Question

Hi i need help in this groups and symmetries question please help.

1a) Let H1, H2, and H3 be subgroups of a group G. If |H1|= 15, |H2| = 12, and |H3|=20 and |G| <100, determine |G|.

1b) Suppose H is a normal subgroup of a group G with |G|=105 and |H| =21. If a is an element of G such that |a| = 7, find the order of the group element aH in G/H.

when i mean |.| those straight lines just mean the order of something for example |G| simply means the order of G

Answer 1

pleasee help @KingGeorge

Answer 2

For 1a, you know that the order of a subgroup must divide the order of a group. Hence, you need to find the least common multiple of 15, 12, 20, and verify that it's the only multiple less than 100.

Answer 3

For 1b, you know that G/H has exactly 5 elements since 105/21=5. Thus, you know that
aHaHaHaHaH=\(a^5\)H=H Thus, you know that the order of \(a\)H must divide 5, yet it must also divide 7. Hence, the order of aH in G/H is....?

Answer 4

hmm alright i shall continue with that

Answer 5

alright i will keep those hints in mind thanks a lot although i have one question

Answer 6

say for U(18) i want to find the order of each element is there any "quick" way to determine the order of each element or do i have to do it seperately

Answer 7

ps just studying for the test and trying to do all the past tests lol

Answer 8

Could you remind what you are using as U(18)?

Answer 9

for U(18) we are looking everything that is relatively prime to 18 so in this case we got {1,5,7,11,13,17} and now it says compute the order of each element

Answer 10

and i know to find the order i have to do 1^1 mod 18, 1^2 mod 18 blah blah all the wat 1^16 mod 18 would be just 1 so the order of 1 is 1

Answer 11

and now i have to do it for 5 and was just wondering if there was any theorem that i could use to find the order of that element right away rather than doing it one by one you know

Answer 12

If you can find a generator for the group, I think you can compute the order of each element quickly. However, remember that the order has to divide 6, so you can only have an order 1, 2, 3, or 6.

Answer 13

So for 5, compute \(5^2\mod 18\), \(5^3\mod 18\) and \(5^6\mod 18\). The first that equals 1 is the order.

Answer 14

what do you mean by it has to divide 6 :S

Answer 15

It's a well known theorem that if \(g\in G\) where \(G\) is a group, then |\(g\)| divides \(|G|\).

Answer 16

yeah i understand that but you said you sauid the order has to divide 6 why 6? is it because of how many elements we have in U(18)?

Answer 17

It has to divide 6 because \(|\{1, 5, 7, 11, 13, 17\}|=6\). So the size of your group is 6.

Answer 18

alright perfect thanks a lot :)

Answer 19

You're welcome.

Real quick, you do understand why the order of aH has to divide 7 correct?

Answer 20

uhh which question are we talking about now the main question?

Answer 21

Back to 1b, you understand why |aH| must divide 7 correct?

Answer 22

because of the lagrange theorem?

Answer 23

I wouldn't use Lagrange's theorem. Rather, note that

aHaHaHaHaHaHaH=\(a^7\)H=1H=H,

so the order must divide 7 as well.

Answer 24

hmm alright makes sense when i first saw that it looked like you were laughing ahahaha lol

Answer 25

but anywho thanks i appreciate it i will keep posting questions here and there

Answer 26

You're welcome.