Is it true that any group of order 30 has a normal (hence unique) Sylow-5 subgroup?(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Groups of order 30

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