Compute the number of Sylow subgroups in Sym(5) and Alt(5)

For p \in \{ 2,3,5 \} find n_p(A_5) and n_p(S_5).

We directly computed the subgroup lattice of A_4, so we know that n_2(A_4) = 1 and n_3(A_4) = 4.

Note that |A_5| = 60 = 2^2 \cdot 3 \cdot 5. By Sylow’s theorem, we have the following.

  • n_2(A_5) \in \{ 1,3,5,15 \}
  • n_3(A_5) \in \{ 1,4,10 \}
  • n_5(A_5) \in \{ 1,6 \}

We know that A_5 is simple, so n_5(A_5) = 6. Now the Sylow 3-subgroups of A_5 have order 3; by this previous exercise, S_5 has 20 elements of order 3, all of which are in A_5. Each of these is in a Sylow 3-subgroup of A_5, and each Sylow 3-subgroup is generated by two elements. Thus n_3(A_5) = 10. Note that the elements of order 4 in S_5 are precisely the 4-cycles, and none of these are contained in A_5. Thus the nonidentity elements of any Sylow 2-subgroup of A_5 have order 2; these elements are precisely the products of two 2-cycles. By this previous exercise, S_5 contains 15 products of two disjoint 2-cycles, all of which are contained in A_5 and each of which is contained in some Sylow 2-subgroup. Note that the Sylow 2-subgroups of A_5 are abelian, and that every element of order 2 in A_5 fixes exactly one element in \{1,2,3,4,5\}. Let \sigma,\tau \in A_5 such that \sigma fixes a and \tau does not fix a. If \sigma\tau = \tau\sigma, then \tau(a) = \sigma(\tau(a)); becuase \sigma fixes only a, we have \tau(a) = a, a contradiction. So \sigma \tau \neq \tau \sigma. Thus, if \sigma,\tau \in A_5 are elements of order 2 that do not fix the same element, then no Sylow 2-subgroup contains both \sigma and \tau. For each k \in \{1,2,3,4,5\}, there are precisely three products of two 2-cycles in A_5 which fix k, corresponding to the image of V_4 (in A_4) under a renaming injection A_4 \rightarrow A_5, and each such image is a Sylow 2-subgroup. Thus n_2(A_5) = 5.

Now consider S_5; note that |S_5| = 120 = 2^3 \cdot 3 \cdot 5. By Sylow’s Theorem, we have the following.

  • n_2(S_5) \in \{ 1,3,5,15 \}
  • n_3(S_5) \in \{ 1,4,10,40 \}
  • n_5(S_5) \in \{ 1,6 \}

Every Sylow 5-subgroup of A_5 is a Sylow 5-subgroup of S_5, so n_5(S_5) = 6. Similarly, every Sylow 3-subgroup of A_5 is a Sylow 3-subgroup of S_5, so that n_3(S_5) \geq 10. Now the nonidentity elements in Sylow 3-subgroups of S_5 are precisely the 3-cycles, and all of these are contained in A_5. So S_5 has no additional Sylow 3-subgroups; hence n_3(S_5) = 10. Note that S_4 \leq S_5, and that every Sylow 2-subgroup of S_4 is a Sylow 2-subgroup of S_5. In this previous exercise, we saw that H_1 = \langle (1\ 2\ 3\ 4), (2\ 4) \rangle, H_2 = \langle (1\ 3\ 2\ 4), (3\ 4) \rangle, and H_3 = \langle (1\ 2\ 4\ 3), (2\ 3) \rangle are the Sylow 2-subgroups of S_4. Then H_4 = \langle (1\ 2\ 3\ 5), (2\ 5) \rangle, H_5 = \langle (1\ 3\ 2\ 5), (3\ 5) \rangle, and H_6 = \langle (1\ 2\ 5\ 3), (2\ 3) \rangle are also distinct Sylow 2-subgroups of S_5, and clearly H_1, H_2, and H_3 are pairwise distinct from H_4, H_5, and H_6. (In particular, the former fix 5 and the latter do not.) So n_2(S_5) \geq 6; thus n_2(S_5) = 15.

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  • Gobi Ree  On December 21, 2011 at 9:04 pm

    Typo: Thus, if \sigma, \tau \in A_4 are elements of order 2 – A_4 should be A_5.

    Also we can directly notice n_2(S_5)=15, since there are 5 ways of embedding S_4 into S_5, so n_2(S_4) \cdot 5 = 15.

    • nbloomf  On January 18, 2012 at 1:42 pm

      Thanks! Also, that argument for n_2(S_5) is slick.

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