# Sylow $5$-subgroup in $S_{16}$

Find two different $5$-Sylow subgroups in $S_{16}$.
Hint: use group multiplication.

Any hints?

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I thought the definition of Sylow-p-subgroup is a subgroup of maximal $p$ power, i.e. maximal $k$ such that $p^k \Big | |G|$. In your question it seems to me there are higher powers of $5$ that divide $|S_{16}| = 16!$. Can someone point out to me what I'm missing? Thanks! – Rudy the Reindeer Feb 16 '11 at 9:44
Does conjugation work here? In this case we just need to find a 5-sylow subgroup of $S_{16}$ which is not normal which will fulfill the condition, don't we? If I miss something, please tell me, thanks. By the way, to type $S_{16}$ is better to use S_{16} instead of S_16 as far as I am concerned. – awllower Feb 16 '11 at 9:54
@Matt: where is the problem? the 5-sylow groups here will be of order $5^3$, cause 16! = ..5..10..15.. – dancing clown Feb 16 '11 at 10:02
my bad, I read your question as if you were looking for subgroups of order $5$. – Rudy the Reindeer Feb 16 '11 at 10:39

Since $16!$ is divisible by $5^3$, but not $5^4$, the $5$-Sylow subgroups of $S_{16}$ are the subgroups of order $5^3$.

An example is $$\langle (1\ 2\ 3\ 4\ 5), (6\ 7\ 8\ 9\ 10), (11\ 12\ 13\ 14\ 15)\rangle.$$ (Since any pair of generators commutes, the group has isomorphism type $\mathbb Z/5\mathbb Z\times\mathbb Z/5\mathbb Z\times\mathbb Z/5\mathbb Z$ and in particular order $5^3$.)

Another example is $$\langle (1\ 2\ 3\ 4\ 5), (6\ 7\ 8\ 9\ 10), (11\ 12\ 13\ 14\ 16)\rangle.$$

Both examples have a unique common fixed point of all elements. In the first example, it is $16$, and in the second example it is $15$. So the examples give two different $5$-Sylow subgroups of $S_{16}$.

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What does a permutation of order 5 look like?

Can you give a condition when two permutations of order 5 commute?

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well it can be a cycle of order 5, when they are disjoint? should I look at the multiplication of 3 disjoint cycles of order 5, hm? – dancing clown Feb 16 '11 at 10:58
You should look at the multiplication of the three (cyclic) groups generated by these cycles, as indicated by your hint, noting that the generators mutually commute. – Myself Feb 16 '11 at 13:08