2
$\begingroup$

These questions are pretty standard, this is the first time I'm trying one on my own, I would like to check my progress:

Using the First Isomorphism Theorem and Lagrange’s Theorem, find all the homomorphisms from $ S_3 \to C_4 $.

My Answer: there are 2, argumentation:

In order to determine all homomorphisms, it is wise to determine the possible sizes of the kernel and the image of such homomorphisms. We know that $|S_3|=3!=6$. Furthermore, for any group homomorphism $f: G \to G'$, corollary 5, enables us to write: $$ |G|=|\text{im}(f)|\times| \ker(f)|$$ $$ |S_3|=6=|\text{im}(f)|\times| \ker(f)|$$ The product of kernel and image can only be 6, this is a good restriction. We will now use the first isomorphism theorem, which tells us: $$\ker(f) \triangleleft S_3 $$ $$\text{im}(f) \leq C_4 $$ Hence the kernel and image are always subgroups, respectively of our orginal group and of the group that is mapped to. We can therefore use Lagrange's theorem, which states that the order of a subgroup must always divide the order of the group. Since $|S_3|=6$ and $|C_4|=4$, this tells us that the possible candidates are: $$ \ker(f)=1, 2, 3, 6 $$ $$ \text{im}(f)=1,2 , 4 $$ We already knew that the product of these two cardinalities/orders had to be $6$. The only possible products that lead to this are for $3 \cdot 2 =6$ and $6 \cdot 1 =6$, so $(|\ker(f)|, |\text{im}(f)|) \in \{(3,2),(6,1)\}$.

The trivial homomorphism that sends every single element of $S_3$ to the identity element in $C_4$ is always a group homomorphism. So, whenever $|\text{Im}(f)|=1$ and $\ker(f)=6$ we have that: for $e\in C_4$ and $\forall x \in S_3$ the trivial homomorphism $f: S_3 \to C_4$ is defined by: $$ f(x)=e$$ The other option is that $|\text{im}(f)|=2$ (and kernel of order 3). Since the image is a subgroup by the isomorphism theorem, and by corollary 3, the order of any element in the group needs to divide the order of the group. Thus we can only have elements of order $2$ or $1$ (identity) in the image. We represent $C_4 = \{e, c, c^2, c^3\}$, with generator $c$. We realise that the only element of order $2$ is $c^2$: $$\text{im}(f) = \langle c^2 \rangle = \{e, c^2\} $$ We also know that the kernel of this homomorphism needs to be of order $3$. Again, the kernel is a subgroup, so the order of the elemens need to divide the order of the group. Elements in the kernel must have orders $1$ or $3$. We know the identity element must always be in the kernel. We first list all elements of $S_3$ and determine the orders so we can decide what can be in the kernel: $$ S_3 = \{e, ( 1 \ 2 \ 3), ( 1 \ 3 \ 2), ( 1 \ 2), ( 2 \ 3), ( 1 \ 3) \} $$ We notice that the identity is of order 1, and the two "cycles" $( 1 \ 2 \ 3), ( 1 \ 3 \ 2)$ are of order 3, they fit the description. We then define the homomorphism $f: S_3 \to C_4$ that maps: $$e, ( 1 \ 2 \ 3), ( 1 \ 3 \ 2) \to e'\in C_4 $$ $$ ( 1 \ 2), ( 2 \ 3), ( 1 \ 3) \to c^2 \in C_4$$

Conclusion: There are 2 homomorphismsm the one above and the trivial homomorphism.

$\endgroup$
  • 1
    $\begingroup$ Or $(1\,2\,3)\in S_3$ is of order $3$, which is coprome to the order of $C_4$, hence $(1\,2\,3)\mapsto e$ for all homomoprhism $S_3\to C_4$. Therefore, each such homomorphism factors over $S_3/\langle (1\,2\,3)\rangle\cong C_2$. Clearly, there are exactly twho homomorphisms $C_2\to C_4$ (the non-trivial element must map to an element of order $1$ or $2$) $\endgroup$ – Hagen von Eitzen Mar 7 '19 at 21:27
3
$\begingroup$

Towards the end your proof is a bit roundabout, but it is entirely valid, except perhaps for one point:

At the very end you define a map $f:\ S_3\ \longrightarrow\ C_4$, but you do not verify that it is in fact a homomorphism. At this point, you have only shown that if there exists a homomorphism $f$ with $|\ker(f)|=3$ and $|\operatorname{im}(f)|=2$, then this must be it.

As for the proof being roundabout towards the end; once you have determined that $$(|\ker(f)|, |\text{im}(f)|) \in \{(3,2),(6,1)\},$$ you could finish the argument more clearly by distinguishing two cases;

  1. Prove that if $|\ker(f)|=6$ then $f$ is the trivial homomorphism.
  2. Prove that if $|\ker(f)|=3$ then $f$ is the sign homomorphism.

Then you can conclude that there are precisely two homomorphisms from $S_3$ to $C_4$.

$\endgroup$
  • $\begingroup$ Yes it behaves a whole lot like the sign homomorphism, thank you for verifying. $\endgroup$ – Wesley Strik Mar 8 '19 at 20:45
3
$\begingroup$

Servaes has already given an effective critique, and one that I can't improve upon, but let me point out a more efficient way to solve this using some observations that are useful in other problems of this form.

First, since the target $C_4$ is abelian, for any $g, h \in S_3$ and any homomorphism $\phi: S_3 \to C_4$ we have $\phi(ghg^{-1}h^{-1}) = 0$, so $\phi$ factors through a homomorphism $\widetilde\phi : S_3 / [S_3, S_3] \to C_4$, i.e., if $\pi : S_3 \to S_3 / [S_3, S_3]$ denotes the canonical quotient map, there is a homomorphism $\widetilde\phi$ such that $\phi = \widetilde\phi \circ \pi$. Here $[S_3, S_3]$ is the commutator subgroup of $S_3$, namely the subgroup generated by all elements of the form $g h g^{-1} h^{-1}$. In particular any element is an even permutation so $[S_3, S_3] \leq A_3 \cong C_3$. On the other hand, since $S_3$ is not abelian, $[S_3, S_3]$ is not trivial, so $S_3 / [S_3, S_3] \cong C_2$ and the kernel of $\pi: S_3 \to S_3 / [S_3, S_3]$ is precisely the subgroup $A_3$ of even permutations.

So, to find all the homomorphisms $S_3 \to C_4$ it suffices to find the homomorphisms $\widetilde\phi : S_3 / [S_3, S_3] \cong C_2 \to C_4$ and then compose them with $\pi$.

Now, $S_3 / [S_3, S_3] \cong C_2$ is generated by a single element, $a$, of order $2$, so $e = \widetilde\phi(e) = \widetilde\phi(a^2) = \widetilde\phi(a)^2$. Thus, $\widetilde\phi(a)$ has order $1$ or $2$.

  • If it has order $1$, then $\widetilde\phi$ is trivial homomorphism and hence so is $\phi = \pi \circ \widetilde\phi$.
  • If it has order $2$, then since the only element of $C_4 = \langle e, b, b^2, b^3 \rangle$ of order $2$ is $b^2$, $\widetilde\phi(a) = b^2$, and one can check directly that this together with $\widetilde\phi(e) = e$ does define a homomorphism, and hence so is $\phi = \pi \circ \widetilde\phi$. Using our above description of $\ker \pi$, we see that $\phi$ is the map that sends even permutations to $e$ and odd permutations to $b^2$.
$\endgroup$

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.