Groups of order 30

1. Jan 19, 2009

math8

Is it true that any group of order 30 has a normal (hence unique) Sylow-5 subgroup?

I know that that the only possibilities for n(5) are 1 or 6.

Now suppose there are 6 sylow 5 subgroups in G. This would yield
(5-1)6=24 distinct elements of order 5 in G. Now there is only 30-24=6
elements left in G and one of these is the identity. This means that there
must be 1 sylow 3 subgroup in G which has 2 distinct elements of order 2
now there is only 4 elements left in G one of them being the identity
so there must be 3 sylow 2 subgroups of G in this case each having 1 distinct
element of order 2

I don't see where is the contradiction here.

The only thing I know is that there must be a cyclic normal subgroup of order 15 in G.

Last edited: Jan 19, 2009
2. Jan 19, 2009

math8

Oh I think I got it.

Every group G of order 30 has a normal cyclic subgroup of order 15 (I can prove this).
Let's call it H.
Now, consider Syl_5(H).
n_5=1, hence if P lies in Syl_5(H), then P is the unique normal subgroup of H of order 5. Hence P char H. Since H is normal in G, it follows P is normal in G.

Now |P|=5 and |G|=(2^2)*3*5. So P is a normal Sylow-5 subgroup of G.