# Centralizer of a group of order 60 with $n_2=15,n_5=6,n_3=10$...

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 $$20$$.

• 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? Nov 9, 2019 at 18:44
• I'm sorry, I thought you guys would understand, $n_p$ is the number of p-Sylows in the group, my bad. Nov 9, 2019 at 18:46
• If these are the numbers of Sylow subgroups, I think you have accounted for all the elements of the group already, which may help. Nov 9, 2019 at 18:48
• You might be counting elements or conjugacy classes, for example. Nov 9, 2019 at 18:49
• Well, what 'im mostly looking for now is the order of the centralizer actually Nov 9, 2019 at 18:50

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 $$G$$.
• @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.) Nov 9, 2019 at 19:05
• 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? Nov 9, 2019 at 19:20
• @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. Nov 9, 2019 at 19:21