- #1

seanthinks

- 9

- 1

- TL;DR Summary
- I'm trying to get a little bit more comfortable with using the Sylow theorems by reworking examples from my abstract algebra textbook.

I'm trying to get a little bit more comfortable with using the Sylow theorems and looking back at some examples from my abstract algebra textbook (we used Fraleigh). In an example involving groups of order 36 (the claim is no such groups are simple), the author writes:

"Such a group ##G## has either one or four subgroups of order 9. If there is only one such group, it is normal in ##G##. If there are four such subgroups, let ##H## and ##K## be two of them. As an in Example 4.3.13, ##H \cap K## must have 3 elements, or ##HK## would, have 81 elements which is impossible. Thus the normalizer of ##H \cap K## has as order a multiple of >1 of 9 and a divisor of 36; hence the order must be either 18 or 36. If the order is 18, the normalizer is then of index 2 and therefore is normal in ##G##. If the order is 36, then ##H \cap K## is normal in ##G##."

I'm trying to rework this example from a different angle and I'm wondering if we could alternatively look at the subgroups of order 4 for which ##G## would have either 1 or 9. If it has 1 then it's normal and we're done. If it has 9, then taking a similar approach, let ##H## and ##K## be two such subgroups. Then if the intersection of ##H## and ##K## is trivial, then ##G## has 27 elements of order 4, leaving us with 8 elements whose order must divide 36 (so either 6, 9, 12, 18, or 36). We know that we have at least 1 Sylow 3-subgroup of order 9 so the remaining elements must be contained in this subgroup. So in fact it's the only Sylow 3-subgroup, so it's normal and we're done. OTOH if ##H \cap K## is nontrivial, then since it's order is a multiple of 4 (greater than 1), and divides 36, it has to have an order of 12 or 36. But both of these are impossible since ##H \cap K## cannot have an order greater than 4. So ##H \cap K## would have to be trivial.

"Such a group ##G## has either one or four subgroups of order 9. If there is only one such group, it is normal in ##G##. If there are four such subgroups, let ##H## and ##K## be two of them. As an in Example 4.3.13, ##H \cap K## must have 3 elements, or ##HK## would, have 81 elements which is impossible. Thus the normalizer of ##H \cap K## has as order a multiple of >1 of 9 and a divisor of 36; hence the order must be either 18 or 36. If the order is 18, the normalizer is then of index 2 and therefore is normal in ##G##. If the order is 36, then ##H \cap K## is normal in ##G##."

I'm trying to rework this example from a different angle and I'm wondering if we could alternatively look at the subgroups of order 4 for which ##G## would have either 1 or 9. If it has 1 then it's normal and we're done. If it has 9, then taking a similar approach, let ##H## and ##K## be two such subgroups. Then if the intersection of ##H## and ##K## is trivial, then ##G## has 27 elements of order 4, leaving us with 8 elements whose order must divide 36 (so either 6, 9, 12, 18, or 36). We know that we have at least 1 Sylow 3-subgroup of order 9 so the remaining elements must be contained in this subgroup. So in fact it's the only Sylow 3-subgroup, so it's normal and we're done. OTOH if ##H \cap K## is nontrivial, then since it's order is a multiple of 4 (greater than 1), and divides 36, it has to have an order of 12 or 36. But both of these are impossible since ##H \cap K## cannot have an order greater than 4. So ##H \cap K## would have to be trivial.

Last edited: