Determining Quotient Group (\mathbb{Z}_2\times\mathbb{Z}_4)/\langle(1,2)\rangle

[Data]

Just because

$$\phi(ab) = \phi(a)\phi(b)$$

for some particular $a$ and $b$ does not imply that it works for every $a$ and $b$. This is the problem. A homomorphism requires it to work for any choices. Thus you didn't prove that anything was a homomorphism or a bijection (although it is certainly possible to find bijections and homomorphisms between the two. It is not possible to find an isomorphism, ie. a function that is both a bijection and a homomorphism. This is what you need to prove).

And actually I spoke a little too soon in my last post (I need to read more carefully!). In $Q$, we find $A^4B^2 =e(-e) = - e \neq e$ but in ${\cal D}_4$ we find $R^4 \rho^2 = ee = e$. So your example doesn't quite work anyways.

The most clear path to prove what you need is that that Muzza suggested:

Prove that if $f(x)$ is an isomorphism between to groups then $\mbox{ord}f(x) = \mbox{ord}x$ and then use the fact that I gave above of unequal numbers of elements of certain orders to get your result.

 

[Oxymoron]

Thanks Data. That is what I wanted to hear. And Muzza's post was most helpful too. I will post back with another attempt later.

Thankyou guys.

By the way Muzza, your "framework" proof, is exactly the path I was trying to take (same idea). Except mine came out all fuzzy.

 

[Oxymoron]

If $$Q$$ and $$\mathcal{D}_4$$ are groups, and $$\phi : Q \rightarrow \mathcal{D}_4$$ is an isomorphism. Then for all $$x \in Q$$,

$$|\phi(x)| = |x|$$

That is, $$\phi$$ maps from exactly one element in $$Q$$ to exactly one element in $$\mathcal{D}_4$$.

Suppose that $$\phi$$ is an isomorphism. Denote $$a$$ to be rotation and $$b$$ to be reflection in $$\mathcal{D}_4$$. Then there are three elements in $$\mathcal{D}_4$$ with order 2:

$$a^2, b, ab^2$$

In $$Q$$ there is only one element of order 2:

$$A^2$$

where A is the 2x2 matrix that I typed earlier.

So if $$|A| = 2$$ then

$$|\phi(A)| = |a^2| = |b| = |ab^2| \in \mathcal{D}_4$$

Thus $$\phi$$ is not an isomorphism.

 

[Oxymoron]

I want to look closer into the dihedral group $$\mathcal{D}_n$$. Where

$$\mathcal{D}_n = \{a,b | a^n = b^2 = e, \, bab = a^{-1}\}$$.

I want to find all normal subgroups and then determine the corresponding quotient groups.


A subgroup $$H$$ of a group $$G$$ is normal if

$$gHg^{-1} = H \quad \forall \, g \in G$$

Take $$n$$ to be odd. If we apply rotation $$a$$ to the n-gon in the positive sense, and proceed to apply it in the negative sense, then we haven't rotated the n-gon at all.

Im talking jibberish. How am I supposed to find the normal subgroups of this group? What are they going to look like? Subsets? I don't know...

 

[Oxymoron]

Note, that if this was a specific dihedral group $$\mathcal{D}_3$$ or something. Then I would write up the multiplication table, work out the subgroups, and then determine which are normal by applying

$$gH = Hg \, \forall \, g\in G$$

But I have no idea how to start doing this problem when $$n$$ is arbitrary.

 

[Oxymoron]

Ok I may have something.

Dihedral groups are special groups that consist of rotations $$a$$ and relfections $$b$$, where the group operation is the composition of these rotations and reflections.

The finite dihedral group $$\mathcal{D}_n$$ has $$2n$$ elements and is generated by $$a$$ (with order $$n$$) and $$b$$ (with order $$2$$). The two elements of the dihedral group satsify

$$ab = ba^{-1}$$

If the order of $$\mathcal{D}_{n}$$ is greater than 4, then the group operations do not commute, ie $$\mathcal{D}_n$$ is not abelian.

The $$2n$$ elements of $$\mathcal{D}_n$$ are

$$\{e, a, a^2, \dots , a^{n-1}, b, ba, ba^2, \dots , ba^{n-1}\}$$

Now, in order to form the quotient groups, I need to find the normal subgroups. The normal subgroups are those which are invariant under conjugation.

I know $$\{e\}$$ and $$\{\mathcal{D}_n\]$$ are going to normal subgroups. But I don't know how to find any others.