So I wanted to check if what I did was correct. I'm not sure if it is and if so what would the correct way to go about this be?

So firstly $|G| = 2^2*3*5$. This confirms there is Sylow 3 - Subgroup's and Sylow 5-Subgroups.

Now for Sylow 3 - Subgroup's, where $n_3$ is the number of Sylow 3-Subgroups, we have, from Sylow's theorems that:

$n_3 \equiv 1$ mod $3$ and $n_3$ | $20$ so $n_3 = \{1,4,10 \} , n_3 \neq 1$ as otherwise we'd have a unique normal group and therefore G would not be simple.

From similar deduction, we arrive at $n_5 = 6$

So we must have that any two Sylow 5 - Subgroups intersect at $\{1 \}$ and so we have:

$60 - ((6-2) * 5) = 40$ elements in the remaining Sylow p - subgroups.

If $n_3 = 4$, similarly:

$40 - ((4-2) * 3) = 34$ elements in the remaining Sylow p - subgroups.

But $(\frac{34}{4}) \notin \mathbb{Z}$ where $4 = 2^2$ for Sylow 2 - subgroups.

Now if $n_3 = 10$ then 16 elements remain and $(\frac{16}{4}) = 4$ Sylow 2 - subgroups and therefore $n_3 = 10$

So upon checking my mark scheme these are the correct answers - but I am not 100% my method is correct or if its rigorous enough. If it is - is there other ways to do it?



You could argue as follows, since Sylow subgroups are permuted transitively under conjugation.

If the number of Sylow $3$-subgroups is $4$, then they are permuted transitively by conjugation, and this gives rise to a non-trivial homomorphism to $S_4$. But $S_4$ has order $24\lt 60$ so such a homomorphism would have a non-trivial kernel.

  • $\begingroup$ Ok that makes perfect sense, thank you! $\endgroup$ – mrhappysmile Jan 11 '15 at 19:53

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