# Thread: A couple more subgroup questions

1. ## A couple more subgroup questions

1. Find all the subgroups of Q8. (Q8 = {I, -I, J, -J, K, -K, L, -L}. Different notation can be used as well). Show that Q8 is an example of a nonabelian group with the property that all its proper subgroups are cyclic.

2. a) Let G be a cyclic group of order n. Show that if m is a positive integer, then G has an element of order m IFF m divides n.
b) Let G be a cyclic group of order 40. List all the possibilities for the orders of elements of G.

3. Let G = <x> be a cyclic group of order 144. How many elements are there in the subgroup <x^26>?

4. Construct a nonabelian group of order 16, and one of order 24.

2. Originally Posted by Janu42
1. Find all the subgroups of Q8. (Q8 = {I, -I, J, -J, K, -K, L, -L}. Different notation can be used as well). Show that Q8 is an example of a nonabelian group with the property that all its proper subgroups are cyclic.
using lagrange's thm, subgroups of Q8 is of order 1, 2, 4 and 8.. get all subgroups then..

Originally Posted by Janu42
2. a) Let G be a cyclic group of order n. Show that if m is a positive integer, then G has an element of order m IFF m divides n.
b) Let G be a cyclic group of order 40. List all the possibilities for the orders of elements of G.
a) for ( $\Rightarrow$) $G$ is cyclic, then $G=$ and $a^n=e$. Let $b\in G$ such that $b^m=e$. then $b=a^k$ for some natural number $k$. thus $b^m = a^{km} = e = a^n$ you can continue from here..

on the other direction, if $m|n$, then $n=mk$ for some integer $k$. thus $e=a^n=a^{mk}=(a^k)^m$...

Originally Posted by Janu42
3. Let G = <x> be a cyclic group of order 144. How many elements are there in the subgroup <x^26>?
$\frac{\mbox{lcm }\{144, 26\}}{26}$

Originally Posted by Janu42
4. Construct a nonabelian group of order 16, and one of order 24.
i think, $Q_8\times \mathbb{Z}_2$ is non-abelian..
$S_4$ is non-abelian

3. I'm still confused on #1 and #4....

Can someone explain to me how #1 works exactly?

4. Originally Posted by Janu42
4. Construct a nonabelian group of order 16, and one of order 24.
In general if for any even number $>4$ there exists an non-abelian group. Just construct $D_n$ - the dihedral group of order $2n$.

5. The dihedral group? I don't know if we've learned that yet....

What exactly is the structure of the dihedral group?

6. Originally Posted by Janu42
What exactly is the structure of the dihedral group?
A geometric way to think of it is all the symmetries of a regular $n$-gon.
There are $n$ rotations which perserve the polygon.
And there are $2$ reflections through each vertex.
So there are a total of $2n$ symmetries, and these symmetries form the dihedral group.

7. i think, $Q_8\times \mathbb{Z}_2$ is non-abelian..
$S_4$ is non-abelian
Is $Q_8\times \mathbb{Z}_2$ a group of order 16? And the group $S_4$ is order 24?

And for #1, the subgroups of Q8, if I'm not mistaken, are the subgroups {I}, {I, -I}, {I, -I, J, -J}, {I, -I, K, -K}, and {I, -I, L, -L}? However, I don't know if these subgroups are cyclic, because the question asks us to show that all the subgroups of Q8 are cyclic, so I don't know, I'm rather confused on this whole question.

8. Originally Posted by Janu42
And for #1, the subgroups of Q8, if I'm not mistaken, are the subgroups {I}, {I, -I}, {I, -I, J, -J}, {I, -I, K, -K}, and {I, -I, L, -L}? However, I don't know if these subgroups are cyclic, because the question asks us to show that all the subgroups of Q8 are cyclic, so I don't know, I'm rather confused on this whole question.
the subgroups of $Q_8$ are {I}, {I, -I}, {I, -I, J, -J}, {I, -I, K, -K}, and {I, -I, L, -L} and they are all cyclic..

Originally Posted by Janu42
Is $Q_8\times \mathbb{Z}_2$ a group of order 16? And the group $S_4$ is order 24?
if $(a,b) \in Q_8\times \mathbb{Z}_2$, then there will be 8 choices for $a$ and 2 choices for $b$ which make it 16 choices for the element $(a,b)$.. since Q_8 is non-abelian, so is $Q_8\times \mathbb{Z}_2$

$|S_n|= n!$