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Let $G = A_5$, the alternating group of degree five,. Let $\pi = \{2,3\}.$ Prove that $M$ is a maximal $\pi$-subgroup of $G$ if, and only if, $M\cong A_4$ or $M\cong S_3$, where $A_4$ is the alternating group of degree four and $S_3$ is the symmetric group $3$.

Jack Schmidt
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    What do you mean of $\pi$-subgroup? – Bobby Jun 04 '13 at 13:25
  • Ok. @Babgen.Let $\pi$ be a set of a prime numbers. A integer $n>0$ is a $\pi$-number if $n = \prod p_i^{\alpha_i}$ where $p_i \in \pi$ for all $i$. –  Jun 04 '13 at 13:33
  • Let $G$ a group. An element $g \in G$ is a $\pi$-element if the order of $g$ is a $\pi$-number. –  Jun 04 '13 at 13:36
  • Let $G$ a group and $S \leq G$. Then, $S$ is a $\pi$-subgroup if $\mid S \mid$ is a $\pi$-number. And the group $G$ is a $pi$-group if all elements of $G$ are $\pi$-elements. –  Jun 04 '13 at 13:37
  • Yet that duplicate question,. @JackSchmidt , has no answer at all... – DonAntonio Jun 04 '13 at 14:07
  • Thank you @JackSchmidt. But you know give me some help on this question? –  Jun 04 '13 at 14:08

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Clearly we have both subgroups $\,S_3\,,\,A_4\;$ embedded in $\,A_5\,$ , and since a $\,\pi$-subgroup here would have to have order$\,\in\{2,3,4,6,12\}\;$ , we at once get that $\,M\cong A_4\,$ is a maximal $\,\pi$-subgroup , and since $\,S_3\rlap{\;\;/}\le A_4\,$ then this one's also a maximal $\;\pi$-subgroup .

DonAntonio
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  • Why the $\pi$-subgroups of order 3 or 4 can not be a maximal $\pi$-subgroups? –  Jun 04 '13 at 16:41
  • In principle, we only know that the subgroups of order six and 12, respectively, have copies of the subgroups of order 3 and 4, respectively, but we have show that each subgroup of order 3 is contained in an order of 6, and the same goes for the order 4. I think this a lack to complete the exercise. Because the fact $A_5$ have copies of the subgroups $A_4$ and $S_3$ is obvious. –  Jun 04 '13 at 16:43
  • They all are contained in a bigger group, @AgenorAndrade. For example, $$\langle (123)\rangle \le \langle (123),,,(12)(45)\rangle$$ In fact, I'm just copying above the argument used to prove that $,S_n,$ can always be embedded in $,A_{n+2},$ ... – DonAntonio Jun 04 '13 at 16:47