Determining subgroups of a finite field and their elements

Question

I'm studying cryptography and while reading some lecture notes I found the following:

$F$*₃₇ has subgroups of order 2 ({2⁰ , 2¹⁸}), 3 ({2⁰ , 2¹² , 2²⁴}), 4, 6, 9, 12, and 18.

How to determine that the subgroups are of order 2, 3, 4, 6, 9, 12, and 18?
How to determine the elements of those subgroups? (for example ({2⁰ , 2¹⁸}) in the first one.) I know in this one that the base is 2 because the field generator is 2, but where did those powers come from?

Thank you.

I'll read Sylow's Subgroups Theorems. I copied the text from professor's notes, so I don't know if he missed an element of order 36. — OneZ, Jan 17 '15 at 04:59
Your teacher is applying the following general fact. Given a cyclic grop of order $n$, generated by $g$, it has a unique subgroup of order $d$ for any divisor of $n$. That subgroup is generated by $g^{n/d}$, so its elements are $g^0,g^{n/d}, g^{2n/d},\ldots,g^{(d-1)n/d}$. Here $n=36$,$g=2$ and $d$ varies. — Jyrki Lahtonen, Jan 17 '15 at 07:11
And I agree with Thomas. This is way more elementary than Sylow. No need for you to go there at this point. — Jyrki Lahtonen, Jan 17 '15 at 07:12

score 2 · Answer 1 · answered Jan 17 '15 at 04:07

You are trying to use too much information!

You are trying to study the cyclic group of 36 elements and you have chosen a generator of the group. Once you know this, the fact it originated from a finite field doesn't matter at all -- you should ignore that piece of information entirely, and just focus on the fact you're studying a cyclic group.

Determining subgroups of a finite field and their elements

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