Group Theory

[uneven lance]

Apart from the non-helpful answer that there are no simple groups of order 8, I'd say first state the definition of what it means for a group to be simple.

[fair umbra]

So A group G is simple if its only normal subgroups are ⟨eG⟩ and G ?

[uneven lance]

Yes. So to show a group is not simple, you have to find a normal subgroup.

Do you remember Lagrange's Theorem?

This will give you some insight on what subgroups can exist.

[fair umbra]

From my notes ive got (Lagrange’s Theorem). Let H be a subgroup of a finite group G. Then,
∣G∣ = [G ∶ H] ⋅ ∣H∣. In particular, the order of H divides the order of G.?

[uneven lance]

Yes. So the possible orders of a nontrivial subgroup is 2 and 4.

[uneven lance]

Try finding some subgroups of order 2 or 4.

[fair umbra]

generally or in relation to my question? because wouldnt the order 2 be the vertical and the hoziontal reflections and order 4 be the rotations of 0,90,180,270?

[uneven lance]

For this particular group. There are several supgroups of order 2 and 4 for the group in your question. You have found 2. See if they are normal or not.

[fair umbra]

so H = {r * f | r in R, f in F} where r and f are order 2 and order 4 respectively?. Now, it can be shown that H is a nontrivial normal subgroup of G. This is because, for any element g in G, the conjugate g * H * g^(-1) is still in H?

therefore g is not simple ?

if this is the case whats the best why to write this out as the proof?

[uneven lance]

That depends on what your professor/teacher expects. I'd say, define your subgroup. Then show it is a normal subgroup. Maybe say "Since H is a nontrivial normal subgroup, then G is not simple."

When proving it is normal, you cou brute force all possibilities, or prove it algebraicly using a representation of arbitrary elements in G and H, or use geometric reasoning.

[fair umbra]

yeah thats far enough , to define the subgroup its just H = {r * f | r in R, f in F} where r and f are order 2 and order 4 respectively is there a more efficient way to right this it seems really long winded to get my point across

[fair umbra]

theres reflections in the diagonals two will that mean my order 2 isnt valid ?

[uneven lance]

I have a bit more time now. You may want to rewrite your definition of the subgroup H. It's not entirely clear what you are referring to. I'd first define two particular symmetry elements r and f for rotations and reflections. Simply state how r and f permute the diagram. Then you can define your subgroup from r and f. Either using set builder notation, or by listing the elements.

[uneven lance]

Can you elaborate?

[uneven lance]

I mean, for this problem, you have to find a normal subgroup. That involves showing the subgroup, and proving it's normal. You can't really short cut it with what you probably know.

[fair umbra]

For all reflections it would be 4 ? Or can you just state for a certain axis

[uneven lance]

The set of reflections do not form a subgroup. The identity is missing for example.

[fair umbra]

I have a feeling I know what I’ve missed one sec

G = {e,p,p^2,p^3,o_1,o_2,o_3,o_4}

[uneven lance]

Those are all 8 elements. Can you find a subgroup?

[fair umbra]

I’m honestly so confused

[fair umbra]

The subgroup is {e,p,p^2,p^3}

[uneven lance]

That is a subgroup. (Subgroup of just the rotations). Can you prove that it is a normal subgroup?

[fair umbra]

A subgroup is said to be a normal subgroup if, for every element g in the group G and every element h in the subgroup H, the conjugate ghg^{-1} is also in H. In other words, a subgroup is normal if and only if gHg^{-1} = H for all g in G and H in H .
Consider an arbitrary element g in G and an arbitrary element h in the subgroup H = {e, p, p^2, p^3} . We want to verify that ghg^{-1} is also in H .

Let g be an arbitrary element in G . We know that G is generated by {e, p, p^2, p^3} , so g can be any element in G

Now, let's consider an arbitrary element h in H = {e, p, p^2, p^3}. The conjugate ghg^{-1} for this case is simply g because ghg^{-1} = g when h is the identity element.

Since g is an arbitrary element in G and ghg^{-1} = g for any h in H , it implies that H is normal in G .

Therefore, {e, p, p^2, p^3} is a normal subgroup of G ?

[uneven lance]

G is not generated by {e, p, p^2, p^3}, as that is a subgroup H. G is generated by {p, o}.

What you have shown is that H is a normal subgroup of H, which is trivial to show.

Instead, you could let g be an arbitrary element in G but not in H, and show that gHg^{-1}=H for the four reflections. That will suffice.

[fair umbra]

So to do that I sub the elements of H into it ?

I thought it should be gHg^{-1} ∈ H not equals ?

[uneven lance]

It does turn out that they are equal. This is a consequence of the fact that elements in a group are invertible. But yes, you only need to show containment in one direction.

Note that ∈ means containment and ⊆ means subset.
So x∈A means that x is an element of A, and A⊆B means every element of A is an element of B.

[uneven lance]

You need to show that gHg^{-1} ⊆ H for every element of G. This statement is true if g∈H, but that is only 4 of the 8 elements of G. There are still 4 others (the 4 reflections) to check.