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Prove that $S_4$ cannot be generated by $(1 3),(1234)$

I have checked some combinations between $(13),(1234)$ and found out that those combinations cannot generated 3-cycles.

Updated idea:
Let $A=\{\{1,3\},\{2,4\}\}$
Note that $(13)A=A,(1234)A=A$
Hence, $\sigma A=A,\forall\sigma\in \langle(13),(1234)\rangle$
In particular, $(12)\notin \sigma A,\forall\sigma\in \langle(13),(1234)\rangle$
So we conclude that $S_4\neq\langle(13),(1234)\rangle$

Wang Kah Lun
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    In the update, $(1~2)\notin\sigma A$ cannot be right, because $\sigma A$ is a partition of ${1,2,3,4}$ (indeed it equals $A$ here), so its elements are subsets, not permutations like $(1~2)$. I think you mean to say $(1~2)A\neq A$ instead. Indeed, $(1~2)A={{2,3},{1,4}}\neq A$. – Marc van Leeuwen Jan 20 '16 at 13:38

3 Answers3

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The partition $\{\{1,3\},\{2,4\}\}$ is invariant under the action of the two proposed generators, but not under all of $S_4$, so they cannot generate all of $S_4$.

Marc van Leeuwen
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  • Can you explain in a simpler way? I can't get your idea. – Wang Kah Lun Jan 19 '16 at 16:38
  • No I cannot explain it in a simpler way. But here is a longer one. The idea is that if one groups together certain elements, there is a "togetherness" relation, and after acting by some permutation, there is a potentially different "togetherness" relation. Here I grouped numbers of equal parity, but if for instance I interchange $1$ with $2$, then it is ${2,3}$ and ${1,4}$ that become together after the permutation. But it happens that permuting by $(1~3)$ or by $(1~2~3~4)$ one gets the same relation back. Composing the latter two never gives a new relation, so one cannot get $(1~2)$. – Marc van Leeuwen Jan 19 '16 at 16:44
  • Apply your first permutation: ${{3,1},{2,4}} = {{1,3},{2,4}}$. Apply your second: ${{2,4},{1,3}} = {{1,3},{2,4}}$. So whatever those two generate fixes that set. But something like $(12) \in S_4$ gives you ${{2,3},{1,4}} \neq {{1,3},{2,4}}$, and so we must not have all of $S_4$. @AlanWang –  Jan 19 '16 at 16:46
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    Another way to use the same idea: label the corners of a square $1,2,3,4$ in a cycle, then the partition ${{1,3},{2,4}}$ describes the diagonals. Both permutations $(1~3)$ and $(1~2~3~4)$ are symmetries of the square, hence preserve the partition into diagonals. But other permutations are not symmetries of the square, and cannot be in the subgroup generated by those two permutations. – Marc van Leeuwen Jan 19 '16 at 16:48
  • I have updated my idea. Take a look to check whether my proof is correct. Thanks. – Wang Kah Lun Jan 20 '16 at 11:55
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If we denote $a = (1234)$ and $b = (13)$, one can easily check that $a^4 = e$, $b^2 = e$ and $ab = ba^{-1}$ which are precisely relations that define dihedral group $D_4$. Thus, subgroup generated by $a$ and $b$ in $S_4$ is isomorphic to quotient of $D_4$, and thus it's order is less or equal than $8$. Since $|S_4| = 4!$, obviously $a$ and $b$ can't generate whole $S_4$. One can easily check that there are $8$ distinct elements in $\langle a,b\rangle$, so it is actually isomorphic to $D_4$.

Ennar
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  • What is meant by "$S_4$ is isomorphic to a quotient of $D_4$" in your answer? Do you mean that it is isomorphic to a subgroup of $D_4$? – Jonathan Huang Nov 08 '24 at 16:30
  • @JonathanHuang, you could argue like that as well, but it feels less natural to me. To get a subgroup of $D_4$, you need to define homomorphism $S_4\supseteq \langle a,b\rangle \to D_4$, where $\langle a, b\rangle$ is a subgroup of $S_4$ generated by $a$ and $b$. If we write a presentation of $D_4 = \langle s, r \mid r^2 = s^4 = e, sr = rs^{-1}\rangle$, you'd want to define homomorphism by mapping $a\mapsto s$, $b\mapsto r$. However, why is that homomorphism? (1/3) – Ennar Nov 09 '24 at 15:06
  • On the other hand, it's easy to define homomorphism $f\colon D_4\to S_4$ by setting $f(s) = a$, $f(r) = b$ by Fundamental theorem on homomorphisms because $D_4 \cong F_2/N$, where $F_2$ is the free group over two generators and $N$ appropriate normal subgroup. (2/3) – Ennar Nov 09 '24 at 15:06
  • Relations $a^4=b^2 = e$ and $ab = ba^{-1}$ then precisely mean that $N$ is contained in the kernel of homomorphism $F_2 \to S_4$, $s\mapsto a$, $r\mapsto b$, so $f\colon D_4\cong F_2/N \to S_4$ is well defined. The image of $f$ is precisely the subgroup $\langle a,b\rangle$ in $S_4$, so it is isomorphic to the quotient $D_4/\ker f$ by the first isomorphism theorem. (3/3) – Ennar Nov 09 '24 at 15:07
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Here we copy/paste/modify another answer:

Define

$\; \tau = (13)$

$\;\sigma = (1234)$

$1^{st} \text {group of calculations:}$

$\;\tau^1 = (13)$

$\;\tau^2 = \tau^0 \quad \text { - the identity permutation}$

$2^{nd} \text{ group of calculations:}$

$\;\sigma^1 = (12)\,(23) \,(34)$

$\;\sigma^2 = (13)\,(24)$

$\;\sigma^3 = (14)\,(24)\,(34)$

$\;\sigma^4 = \sigma^0 \quad \text { - the identity permutation}$

$3^{rd} \text{ group of calculations:}$

$\tau\sigma = (12) \,(34)$

$\tau\sigma^2 = (24)$

$\tau\sigma^3 = (14) \,(23)$

$4^{th} \text{ group of calculations:}$

$\sigma\tau = (14) \,(23)$

$\sigma^2\tau = (24)$

$\sigma^3\tau = (12)\,(34)$


So far we've identified exactly $8$ permutations that are in the group generated by $\tau$ and $\sigma$.

When we run the $4^{th} \text{ group of calculations}$ we get the same permutations as the $3^{rd} \text{ group of calculations}$, and we can now write these symbolic (defining) rules,

$\tag 1 \tau^2 = 1_d$ $\tag 2 \sigma^4 = 1_d$ $\tag 3 \sigma\tau = \tau\sigma^3$

Given any word (string) in the letters $\tau$ and $\sigma$ we can 'move' all the $\tau$ letters to the left and standardize (present/represent) the permutation to have the form

$\tag 4 \tau^n \sigma^m \quad n \in \{0,1\} \text{ and } m \in \{0,1,2,3\}$

We conclude that the group generated by $\tau$ and $\sigma$ has exactly $8$ elements.

Since

$\; \sigma \tau = \tau \sigma^3$

and

$\; \tau \sigma \ne \tau \sigma^3$

we also know that this a non-abelian subgroup of $S_4$.

CopyPasteIt
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