Question

If g generates G, then give generators for a typical subgroup of each order (in terms of g).

is this just g to each odd power? i.e. g^3, g^5...g^15

Answer 1

What a weird question... what does it mean, "typical subgroup of each order" ?

Answer 2

oh sorry, G is order 16

Answer 3

Much better :)

Answer 4

Let's start with the generators which generate a subgroup of order 16 (in other words, the OTHER generators of G)

Can you find them?

Answer 5

g, g^2, g^4, g^8, g^16

Answer 6

the orders for those are 16, 8, 4, 2, 1 i believe

Answer 7

Let's do this systematically :D
First, find the elements of G which generate G itself, can you find them?

Answer 8

hmm, I thought I had. Wouldnt all the g's to odd powers do the trick, since they would hit the even powers, but the even powers wouldnt hit the odds

Answer 9

Okay, you're right, all the odd powers of g would generate G.

Now, which powers of g would generate a subgroup of order 8?

Answer 10

Am I limiting my candidates to the odds at this point? Otherwise, I think my answer to your question is g^2

Answer 11

We've already established that the odd powers of g would generate the entire group G.
That takes care of the subgroup of order 16.

Now what about the generators of the subgroup of order 8?

Answer 12

ahah I see so it is g^2

Answer 13

g^2 among others...

Answer 14

So, first I should have found all the possible orders of G, which I believe are 0,1,2,4,8,16, and then start thinking about which generators create a group of these orders?

Answer 15

Also, what other generators create a subgroup of order 8? , I assume that since the odds will all generate one of order 16, then it is an even power I am looking for. But since evens will just repeat, how can there be more than 1 generator for a subgroup of order 8?

Answer 16

You underestimate the cunning of our dear generators :D
Try \(\large g^6\)
see what it generates ^_^

Answer 17

Ahhhhh, I was not consider even that were not factors of 16. Silly me. g^6 gives 6, 12, 2, 8, 14, 4,10,16

Answer 18

I don't think so...

Answer 19

It gives 8, not 6 :P

Answer 20

Did you even count them? XD

Answer 21

I listed 8: 6,12,2,8,14,4,10,16

Answer 22

See? \(\large g^2 \) is not the only generator of a subgroup with 6 elements :P

Answer 23

Sorry, I just really poorly wrote what I mean. I meant to say that I hadn't considered the even powers below 16 that were not factors of 16. i.e. 6, 10, 12, and 14. Testing those should let me flush out my list

Answer 24

Here, let me give you a little shortcut, you interested?

Answer 25

Sure!

Answer 26

Consider \(\Large g^n\)

To find out its order (the order of the subgroup it generates) simply get this:
\[\LARGE \frac{16}{gcf(16,n)}\]

Answer 27

You'll see it works... with all the odd powers, with 2, with 6... with EVERYTHING ^^

Answer 28

so the gcf of 16 and 6 for example, is 2, and 16/2 is 8... holy crap! I love that!

Answer 29

mhmm :P
Check all the odd powers... the gcf would be 1 XD

Answer 30

thats a powerful tool for larger order groups. Thanks a ton for your help explaining this :)

Answer 31

Yupperz ^_^