1
$\begingroup$

Let $G=A_5$, the alternating group of degree 5. Please help me show that for $\pi = \{2,3\}$ we have that $M \leq G$ is a maximal $\pi$-group if and only if $M \cong A_4$ or $M \cong S_3$ (the symmetric group).

Here a $\pi$-subgroup is a subgroup all of whose elements are $\pi$-elements, a $\pi$-element is an element whose order is a $\pi$-number, and a $\pi$-number is a number all of whose prime factors are in $\pi$.

$\endgroup$
4
  • 3
    $\begingroup$ Are you asking how to prove this statement? $\endgroup$ May 10 '13 at 17:03
  • $\begingroup$ Yes, but there are two different kinds of copies of $S_3$ in here, namely $\langle (1, 2, 3), (1, 2) \rangle$ and $\langle (1, 2, 3), (1, 2) (4, 5) \rangle$. They are both isomorphic to $S_3$, but not conjugate. $\endgroup$ May 10 '13 at 17:14
  • 1
    $\begingroup$ But $\langle(1,2,3),(1,2)\rangle$ does not lie in $A_5$. $\endgroup$
    – Derek Holt
    May 10 '13 at 20:49
  • $\begingroup$ The correct question is: Prove that the M is maximal $\pi$-subgroups if and only if $M \cong A_4$ or $M \cong S_3$ ($\pi = \{2,3\}$ $\endgroup$
    – user59969
    May 12 '13 at 14:54
1
$\begingroup$

Each $M$ is a $\pi$-subgroup. $A_4$ has index 5, so is maximal by Lagrange's theorem and the primality of 5. $S_3 = \langle (1,2,3)(4,5), (1,2)(4,5) \rangle$ has index 10, but any $\pi$-subgroup containing it would contain $S_3$ as a normal subgroup. However, this $S_3$ is its own normalizer in $A_5$ [ (proof a) direct computation, (proof b) it is a Sylow 3-normalizer, (proof c) its normalizer in S5 has an odd permutation but contains this S3 as index 2 ], hence $S_3$ is maximal amongst $\pi$-subgroups. Indeed $S_3$ is maximal in $A_5$ since any other subgroup containing it would be of order 30 and $A_5$ has no such subgroup [ $A_5$ is simple, order 30 is index 2 and normal ].

If $M$ is a $\pi$-subgroup of $A_5$, then it has order in $\{1,2,3,4,6,12\}$. By Sylow's theorem, order 1,2,3,4 are ruled out, since such a group is contained in the normalizer of a Sylow (2 or 3) subgroup containing it. A subgroup of order 6 has a normal Sylow 3 subgroup, so is contained in the normalizer within $A_5$ of its Sylow 3-subgroup (which is just $S_3$). A subgroup of order 12 either has a normal Sylow 2-subgroup, and so is contained in the normalizer within $A_5$ of its Sylow 2-subgroup (which is just $A_4$), or has a normal Sylow 3-subgroup, and so is contained in a subgroup of order 6 [ contradiction ].

Hence we have listed all maximal $\pi$-subgroups and all maximal subgroups that are $\pi$-subgroups up to conjugacy within $A_5$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy