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This section says:

There is a subgroup (indeed, $6$ conjugate subgroups) of $S_6$ which are abstractly isomorphic to $S_5$,

At this point I'm thinking: certainly: the group of all permutations of $\{a,b,c,d,e,f\}$ that leave the letter $a$ fixed is isomorphic to $S_5$. And there are six groups like it, since one can choose any of the six letters as the one that will remain fixed. But the section continues:

There is a subgroup (indeed, $6$ conjugate subgroups) of $S_6$ which are abstractly isomorphic to $S_5$, and transitive as subgroups of $S_6$.

But the groups I identify above do not act transitively on $\{a,b,c,d,e,f\}$, so this must be about some other subgroups. What are they? Are they images of the six groups I mention above under an outer automorphism?

Michael Hardy
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  • Maybe I'll post an affirmative answer to this question below after working this through. (I see that I am not allowed to use a "permutation groups" tag.) ${}\qquad{}$ – Michael Hardy May 19 '15 at 00:53

2 Answers2

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I will show you in principle how this is done, but the actual mechanics are tedious. Let:

$$P_1 = \{e,(1\ 2\ 3\ 4\ 5),(1\ 3\ 5\ 2\ 4), (1\ 4\ 2\ 5\ 3),(1\ 5\ 4\ 3\ 2)\}\\ P_2 = \{e, (1\ 2\ 3\ 5\ 4),(1\ 3\ 4\ 2\ 5),(1\ 5\ 2\ 4\ 3),(1\ 4\ 5\ 3\ 2)\}\\ P_3 = \{e,(1\ 2\ 4\ 3\ 5),(1\ 4\ 5\ 2\ 3), (1\ 3\ 2\ 5\ 4), (1\ 5\ 3\ 4\ 2)\}\\ P_4 = \{e, (1\ 2\ 4\ 5\ 3),(1\ 4\ 3\ 2\ 5), (1\ 5\ 2\ 3\ 4), (1\ 3\ 5\ 4\ 2)\}\\ P_5 = \{e,(1\ 2\ 5\ 3\ 4),(1\ 5\ 4\ 2\ 3), (1\ 3\ 2\ 4\ 5), (1\ 4\ 3\ 5\ 2)\}\\ P_6 = \{e,(1\ 2\ 5\ 4\ 3), (1\ 5\ 3\ 2\ 4), (1\ 4\ 2\ 3\ 5), (1\ 3\ 4\ 5\ 2)\}$$

If we pick an element $\sigma$ of $S_5$, say $(1\ 2\ 3)(4\ 5)$, we find that:

$$\sigma P_1\sigma^{-1} = P_5\\ \sigma P_2\sigma^{-1} = P_3\\ \sigma P_3\sigma^{-1} = P_6\\ \sigma P_4\sigma^{-1} = P_2\\ \sigma P_5\sigma^{-1} = P_4\\ \sigma P_6\sigma^{-1} = P_1$$

That is, if $\phi:S_5 \to S_6$ is our (exotic) embedding, then:

$\phi((1\ 2\ 3)(4\ 5)) = (1\ 5\ 4\ 2\ 3\ 6)$ (note this indeed takes an element of order $6$ to an element of order $6$). It should be clear that if $\sigma \in S_5$ is a $5$-cycle, it fixes the sylow $5$-subgroup it belongs to, and permutes the rest, pick one at random, and verify it creates a $5$-cycle in $S_6$.

Note that we only need the $3$-cycles of $S_5$ to show this action is indeed transitive; for example, to send $P_1 \to P_5$, we can conjugate by the $3$-cycle $(3\ 5\ 4)$ (I deliberately chose the $5$-cycle generators to start with $(1\ 2\ \dots)$ to make this clear).

Michael Hardy
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David Wheeler
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    This is a really great explanation! I've read a bit about this but never really "got" what was going on. This is a fantastic illustration that $S_5$ does indeed act transitively on its set of Sylow $5$-subgroups, and exactly how it yields a nice embedding $S_5 \hookrightarrow S_6$. – pjs36 May 19 '15 at 16:31
  • Very interesting. Is there some actual pattern you're following in describing your six sets $P_1,\ldots,P_6$? ${}\qquad{}$ – Michael Hardy May 19 '15 at 19:45
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    Two possibilities suggest themselves: (1) Add to the cited section of the Wikipedia article both this construction and the one in the answer by @JoshB , and (2) instead, put a "main article" link in that section, linking to a new Wikipedia article we would create, precisely about these objects. ${}\qquad{}$ – Michael Hardy May 19 '15 at 20:44
  • @MichaelHardy, yes-first I listed the 3-cycles of $S_5$ in "dictionary order", which leaves $2$ more elements to move to make a $5$-cycle, the pair of which I appended, for each $3$-cycle, in increasing, then decreasing order. I then took the first $5$-cycle on my list, and generated its cyclic subgroup. Then I moved down the list to my next "untaken" $5$-cycle, checking carefully that no $5$-cycle was "double-listed". – David Wheeler May 19 '15 at 23:27
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Note that $S_5$ contains a subgroup of order $20$ (generated by, say, $(1,2,3,4,5)$ and $(1,3,4,2)$ ). The action of $S_5$ on the $6$ cosets of a subgroup of order $20$ provides a permutation representation of $S_5$ on $6$ points. And yes, an outer automorphism maps this sort of $S_5$ to the first type you were thinking of.

Josh B.
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  • Clearly the order of the subgroup of $S_5$ generated by $(1,2,3,4,5)$ and $(1,2,3,4)$ is a multiple of $20$ and must be either $20$, $40$, $60$, or $120$, but I'll need to check the arithmetic with permutations to be sure of which one. To be continued${},\ldots\qquad{}$ – Michael Hardy May 19 '15 at 19:37
  • Note that I have $(1,3,4,2)$ not $(1,2,3,4)$. This group of order 20 is a semidirect product. It contains a normal subgroup isomorphic to the cyclic group of order 5 and 5 cyclic subgroups of order 4. Note that the automorphism group of $C_5$ is isomorphic to $C_4$. – Josh B. May 19 '15 at 21:40