# Finding all groups $H$ (up to isomorphism) such that there is a surjective homomorphism from $D_8$ to $H$

Here, $$D_8$$ is the dihedral group of order $$8$$.

From the First Isomorphism Theorem, I'm trying to find a solution for this by considering the quotient groups of $$D_8$$. If $$\phi$$ is a surjective homomorphism from $$D_8$$ to $$H$$ then it's image is isomorphic to the quotient group of $$D_8$$ with the kernel of $$\phi$$.

I'm considering what order $$H$$ could have: 1, 2, 4 and 8.

So far I have:

If $$H$$ has order 4, then the normal subgroup (kernel of $$\phi$$ ) must have order 2. This cannot involve any reflections, since if it were a subgroup then the inverse reflection would have to be there, and a reflection along with its inverse does not form a normal subgroup. So we are left with it being a subgroup containing rotations only and being order 2. This is $$\{ R_{2 \pi}, R_{\pi} \}$$ (full and half rotations respectively), and we can see this is a normal subgroup.

I'm then told:

This gives a quotient isomorphic to $$C_2$$ x $$C_2$$.

My questions:

1. Why is this isomorphic to $$C_ 2$$ x $$C_2$$?
2. How can we say this without saying what the surjective homomorphism is? I find it very tricky to deal with isomorphisms between cosets and set elements.
3. In general, how would you go about finding such groups for any group ($$D_{12}, S_4$$, say)?

1. We can see that your quotient is isomorphic to $C_2 \times C_2$ by looking at what is remaining in the group after you take the quotient and see how their corresponding cosets behave. First you know that the subgroup must have order $8/2 = 4$, so this gives us only two options. Notices that the reflections (which have order two) remain in the group, and their cosets will have order two in the quotient group too. Also the quarter-rotations are still in the group and a quarter-rotation's coset has order two in the quotient since two quarter-rotations is a half rotation, which is now identity. So since our group of order four has two (cyclic) subgroups of order two, it can't be $C_4$, so it must be $C_2 \times C_2$.

2. Here's a topological way to think about it. By quotienting by a half-rotation, we're kinda saying that we want a half-rotation to be the same as a full rotation (identity). So thinking of the square, we are saying that we want each pair of opposite vertices and edges to be basically the same vertex and edge. So take the square and fold it so that opposite vertices are now the same vertex, and opposite edges are now the same edge. The result should have two vertices with two edges connecting them. We can call this a two sided polygon, and the quotient group will be the symmetries of this shape. A topologist would say that the square is a two-fold cover of this shape.

3. In general, this is the correct approach for finding normal subgroups of any group. Just start looking at cyclic subgroups and see if they are normal. Then look at subgroups generated by two elements, and so on. And of course there are a few special subgroups that you always know are normal, like the center and the commutator.

If $$D_8/H$$ were cyclic of order $$4$$, then there would exist a coset $$gH$$ of order $$4$$, namely an element $$g\in D_8$$ such that $$g,g^2,g^3\notin H$$ and $$g^4\in H$$. So, there are two cases only to parse:

1. $$g^4=R_{2\pi}(=1)$$: if $$g$$ has order $$4$$, then $$g=R_{\pi/2}$$, but then $$g^2\in H$$: contradiction; if $$g$$ has order $$2$$, then again $$g^2\in H$$: contradiction;
2. $$g^4=R_\pi$$, then $$g$$ has order $$2$$, hence $$g^2\in H$$: contradiction.

Therefore, $$D_8/H\cong C_2^2$$.

If the homomorphism were injective, we'd have an isomorphism.

That leaves homomorphisms into groups of order $$1,2,$$ or $$4;$$ in particular abelian ones. Group cohomology, in particular the first and second, deal with these, but is not necessary here. More generally, it might be.

The commutator of $$D_8$$ is $$D_8'=\langle r^2\rangle ,$$ and since the quotients are abelian, it's contained in the kernel. As a result the quotient/image cannot have any elements of order $$4.$$

Quotienting by the commutator gets us $$C_2^2.$$

You can also quotient by $$C_4=\langle r\rangle,$$ since it's normal. Of course that gets us $$C_2.$$