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There is a $2$ step task.

  1. Given the symmetric $S_5$ group. Find all $S_5$ Sylow subgroups.

  2. For each subgroup found check whether the subgroup is normal or not.

For the first step I have found using the Sylow theorem that these subgroups are:

  • $D_8$ dihedral group for which $p=2$ and $s=15$
  • $Z_3$ cyclic group for which $p=3$ and $s=10$
  • $Z_5$ cyclic group for which $p=5$ and $s=6$

However, I have no idea how to correctly check whether these subgroups are normal or not. I could do that "by hand" or automated using python but definitely that's not the case. Doing that by hand is too complicated since $\vert{S_5}\vert=120$ and calculating all the left and right cosets is overcomplicated. There definitely should be a way. Thank you in advance!

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First of all, Sylow theorems say that all Sylow $p$-subgroups are conjugate in $G$, so as soon as you discover that the number $s_p$ of Sylow $p$-subgroups is greater than 1, you know they are not normal.

Another way to see that none of these are normal is to note that $A_5$ is a simple group. In particular, if $N\lhd S_5$ is normal, then $N\cap A_5$ is normal in $A_5$. If follows that $N\cap A_5=A_5$ or $N\cap A_5=\{1\}$. In the former case, you may conclude that $N$ is either $A_5$ or $S_5$. In the latter case, use the fact that $NA_5=S_5$ (since it properly contains $A_5$) and $$5!=|NA_5|=\frac{|N||A_5|}{|N\cap A_5|}=\frac{5!}{2}|N|$$ to conclude that $|N|=2$ (no such $N$ exists, but never mind). In any case, none of the Sylow subgroups are normal.

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  • $\begingroup$ And there is one more question, if $D_8$ is not normal it means that there are more than $1 $group for which $p=2$. But how many? Both 3 and 5 $=1 (mod p)$ and divide $\vert{G}\vert/\vert{D_8}\vert = 15$. So how many 2-sylow subgroups there are? $\endgroup$ – tna0y Dec 14 '16 at 13:38
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    $\begingroup$ Well, for this I would note that each copy of $D_8$ contains exactly two elements of order 4 (i.e. 4-cycles) and no two copies have a 4-cycle in common. My count gives $5!/4=30$ such 4-cycles. Pairing each 4-cycle with its inverse yields $30/2=15$ copies of $D_8$. $\endgroup$ – David Hill Dec 14 '16 at 19:26
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One of the Sylow theorems states that all $p$-Sylow subgroups are conjugate to each other. This implies that a Sylow $p$-subgroup is normal iff it is the unique such Sylow $p$-subgroup. For each of the Sylow $p$-subgroups you've written, you also have the size $s$ of their respective conjugacy classes. Since none of these conjugacy classes has size $1$, none of these Sylow subgroups is normal.

In fact, the only nontrivial normal subgroup of $S_5$ is $A_5$, which is of order 60, hence not a Sylow subgroup.

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The whole idea is to leverage theory like the Sylow problems as much as possible. There shouldn't be any need to do anything by hand!

Here, we don't even need much Sylow knowledge: just a little knowledge of the symmetric groups. There is exactly one proper normal subgroup of $S_5$, namely the alternating subgroup $A_5$ of order $60$. Thus there is no normal subgroup of size $3, 5$ or $8$: no normal Sylow $p$-subgroups.

By the way (regarding your computations using Sylow theory) you didn't mention how you ruled out $4$ or $40$ Sylow $3$-subgroups, or $3$ or $5$ sylow 2-subgroups...

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