# Thread:

1. ## Group Proof

Hi. I'm stuck on a university level algebra question in my textbook and I was hoping to receive some help on it.
The problem is: Let G be a group and let H1, H2 be subgroups of G. Prove that if H1 union H2 is a subgroup then either H1 is a subset of H2 or H2 is a subset of H1. Sorry about the notation or lack thereof. I don't know all the programming stuff.

Plus I'm just reading something and I was wondering. Are the symbols S3, Q8, and D4 notations for specific groups? I have never seen them before and a problem I have seems to assume I should know them. Anyway, all the numbers after the letters are subscripts.

Thanks for any help. Pvt Bill Pilgrim.

2. Originally Posted by PvtBillPilgrim

Plus I'm just reading something and I was wondering. Are the symbols S3, Q8, and D4 notations for specific groups? I have never seen them before and a problem I have seems to assume I should know them. Anyway, all the numbers after the letters are subscripts.
Yes,
$S_n$ denotes the symettric group, also known as the group of premutations.

$D_4$ denotes the dihedral group on a vertices of a polygon.

I never seen $Q_8$, maybe the octions

3. Originally Posted by PvtBillPilgrim
Hi. I'm stuck on a university level algebra question in my textbook and I was hoping to receive some help on it.
The problem is: Let G be a group and let H1, H2 be subgroups of G. Prove that if H1 union H2 is a subgroup then either H1 is a subset of H2 or H2 is a subset of H1. Sorry about the notation or lack thereof. I don't know all the programming stuff.
I never considered that, it is interesting.
Okay the problem is that when you take the union of two subgroups of a group not necessarily a subgroup, why? Well, there is an identity element, there is an inverse for each element and multiplication is associate. The problem is it fails to be closed.

Let us do this by contradiction...

Assume, that neither $H_1\leq H_2$ and $H_2\leq H_1$ is true. (Note the meaning of $\leq$ is the same as $\subseteq$ because these are groups).
Thus there exists $x,y\in G$ such as,
$x\in H_1 \mbox{ and }x\not \in H_2$
$y\not \in H_2 \mbox{ and }y \in H_1$
But the condition of the problem,
$H_1\cup H_2$ is a group, and $x,y\in (H_1\cup H_2)$
Thus,
$xy\in (H_1\cup H_2)$
Thus,
$xy\in H_1 \mbox{ or }xy\in H_2$
Thus,
$xy=h_1 \, \exists h_1\in H_1$
$xy=h_2 \, \exists h_2\in H_2$
Using group properties we have,
$y=x^{-1}h_1$
$x=h_2y^{-1}$
But,
$x\in H_1$ implies $x^{-1}\in H_1$ and $h_1\in H_1$ thus, $x^{-1}h_1\in H_1$ by closure. Thus, $y\in H_1$
But,
$y\in H_2$ implies $y^{-1}\in H_2$ and $h_2\in H_2$ thus, $h_2y^{-1}\in H_2$ by closure. Thus, $x\in H_2$.

Thus,
$y\in H_2\mbox{ and }x\in H_1$---> Contradiction.

Thus, (de Morgan's negation)
$H_1\leq H_2$ or $H_2\leq H_1$

4. I really appreciate the proof.

Now just a few questions:
How exactly do you define these particular groups, say Q8, S3, D4?
What would be the order of each element of these groups?
What would be the subgroups of these groups (which ones are cyclic?)

I must have missed something in class and now I'm behind. All my notes seem to assume that these are known.

5. Originally Posted by PvtBillPilgrim
I really appreciate the proof.

Now just a few questions:
How exactly do you define these particular groups, say Q8, S3, D4?
I cannot give you a lecture on this. But you should got to Wikipedia and read.
$S_3$ is a group of symettries on $\{1,2,3\}$
Dihedral group is more complicated to explain. $Q_8$ are the octions. These are,
$\{1,-1,i,-i,j,-j,k,-k\}$
And,
$i^2=j^2=k^2=-1$
And, $ijk=-1$
This is non-abelian (in fact all 3 that you mentioned are non-abelian).

What would be the order of each element of these groups?
The order of an element is the smalles positive $n$ (it exists for finite groups) such that,
$a^n=e$.

For example, in the octions
$1^1=1$ Order of 1
$(-1)^2=1$ Order of 2
$(i)^4=(-i)^4=(j)^4=(-j)^4=(k)^4=(-k)^4=1$---> Order of 4
What would be the subgroups of these groups (which ones are cyclic?)
Let us look at the octions.
The Klein 4 group is its subgroup,
$\{1,-1,i,-i\}$
It happens to be cyclic,
$<i>$ is a generator.

6. Just one more question:

What are the subgroups of S3, the symmetric group and which are cyclic?

I figured out everything else, but stuck on this.

7. Have you considered searching for a group table online?

The group $S_3$ is the same as $D_3$ (or $D_6$ depending on your style).

Here is the Dihedral Group D3

8. Here is the group $S_3$.

There are going to be $3!=6$ elements.
Here there are.... (note I given them a name).

$\rho_0=\left( \begin{array}{ccc}1&2&3\\1&2&3 \end{array} \right)$
$\rho_1=\left( \begin{array}{ccc}1&2&3\\2&3&1 \end{array} \right)$
$\rho_2=\left( \begin{array}{ccc}1&2&3\\3&1&2 \end{array} \right)$
$\mu_1=\left( \begin{array}{ccc}1&2&3\\1&3&2 \end{array} \right)$
$\mu_2=\left( \begin{array}{ccc}1&2&3\\3&2&1 \end{array} \right)$
$\mu_3=\left( \begin{array}{ccc}1&2&3\\2&1&3 \end{array} \right)$
These will form a group shown below.

We can see that the set,
$\{\rho_0,\rho_1,\rho_2\}$
Is a subgroup of this group.
Furthermore, since the order of this subgroup is 3, a prime, it is cyclic.

While we have this diagram let us see if we can find any subgroups.
Of course,
$\{\rho_0\}$ is a subgroup, it is the trival subgroup for it contains the identity element.
$S_3$ is the improper subgroup it contains all of these elements.
But are there any other except for the one I mentioned above?

By Lagrange's theorem the order of a subgroup divides the order of the finite group. So if they exist they must be either 2 or 3. (In fact they must exist, it is called Cauchy's theorem but that might be too advanced for you since you are just starting to learn group theory). But still even if you never learned Cauchy's theorem you can still come to a conclusion by looking at the group diagram.

By observing this group we see that $\{\rho_0,\mu_1\},\{\rho_0,\mu_2\},\{\rho_0,\mu_3\}$ are subgroups. It can be shown (again might be too difficult for you yet) that there cannot exist more subgroups of order 2. Or you can just look at all the possibilities and see that these are the only ones.
And all of these are cyclic! Because 2 is a prime.

We have succesfully given all the subgroups and shown they are all cyclic except for the improper subgroup, the group itself.
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