Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

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, $\,kp\leq n\,$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.