# Finding all groups H (up to isomorphism) such that there is a surjective homomorphism from D8 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 $C_2 \times C_2$.