In $Z_{24}$, list all generators for the subgroup of order 8

Question

Q. In $\Bbb Z_{24}$, list all generators for the subgroup of order $8$.

So, I know that the divisors of $24$ which are $1,2,3,4,6,8,12 $ and $24$ are the orders of the sets in the subgroup.

I'm not sure if this is a trick question but I was only able to find one generator which is $\langle 3\rangle$, so was the plural in generators unnecessary? Or am I missing a generator.

The group of order $8$ is cyclic, hence it has the same number of generators as $\Bbb Z/8\Bbb Z$. — PVAL-inactive, Mar 11 '15 at 00:17

score 8 · Answer 1 · edited Dec 17 '19 at 19:52

8

Yes, $9,15$, and $21$ are also generators. In any cyclic group of order $n$ then generator of a subgroup of order $d$ has to form $g^m$ where the g.c.d $(m,n)=n/d$.

So, with $n=24$ and $d=3$, we want to find all numbers $m$ such that $1\le m\le 24$. $(m,24)=24/8=3$.

Running through the multiples of $3$ which are less than $24$ gives the generators $3,9,15,21$.

edited Dec 17 '19 at 19:52

gsgx

789

answered Mar 11 '15 at 00:20

Tim Raczkowski

20,046

what is m in this case? – kero Mar 11 '15 at 00:25
I'll add to my answer. – Tim Raczkowski Mar 11 '15 at 00:26

score 6 · Accepted Answer · answered Mar 11 '15 at 00:19

6

a cyclic group of $n$ elements has $\varphi(n)$ divisors, so you should have $\varphi(8)=4$ generators.

The subgroup of order $8$ is : $\{0,3,6,9,12,15,18,21\}$. The generators are: $\{3,9,15,21\}$

answered Mar 11 '15 at 00:19

Asinomás

105,651

How did you get that set {3,9,15,21}? – kero Mar 11 '15 at 00:23
1

you can look at your set as ${0,3,2\cdot3,3\cdot3,4\cdot3,5\cdot3,6\cdot3,7\cdot 3}$, you want the ones that have coefficients relatively prime the the order of the subgroup (Which is $8$) – Asinomás Mar 11 '15 at 00:28
got it! Thanks!! – kero Mar 11 '15 at 01:32

In $Z_{24}$, list all generators for the subgroup of order 8

2 Answers2