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Let $G=A_4$. Prove that does not exist subgroup $H\le G$ s.t $|H|=6$.

I don't know from where to start (maybe I need to prove that if so $H\triangleleft A_4 $?) Any hint will be appreciated.

Mikasa
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    A subgroup of $A_4$ whose order is a multiple of $6$ must contain a $3$-cycle, and a product of two disjoint $2$-cycles. – Daniel Fischer Aug 31 '13 at 16:49
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    A group of order $6$ contains an element of order $3$: what are the elements of order $3$ in $A_4$? Similarly, a group of order $6$ contains an element of order $2$: what are the elements of order $2$ in $A_4$? – Mariano Suárez-Álvarez Aug 31 '13 at 16:53
  • Notice that the title of your question and the question itself as different questions! (Also, there is never any need to abbreviate such that) – Mariano Suárez-Álvarez Aug 31 '13 at 16:54
  • So I need to find 3-cycle and product of 2-disjoint-cycles which their product is not in the subgroup? I can't find them... –  Aug 31 '13 at 16:59
  • Can I say that for example $(34)(134)(124)=(12)\notin A_4$? –  Aug 31 '13 at 17:25
  • You can say that, @CoarguAliquis...but I can't see how that'll help you prove what you want. – DonAntonio Aug 31 '13 at 17:27
  • o.k but can I say that we know type of cycles in $A_4$, if we take their squares they are in $H$ but since we have 8 cycles of 3, that's a contradiction? –  Aug 31 '13 at 17:34
  • Polishing that argument may help, @CoarguAliquis...but you need the product of an element of order two and one of order three. – DonAntonio Aug 31 '13 at 17:35

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A subgroup of order $\;6\;$ has an element of order three and an element of order two, which then must be a $\;3-$cycle and the product of two disjoint $\;2-$cycles (why?). WLOG, suppose that these are

$$(123)\;,\;\;(1i)(jk)\implies (123)(1i)(jk)=\begin{cases}(134)&,\;\;i=2\\{}\\(243)&,\;\;i=3\\{}\\(142)&,\;\;i=4\end{cases}$$

And in any case you already have two different $\;3-$cycles and at least one product of two disjoint transpositions...and this can't be (why? Check you will already have more than 6 elements and thus the subgroup is the whole group $\,A_4\,$)

DonAntonio
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