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Describe all ways in which $S_3$ can operate on a set of four elements.

My approach: This question can be broken down into: How many homomorphisms exist from $S_3$ to $S_4$. Say $\varphi : S_3 \to S_4$ is a homomorphism. Then we have three possibilities for $\text{ker }\varphi$: $\{1\}, \{1, (1\ 2\ 3), (1\ 3\ 2)\}$, and $S_3$.

The case in which $\text{ker }\varphi = S_3$ is the trivial homomorphism that maps everything to the identity.

Now, the case in which $\text{ker }\varphi = \{1\}$ is the same as saying that the mappings are injective. This comes down to picking three of the four elements and permutating them and leaving the fourth one fixed. There are $\binom43 = 4$ ways of doing this.

Say $\text{ker }\varphi = \{1, (1\ 2\ 3), (1\ 3\ 2)\}$. This means that $\varphi((1)) = \varphi((1\ 2\ 3)) = \varphi((1\ 3\ 2)) = (1)$. Morover we can observe the following two properties immediately:

  1. $\varphi((1\ 2\ 3)) = \varphi((1\ 3))\varphi((1\ 2)) = (1)$. Equivalently $\varphi((1\ 2)) = \varphi((1\ 3))$.
  2. $\varphi((1\ 3\ 2)) = \varphi((1\ 3))\varphi((2\ 3)) = (1)$. Equivalently $\varphi((1\ 3)) = \varphi((2\ 3))$.

and thus $\varphi((1\ 2)) = \varphi((1\ 3)) = \varphi((2\ 3))$. But we know by properties of homomorphisms that $\vert \varphi((1\ 3)) \vert \mid \vert (1\ 3) \vert = 2$. So $\vert \varphi((1\ 3)) \vert$ is 1 or 2. But if the order of $\varphi((1\ 3))$ were 1 it would be in the kernel, which would be a contradiction to the kernel we chose, so it must be 2. We can map $(1\ 3)$ to any 2-cycle in $S_4$, of which there are 6, as well as any product of disjoint 2-cycles, of which there are 3. Hence we have 9 possible homomorphic mappings given this kernel.

Adding up all the possible homomorphisms from $S_3$ to $S_4$ that we counted, we get 14 different ways in which $S_3$ can act on four elements, as described above.

Is this correct? Are there 14 homomorphisms from $S_3$ to $S_4$? Is my reasoning correct? Or are there any hidden assumptions I made that I shouldn't have made?

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2 Answers 2

up vote 5 down vote accepted

For the "medium-sized" kernel $\{1,(1\,2\,3),(1\,3\,2)\}$, you might simply observer that such an operation is indeed equivalent to a faithful operation $S_3/\ker \phi\cong S_2$. And yes, such an operation is given exectly by choosing an element of order two in $S_4$.

However, I am not satisfied with your argument for trivial kernel, i.e. that an injection $S_3\to S_4$ automatically means that one point is fixed. For example, with six instead of four elements this would obviously not be true: One way to operate would be to perform the same permutations on the first and on thel last three elements. With four elements as actually given here, you are right, but it has to be shown! (Hint: Consider the operation of a 3-cycle.) Moreover, merely selecting three out four elements does not uniquely determine the operation. Ultimately, you should find 24 instead of 4 faithful operations.

After so many comments let's start afresh counting all homomorphisms $f\colon S_3\to S_4$. Note that $S_3$ is generated by $\sigma =(1\,2)$ and $\tau=(1\,2\,3)$, hence $f$ is already determined if we know $f(\sigma)$ and $f(\tau)$. The order of $f(\tau)$ must be a divisor of the order of $\tau$, i.e. either $f(\tau)=1$ or $f(\tau)$ is one of the 8 elements of order 3 in $S_4$. Similarly, $f(\sigma)=1$ or it is one of the 9 elements of order 2.

(i) If $f(\tau)=1$ nothing restricts our choice for $f(\sigma)$, so we find $10$ homomorphisms of this kind.

(ii) If $f(\tau)$ is of order three, say $f(\tau)=(a\,b\,c)$, then our choice for $f(\sigma)$ is somewhat restricted because $\sigma\tau=(2\,3)$ is of order 2. This directly forbids $f(\sigma)=1$. If $f(\sigma)$ affects the fourth element $d$, then wlog. $f(\sigma)(d)=a$ and hence $f(\sigma\tau)(d)=b$. Therefore we must have $f(\sigma\tau)(b)=d$, i.e. $f(\sigma)(c)=d$. But if $f(\sigma)$ permutes $c\mapsto d\mapsto a$, it cannot be of order 2. Hence $f(\sigma)$ merely is a permutation of order two of the set $\{a,b,c\}$. Wlog. $f(\sigma)=(a\,b)$. We have 4 choices for $d$ (the fixpoint) and then 3 choices for $c$ (the other fixpoint of $f(\sigma)$), two choices for $a$ (the image of $c$ under $f(\tau)$). (And in fact any such choice is valid: We simply replace the elements $1, 2, 3$ with $a, b, c$ in that order and leave $d$ fixed, thus this is essentially the canonical operation of $S_3$ on $\{1,2,3\}$). Thus there are exactly $24$ homomorphisms of this kind.

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So your saying that there are 34 possible homomorphisms, but I first have to show that if the mapping is injective, There are 24 unique homomorphisms that arise from it, totaling 24 + 9 + 1 = 34 – Robert Cardona Oct 22 '12 at 6:19
I have two 3-cycles in $S_3$. For the first one, I have 4 possibilities of where I can map it to, so for the second one I would have three possibilities of where I could map it to. Is that correct? Or by choosing where the first one is being mapped to, don't I affect where the second one can be mapped to? – Robert Cardona Oct 22 '12 at 7:01
My bad. Given a three cycle in $S_3$ I have 8 3 cycles I can map it to, the mapping of the second three cycle in $S_3$ follows directly from your choice of the first. Somehow I have to show that one of the 2 cycles in $S_3$ can have 3 possible choices to be mapped to, while the other 2 2-cycles are fixed by your mapping of the first 3 cycle? Please work out the injective cases for $S_3 \to S_4$ and as you mentioned $S_3 \to S_6$. These have been driving me crazy all day :/ – Robert Cardona Oct 23 '12 at 2:20
Another approach I thought of was to pick two 2-cycles in $S_3$ that generate $S_3$ and find out how many different ways you could map them. Like $(1\ 2)$ and $(2\ 3)$. But I don't know what do do once I pick one mapping, how does that affect where I can map the second one to. There are 9 elements of order 2 in $S_4$ but I'm guessing the disjoint 2-cycles wouldn't count, so can map $(1\ 2)$ to 6 2-cycles, how many places can I map $(2\ 3)$ to? given the first constraint? I'm having trouble figuring this out :/ – Robert Cardona Oct 23 '12 at 3:59

We work instead in terms of the corresponding permutations representations (homomorphisms) $\phi: S_3\to S_4$. To avoid confusion we work with $S_3(\{1,2,3\})$ and $S_4(\{a,b,c,d\})$. Let $x = (1,2)$ and $y = (1,2,3)$, so that $$x^2=y^3=1,\text{ }y^2 = (1,3,2),\text{ }xy = (2,3) = yx^2,\text{ }xy^2 = (1,3) = yx.$$

By the First Isomorphism Theorem, $\text{Ker}\,\phi$ is normal in $S_3$ and $\text{Im}\,\phi \cong S_3/\text{Ker}\,\phi$, with isomorphism $\phi(g\text{Ker}\,\phi) = \phi(g)$. But$$yxy^{-1} = xy^2y^{-1} = xy\ne 1,\,x,$$so $S_3$ has no normal subgroups of order $2$, and $\text{Ker}\,\phi$ must be one of $\{1\}$, $A_3$, $S_3$.

If $\text{Ker}\,\phi = S_3$, then$$\text{Im}\,\phi\cong S_3/S_3\cong S_1,$$so $\phi$ is the trivial homomorphism, i.e. it always maps to $(a)(b)(c)(d)$.

If $\text{Ker}\,\phi = A_3$, then$$\text{Im}\,\phi\cong S_3/A_3\cong S_2,$$ $$\phi(1)=\phi(y)=\phi(y^2) = (a)(b)(c)(d),$$while$$\phi(x)=\phi(xy)=\phi(xy^2)$$ has order $2$ and takes the form $(p,q)$ or $(p,q)(r,s)$, for distinct $p,q,r,s\in\{a,b,c,d\}$.

If $\text{Ker}\,\phi = \{1\}$, then$$\text{Im}\,\phi\cong S_3/\{1\} \cong S_3.$$So $\phi(x)$ takes the form $(p,q)$ or $(p,q)(r,s)$, and $\phi(xy)$ takes the form $(p',q')$ or $(p',q')(r',s')$. Without loss of generality, $p' = p$.

  • If $\phi(x) = (p,q)$ and $\phi(xy) = (p,q')$, then $q\ne q'$, or else$$\phi(y) = \phi(x)\phi(xy) = (p,q)(p,q')$$becomes the identity, so $\phi(y) = (q,p,q')$. Because $\phi(x) = (q,p)$, $\phi$ simply maps $\pi\in S_3(\{1,2,3\})$ to the corresponding permutation with $1$, $2$, $3$ replaced by $q$, $p$, $q'$, respectively.
  • If $\phi(x) = (p,q)$ and $\phi(xy) = (p,q')(r',s')$ (or the other way around; these two cases are interchangeable by "switching" $1$ and $3$, since $x$, $xy$ are both transpositions), then $q\ne q'$, or else$$\phi(y) = \phi(x)\phi(xy) = (p,q)(p,q')(r',s')$$becomes a transposition and has order $2$ instead of $3$, so without loss of generality, assume $r' = q$. But then$$\phi(y) = (q,p,q')(q,s') = (q,s',p,q')$$has order $4$ instead of $3$, so this case is impossible.
  • If $\phi(x) = (p,q)(r,s)$ and $\phi(xy) = (p,q')(r',s')$, then $q\ne q'$, or else $\{r,s\} = \{r',s'\}$ and $\phi(x)=\phi(xy)$ forces $\phi(y)$ to be the identity. Without loss of generality, assume $q' = r$, so $\{r',s'\} = \{q,s\}$ means $\phi(xy) = (p,r)(q,s)$. Thus$$\phi(y) = (p,q)(r,s)(p,r)(q,s) = (p,s)(q,r)$$ has order $2$ instead of $3$, so this case is also impossible.

Remark. The only "unexpected" permutation here is $\phi$ taking transpositions to $(a, b)(c, d)$ (or $(a, c)(b, d)$ or $(a, d)(b, c)$) and other elements to the identity $(a)(b)(c)(d)$.

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