So I'm studying a simple group of order $|G|=60$, with $n_2=15,n_5=6,n_3=10$ and I'm trying to show that the centralizer $C_G(x)=\{g \in G | xg=gx\}$ of a $x \in P \cap Q$ where $P$ and $Q$ are 2-Sylow groups that intersect non-trivially such that $|P \cap Q|=2$. I would like to show that $|C_G(x)|= 12$ or $20$, and consequently show that it is isomorphic to a subgroup of $S_5$ using an action of $G$ on $G / H$. (for then show that the only subgroup of order $60$ in $S_5$ is $A_5$)

I was able to conclude this same thing with $n_2=5$ before, but I don't really know how to do it for this case.

Would you have any tips? Thanks.

edit: I found something related in https://coolnumbers.wordpress.com/2011/12/28/a_5-is-the-only-simple-group-of-order-60/ but is not quite the same.

edit: right now I'm trying to find $f$ such as $Ker(f)=C_G(x)$ et $Im(f)=5-Sylow$ ou $3-Sylow$ that would show that $G/C_G(x) \cong Im(f)$ therefore $[G:C_G(x)]=12$ or $20$.

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    $\begingroup$ You could usefully tell us what $n_2$ etc means. Also there is only one simple group of order $60$ up to isomorphism - are you trying to identify all such groups and/or prove that only one exists? $\endgroup$ Nov 9, 2019 at 18:44
  • $\begingroup$ I'm sorry, I thought you guys would understand, $n_p$ is the number of p-Sylows in the group, my bad. $\endgroup$ Nov 9, 2019 at 18:46
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    $\begingroup$ If these are the numbers of Sylow subgroups, I think you have accounted for all the elements of the group already, which may help. $\endgroup$ Nov 9, 2019 at 18:48
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    $\begingroup$ You might be counting elements or conjugacy classes, for example. $\endgroup$ Nov 9, 2019 at 18:49
  • $\begingroup$ Well, what 'im mostly looking for now is the order of the centralizer actually $\endgroup$ Nov 9, 2019 at 18:50

1 Answer 1


Note that there is one element of order $1$, fifteen elements of order $2$, $24$ elements of order $5$ and $20$ elements of order $3$ within the (supposed) Sylow Subgroups accounting for all $60$ elements of $G$.

Since the Sylow subgroups of given order are permuted transitively by conjugation only the identity element can be central, so the centre must be trivial.

  • $\begingroup$ I'm sorry, why 24 elements of order 5 and 20 elements of order 3? Normally we have only 6 elements of order 5 and 10 of order 3 which are the 5-Sylows and the 3-Sylows, isn't it? $\endgroup$ Nov 9, 2019 at 19:01
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    $\begingroup$ @LucasTonon There are six subgroups of order $5$. Since any non-identity element generates such a group, they can't share any elements. Each Sylow subgroup of order $5$ contains four distinct elements of order $5$. (That's why I wanted to make sure what you were counting.) $\endgroup$ Nov 9, 2019 at 19:05
  • $\begingroup$ Alright, so the argument for the 2-Sylows of order 4 is the same, we have four elements which 1 is of order 2, $\{e,x_1,x_1^{-1},x_2=x_2^{-1}\}$ then $x_2$ has order 2. But there are 2-Sylows such as P and Q that intersect not trivally with an element of order 2 more specifically, so howcome there can be 15 of order 2? $\endgroup$ Nov 9, 2019 at 19:20
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    $\begingroup$ @LucasTonon Sorry, I'm counting wrong - the 2-subgroups of course have order $4$ and each have two elements of order $4$ so you can't have this combination of Sylows. $\endgroup$ Nov 9, 2019 at 19:21
  • $\begingroup$ What I've been trying to do is to show that the index of this centralizer is either 3 or 5, that might be easier $\endgroup$ Nov 9, 2019 at 20:03

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