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Ok, this question has been giving me trouble for the last few hours. I need help for a test tomorrow.

Let $\Bbb Z _{20}$ 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 $K$.

$\Bbb Z_{20}= \{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19 \}$

where $20=e=0$

When I attempted this, i found the groups generated by:
$ \langle 1 \rangle , \langle 2 \rangle, \cdots \langle 20 \rangle $.
in which some subgroups were identical to $\Bbb Z_{20}$ and other subgroups were the same other subgroups, i.e. $\langle 5 \rangle = \langle 15 \rangle $.

so my question is, is there a way to know if a certain generator like $\langle 5 \rangle $ will be equal to $\langle 15 \rangle $ 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 \leq K $ for a group $ K $

Edit: I've weeded out all of the same subgroups generated by $ \langle 1 \rangle , \langle 2 \rangle, \cdots \langle 20 \rangle $.. and found that the subgroups that do not contain 4 are:

$ \langle 17 \rangle = \{ 17,14,11,8,15,12,9,6,3,0 \} \\ \langle 5 \rangle= \{5,10,15,0\} \\ \langle 10 \rangle = \{10,0\} $

all of the other subgroups are either equal to $\Bbb Z_{20}$ or contain a $4$. is this right? is there an easier way to find the proper nontrivial subgroups without computing all of $ \langle 1 \rangle , \langle 2 \rangle, \cdots \langle 20 \rangle ?$

Sebastiano
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user126865
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  • I think there is a mistake in your calculation since $\langle 17 \rangle$ should contain 4. – AG. Feb 07 '14 at 03:31
  • youre right, also, <17>=Z20, right? – user126865 Feb 07 '14 at 16:55
  • yes, and that's because gcd(17,20)=1. Similarly, $\langle 8 \rangle$ equals $\langle 4 \rangle$. In general $x$ generates the same subgroup as $gcd(x,20)$. – AG. Feb 08 '14 at 08:33

2 Answers2

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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.

user76568
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The subgroups of a cyclic group $\{1,x,x^2,\ldots,x^{n-1}\}$ of order $n$ are exactly subsets of the form $\langle x^d \rangle$, where $d$ divides $n$. Since $n=20$, the distinct subgroups are $\langle x^d \rangle$, where $d=1, 2, 4, 5, 10, 20$. Among the six subgroups of $Z_{20}$, the element $x^4$ is contained in $\langle x \rangle, \langle x^2 \rangle$ and $\langle x^4 \rangle$, and is not contained in the three other subgroups.

To reduce the amount of computations, use the fact that the subgroups of $Z_n$ are in 1-1 correspondence with the positive divisors of $n$.

AG.
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