Ok, this question has been giving me trouble for the last few hours. I need help for a test tomorrow.
Let Z20 be the cyclic group of integers mod 20 with addition. Let H and K be distinct nontrivial proper subgroups of G such that H is also a nontrivial subgroup of k and 4 is not in K. Describe H and K.
Z20={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19}
where 20=e=0
When I attempted this, i found the groups generated by:
<1>, <2>,…,<20>.
in which some subgroups were identical to Z20 and other subgroups were the same other subgroups, i.e. <5>=<15>.
so my question is, is there a way to know if a certain generator like <5> will be equal to <15> without actually computing it? how do i know which generators will be unique?
also what is the simplest way of finding non trivial subgroups of a given group?
a subgroup is a set in which is closed, associative, has an identity and an inverse.
nontrivial excludes subgroups such as {e}.
and proper means that H <= K for a group K
edit: Ive weeded out all of the same subgroups generated by <1>,<2>,…<20>. and found that the subgroups that do not contain 4 are:
<17> = {17,14,11,8,15,12,9,6,3,0}
<5> = {5,10,15,0}
<10> = {10,0}
all of the other subgroups are either equal to Z20 or contain a 4. is this right? is there an easier way to find the proper nontrivial subgroups without computing all of <1>,2,…,<20>?
Best Answer
Notice that $15=5^{-1}$.
In general $<x>$ = $<x^{-1}>$ can be easily shown for cyclic groups.
I think the easiest subgroups of a group are exactly It's cyclic subgroups.