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Describe explicitly all Sylow 5= subgroups in $S_{10}$. Prove that every Sylow 5-subgroup is isomorphic to $C_5 \times C_5$. Prove that every two Sylow 5-subgroups are conjugate (explictiley, not using the Sylow theorem). HINT: Look at the cycle decomposition of permutations $\sigma \in S_{10}$.

I know that a group, $G$, is a $p$-group, where $p$ is a prime number, is $|G| = p^k$. And, if you have $|G| = p_1^{k_1} \times ... \times p_n^{k_n}$, then the Sylow $p$-subgroup is given by $|H| = p^k$, i.e the largest $p^k$ dividing $|G|$.

So from my question, I get the order of the group to be $10! = 2^8 \times 3^4 \times 5^2 \times 7$. Clearly this is a 5-subgroup as the largest $p^k$ is $5^2$. From here, I get that this subgroup has $|p| = 5^2 = 25$ elements, which is the same number of elements in $C_5 \times C_5$ and so they are isomporphic.

As the biggest $p$ dividing 10! is 5, I get that my subgroup will contain the permutation of elements

$$\{ (a_1 .... a_5)^i(b_1 ... b_5)^j\} | \{a_1, ... a_5, b_1, ... ,b_5\} = \{1 , ... , 10\}, i,j = 0, ... 4, $$

However from here, I don't get how to do the last conjugate bit. Is my previous stuff all correct aswell?

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Just because two groups have the same number of elements, they need not be isomorphic –  Tobias Kildetoft Jan 7 '13 at 13:52
The group could be isomorphic to $C_{25}$ for example –  Amr Jan 7 '13 at 13:53
There are several "fishy" claims in your question. One was already pointed out by Tobias above, other one is when you say that "the biggest p dividing 10! is 5"...what did you mean by that? Because the biggest prime dividing $\,10!\,$ is $\,7\,$... –  DonAntonio Jan 7 '13 at 13:55
Do you know the fact that every group of order $p^2$ is abelian ? –  Amr Jan 7 '13 at 13:58
Yeah but $C_{25}$ is also isomprphic to $C_5 \times C_5$. If all groups with the same number of elements are isomprohic to each other, then I can list every group with 25 elements and eventually ge $C_5 \times C_5$ can't I? With the biggest $p$ dividing $10!$, I meant that that $5$ is the largest prime number which has a power greater than $1$ in $10!$. I know that $p^2$ claim, so $S_{10}$ is abelian. I don't see how that comes into it though –  Kaish Jan 7 '13 at 14:14

1 Answer 1

Hints: (Be sure you know or can prove the following):

1) Every permutation in $\,S_n\,$ can be written as a product of disjoint cycles. The lengths $\,(a_1,a_2,...,a_n)\,$ of the cycles appearing in the decomposition of a permutation is called (by me, at least) the cycle-type of the permutation

2) If $\,a,b\,$ are two cycles of order $\,\alpha,\beta\,$ resp., the order of $\, ab\,$ is $\,l.c.m.(\alpha,\beta)\,$ (the least or minimal common multiple)

3) Very important!: two cycles are conjugated in $\,S_n\,$ iff they have exactly the same length, and from here: two permutations are conjugated in $\,S_n\,$ iff they have the very same cycle-type.

4) The only way to get an element of order a prime $\,p\,$ in $\,S_n\,\,,\,p\leq n\,$ , is by means of the products $\,t_1\cdot\ldots\cdot t_k\,$ , where each $\,t_i\,$ is a cycle of order $\,p\,$ and, of course, $\,kp\leq n\,$

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Yes @amWhy, it is. I'm more used to minimal common multiple than to lower (or least or whaever with $l$) common multiple. Again, this could be a language mistake by me –  DonAntonio Jan 7 '13 at 14:17
Fixed as there're more sites in google with lcm than with mcm. Thanks –  DonAntonio Jan 7 '13 at 14:19
@DonAntonio So cycle type of $a$ and $b$ is different to the order of $ab$? –  Kaish Jan 7 '13 at 14:29
That previous comment is silly. What I want to know is if from point three I don't know what the permutation for $S_{10}$ is, how can I write it in disjoint cycles and show that as they're lengths are the same, they are conjugated. –  Kaish Jan 7 '13 at 14:35
@Kaish, you already know the Sylow $\,5-$subgroup of $\,S_{10}\,$ has order $\,25=5^2\,$, so there're only two options: either $\,C_{25}\,$ or $\,C_5\times C_5\,$ . Thinking of the points in my answer you must deduce that the former possibility is imposible (as there's no element of order $\,25\,$ in $\,S_{10}\,$) , so that it must be the latter. Well, now think what the generators of such a group can be and how can you find out all their conjugates (and, thus, the Sylow subgroup's) –  DonAntonio Jan 7 '13 at 14:56

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