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Find all the Sylow $2$ and the Sylow $3$ subgroups of $A_5$.

My try:

$|A_5|=60$.Number of Sylow $2$ subgroups of $A_5=n_2=1,3,5,15$.

Number of Sylow $3$ subgroups of $A_5=n_3=1,4,10$.Neither $n_2$ nor $n_3=1$ as $A_5$ is simple.

Now the Sylow $2$ subgroups of order $4$ may be:

  • $\{e,(12)(34),(14)(23),(13)(24)\}$,
  • $\{e,(15)(34),(13)(45),(14)(35)\}$
  • $\{e,(23)(45),(24)(35),(25)(34)\}$
  • $\{e,(12)(35),(13)(25),(15)(23)\}$
  • $\{e,(12)(45),(15)(24),(14)(25)\}$

I am wondering if there are any other Sylow 2 subgroup of order $4$.

Regarding the Sylow $3$ subgroups ;

Number of $3$ cycles in $A_5=20$ .Each of these cycles gives rise to a subgroup of order $3$ given by

  • $\{e,(123),(132)\}$
  • $\{e,(143),(134)\}$ and so on .As each subgroup has two elements of order $3$,so we get $10$ subgroups.

I think it is correct in case of $A_5$.But I am not sure in case of $A_4$ .Please help.

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  • $\begingroup$ Why would you wonder whether there are more Sylow $\;2\,-$ subgroups? You know their number fulfills $\;n_2=1\pmod2\;,\;\;n_p\,\mid\,\frac{|A_n|}4=15\;$ , so there aren't many options, are there? $\endgroup$ – DonAntonio Oct 7 '16 at 7:44
  • $\begingroup$ @DonAntonio;I was wondering if $n$ can be equal to $15$ $\endgroup$ – Learnmore Oct 7 '16 at 9:38
  • $\begingroup$ @S Well, it should be more or less clear that it cannot be, since if you get a Sylow $\;2\,-$ subgroup, then all the other Sylow $\;2\,-$subgroups will be isomorphic to that one. Since you already have one that is isomorphic to the Klein viergruppe $\;\Bbb Z_2\times\Bbb Z_2\;$, then all must be isomorphic to that (in fact, even conjugate...), and the only way to obtain groups like this in this case is as you did: direct product ot two subgroups of order two with commuting generators from $\;A_5\;$ ... $\endgroup$ – DonAntonio Oct 7 '16 at 9:44
  • $\begingroup$ @S Your calculation is flawed since there could possibly be two different Sylow $\;2\,-$ subgroups with non-trivial intersection $\endgroup$ – DonAntonio Oct 7 '16 at 9:46
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Note that as $A_5$ is simple, there are more than one Sylow 5-subgroups. Namely there are $6$. As we know that there are $10$ Sylow 3-subgroups. Then at least $44$ permutations have a order $5$ or $3$. Hence we're left with $15$ more element. As every Sylow 2-group has at least $2$ unique elements it's impossible to have $15$ Sylow 2-subgroups. As you've listed all $5$, then number of Sylow 2-subgroups is exactly $5$.

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  • $\begingroup$ Why can't be $n=15$?If $n=15$ then we get at least $15*2=30$ elements of order $2$,already we have $20$ elements of order $3$ and also $20$ elements of order $5$ adding to 70 elements in $A_5$ which is false: $\endgroup$ – Learnmore Oct 7 '16 at 9:44
  • $\begingroup$ Is this correct? $\endgroup$ – Learnmore Oct 7 '16 at 9:45
  • $\begingroup$ @S.Bandopadhaya Almost yes. Everything is OK, except that we have at least $24$ elements of order $5$. Also for the sake or arguement we have one of order $1$. In any case it's impossible as the number of elements in $A_5$ would be greater than $60$. $\endgroup$ – Stefan4024 Oct 7 '16 at 9:49
  • $\begingroup$ Yes I got it thank you very much $\endgroup$ – Learnmore Oct 7 '16 at 9:57

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