Determine all homomorphic images of $D_4$ up to isomorphism.

Determine all homomorphic images of $D_4$ up to isomorphism.

What exactly is a homomorphic image without mentioning a second group? Isn't a homomorphism a map between two groups?

Solution 1:

Let's use the fact that homomorphic images of $D_4$ are quotients of $D_4$ by normal subgroups.

$D_4=\{e,r,r^2,r^3,s,sr,sr^2,sr^3\}$ has $4$ proper normal subgroups. $3$ have index $2$. It's pretty easy to see these are: $\{e,r,r^2,r^3\},\{e,sr,r^2,sr^3\}$ and $\{e,r^2,s,sr^2\}$. In this case the homomorphic image is $\mathbb Z_2$.

There is one index $4$ normal subgroup: $\{e,r^2\}$ and in this case the homomorphic image is the Klein $4$ group, $\mathbb Z_2×\mathbb Z_2$, because there are two (actually three) elements of order $2$ in the quotient.

Of course, there are also $\{e\}$ and $D_4$, trivially.

Solution 2:

Yes, you need a second group $G$ to have a homomorphism $\phi\colon D_4\to G$. The question asks only for the image of such a homomorphism, hence we may as well assume that $\phi$ is onto. So to rephrase the question:

Find all groups $G$ (up to isomorphism) such that there exists an onto homomorphism $D_4\to G$.

Solution 3:

Lots of the answers here talk about normal subgroups, so I'll contribute one without. For bigger cases, normal subgroups are a good way of approaching the problem, but for a simple case like this it's not too bad without.

Firstly, suppose $\phi: D_4 \to H$ is surjective. Then, $|H|$ must be a factor of $8$, so it must be $1,2,4,8$. Write $D_4 = \langle r,s \mid r^4 = s^2 = e, rsr = s^{-1} \rangle$.

• If $|H| = 8$, we have a surjection between sets of the same size, so $\phi$ is bijective. Then, looking at the images of the generators of $D_8$, we can see they must satisfy the same relations, so $H \cong D_8$.

• If $|H| = 4$, there are only two options for $H$ as there are only two groups of order $4$, namely $H \cong C_2 \times C_2$ or $C_4$. But, we must satisfy the relations of $D_4$, so $\phi(rsr)=\phi(s)$. Both candidates are abelian, so we must have $\phi(r^2)=e$. We know $\phi(r)$ is non trivial, otherwise there's not enough in the image, so must be of order 2 and similarly $\phi(s)$ is order 2. So, the only homomorphism that works is $$\phi(r) = (1,0)\in C_2 \times C_2\\\phi(s) = (0,1) \in C_2 \times C_2. $$ so both options are possible.

• If $|H| = 2$, we must have $H \cong C_2$. This time, there are two homomorphisms we could choose, I'll pick $\phi(s) = 1$, $\phi(r) = 0$. It's worth thinking about the other.

• Finally, if $|H|=1$, $H$ is the trivial group, which clearly works.

Again, normal subgroups are the cleaner approach, and you don't have to get your hands dirty as much. But, this direct approach may help understanding.