicative notation for the group $G$) $$ g^3=\varphi(r)^3 = e_G, \ h^2=\varphi(f)^2 = e_G, \ ghgh=\varphi(r) \varphi(f) \varphi(r) \varphi(f)= e_G.$$ In our case, this means (now in additive notation) in $\mathbb{Z}/2\mathbb{Z}$ $$ 3g = 0, \ 2h =0, 2g + 2h =0 $$ This implies $g=3g=0$ and the remaining equations are satisfied no matter what the choice of $g,h\in \mathbb{Z}/2\mathbb{Z}$. Hence, we get exactly the two choices you wrote.

Answer 2

Any nontrivial homomorphism to the group $\mathbb{Z}/2\mathbb{Z}$ (or any cyclic group of prime order) must be surjective. Also, the group $\mathbb{Z}/2\mathbb{Z}$ has only the identity automorphism. Hence, it suffices to determine all the subgroups of $S_3$ of index 2 (which would then automatically be normal, because any subgroup of index 2 in any group is normal). Equivalently, we need to determine the elements of order 3.

The following shows the order of each element of $S_3$:

• 1, 2, 3: This is the identity, so it has order 1.
• 1, 3, 2: This interchanges 2 with 3, so it has order 2.
• 2, 1, 3: This interchanges 1 with 2, so it has order 2.
• 2, 3, 1: This is the 3-cycle $\begin{pmatrix}1&2&3\end{pmatrix}$, so it has order 3.
• 3, 1, 2: This is the 3-cycle $\begin{pmatrix}1&3&2\end{pmatrix}$, so it has order 3.
• 3, 2, 1: This interchanges 1 with 3, so it has order 2.

The 2 permutations of order 3 are inverse to each other, so they generate the same cyclic subgroup of order 3, which is $\{e, \begin{pmatrix}1&2&3\end{pmatrix}, \begin{pmatrix}1&3&2\end{pmatrix}\}$. This is exactly the alternating group $A_3$.

The two homomorphisms are therefore the trivial one and the parity homomorphism (mapping $A_3$ (even permutations) to zero and the 2-cycles (odd permutations, and also involutions other than the identity) to one).