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.