# Thread: Problem: Intersection of 2-Sylow-groups

1. ## Problem: Intersection of 2-Sylow-groups

Hi,

I am stuck proving that a simple group $G$ of order $60$ is isomorphic to $A_5$.

In particular:

- I have shown $|Syl_5(G)|=6$ and $|Syl_3(G)|=10$, so there must be $6\cdot(5-1)=24$ elements of order $5$ and $10\cdot(3-1)=20$ elements of order $3$.

- I could further show $|Syl_2(G)|\notin\{1,3\}$.

- Now, where I get stuck: I need to show $|Syl_2(G)|\neq15$ and therefore $|Syl_2(G)|=5$.
I know I have to use the fact that there are already $20+24=44$ elements of order coprime to $2$. That leaves only $60-44-1=15$ elements to have order 2 or 4, which somehow has to lead to a contradiction to $|Syl_2(G)|=15$. But the 2-Sylow-groups have order $2^2$ and can thus intersect non-trivially. So how can I argue from here?

Any help would be much appreciated!

2. ## Re: Problem: Intersection of 2-Sylow-groups

Could you possibly use the fact that all sylow p subgroups are conjugate to each other (with p=2)?

3. ## Re: Problem: Intersection of 2-Sylow-groups

I could not yet figure out how to use that, but I will try for some hours. Thanks for the input!

4. ## Re: Problem: Intersection of 2-Sylow-groups

Had vague thoughts about the conjugacy class of 2-Sylow-subgroups generating a normal subgroup in G, but I couldn't see how that would lead me further. Can you help me a little?

5. ## Re: Problem: Intersection of 2-Sylow-groups

Hi,
I believe this is what you need to finish the proof that G is isomorphic to A5; namely that there are 5 Sylow 2 subgroups of G. After this determination, I left the rest of the proof to you:

6. ## Re: Problem: Intersection of 2-Sylow-groups

Thank you!
That was all I need to conclude my proof.