# Subgroups

• Feb 1st 2013, 11:47 AM
lovesmath
Subgroups
How many distinct subgroups does Z/36Z have? How would I begin solving this?
• Feb 1st 2013, 01:01 PM
Deveno
Re: Subgroups
things to remember with cyclic groups:

1) every subgroup is cyclic.

2) if G is cyclic of order n, and k divides n, G has a UNIQUE subgroup of order k (so it has one, and only one, for each divisor k of n).

in particular, if G = <a>, with |a| = n, and k|n, so that n = kd, then the subgroup of order k is <ad>.

for ABELIAN groups, ad is usually written da. since Z/36Z is generated by 1+36Z = [1], we need only consider <d[1]> = <[d]> for each divisor d of 36.

the divisors of 36 are:

1,2,3,4,6,9,12,18, and 36.

• Feb 1st 2013, 01:02 PM
ILikeSerena
Re: Subgroups
Hi lovesmath! :)

The size of a subgroup divides the size of the group.
How many dividers of 36 can you find?
Pick one. Is it a subgroup?
• Feb 1st 2013, 06:46 PM
lovesmath
Re: Subgroups
Okay, so the divisors are 1, 2, 3, 4, 6 9, 12, 18, 36. When the order is 36, the generator is <1>. When the order is 18, the generator is <2>.
So, we have the following:
order 36, <1> = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 0}
order 18, <2> = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 0}
order 12, <3> = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 0}
order 9, <4> = {4, 8, 12, 16, 20, 24, 28, 32, 0}
order 6, <6> = {6, 12, 18, 24, 30, 0}
order 4, <9> = {9, 18, 27, 0}
order 3, <12> = {12, 24, 0}
order 2, <18> = {18, 0}
order 1, <0> = {0}

Am I on the right track?
• Feb 1st 2013, 08:22 PM
Deveno
Re: Subgroups
those are them!
• Feb 6th 2013, 03:30 PM
lovesmath
Re: Subgroups
But is each one of those subgroups distinct?
• Feb 6th 2013, 04:02 PM
jakncoke
Re: Subgroups
yes, there is no other subgroups of those sizes which are different.