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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?

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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.

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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 $D_4\to G$.

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  • $\begingroup$ What exactly does "up to isomorphism" mean if the question is rephrased in this way? Because if a finite set is onto another set doesn't that imply it's already a bijection, hence isomorphic? I see that $|\phi(D_4)| = 1, 2, 4, $ or $8$, so it would have to be sets filled with those number of elements. $\endgroup$ – Oliver G Aug 26 '16 at 12:55
  • $\begingroup$ No, an onto function between finite sets is not necessarily a bijection. For example, the function from $\{1,2\}$ to $\{1\}$ given by $f(x)=1$ is not injective. $\endgroup$ – Michael M Aug 26 '16 at 16:13
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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.

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  • $\begingroup$ The kernel of a map is normal and a normal subgroup of order $2$ is central (i.e., contained in the center). As $D_4$ has a center of order $2$, it has exactly one normal subgroup of order $2$. Hence it has exactly one quotient of order $4$, not two like you wrote. $\endgroup$ – j.p. Apr 16 '18 at 5:56
  • $\begingroup$ My mistake, I have edited. $\endgroup$ – B. Mehta Apr 16 '18 at 10:49
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Yes— and we're supposed to find all possible second groups. There are some groups $G$ that are the images of a homomorphism $g:D_4\rightarrow G$, and other groups $H$ where there isn't any homomorphism $h:D_4\rightarrow H$ at all. So, knowing about "the homomorphic images of $D_4$" tells you something about the nature of $D_4$ itself.

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$D_{4}$ had order 8. Its homomorphic images therefore have order dividing 8, and the ones of order 1 and 8 are trivial (the trivial group, and $D_{4}$ itself.

Let $D_{4}$ be presented by $<\sigma,\tau | e=\tau^{2}=\sigma^{4},\tau\sigma\tau=\sigma^{-1}>$. (You can think of $\sigma$ as the 90 degree counterclockwise rotation and $\tau$ as a reflection across the x-axis, for example.) Since reflections (the elements not in $<\sigma>$) are not central, they do not generate normal subgroups of order 2, so the only subgroup of order 2 is <\sigma^{2}>. Since all squares in $D_{4}$ are in <\sigma^{2}>, the quotient has all its elements of order 2 and so must be the Klein four-group V (aka $C_{2}XC_{2}$, aka $2^{2}$).

There are three normal subgroups of order four all must have the cyclic group of order 2 as the quotient, since that's the only group of order 2 = 8/4. However, it is important to note that one of these arises in a somewhat different way. Namely, the one with kernel $<\sigma>$ encodes in the homomorphic image whether the source symmetry was a reflection or a rotation. The other subgroups are $<\sigma^{2},\tau>$ and $<\sigma^{2},\sigma\tau>$ Since these are carried to each other by an automorphism of $D_{4}$, they are substantially the same thing. Thus, "up to automorphism" of the target group, there is only one quotient of order 2, but there are three homomorphisms to that target, two of which are the same up to relabeling $\tau$ as $\sigma\tau$, but the third of which is fundamentally different.

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A homomorphic image of $D_4$ is a group $G$ such that there is a surjective homomorphism $D_4 \to G$.

Every homomorphic image $G$ of $D_4$ is isomorphic to $D_4/N$, where $N$ is a normal subgroup of $D_4$. Therefore, the order of $G$ divides the order of $D_4$, which is $8$. Thus:

  • If the order of $G$ is $8$, then $G \cong D_4$.

  • If the order of $G$ is $4$, then $G \cong C_4$ or $G \cong C_2 \times C_2$.

  • If the order of $G$ is $2$, then $G \cong C_2$.

  • If the order of $G$ is $1$, then $G \cong C_1$.

These are the possibilities. You need to check whether they can be realized. This is where knowledge of the normal subgroups of $D_4$ enters. The only real work is when the order of $G$ is $4$ because $C_4$ is a normal subgroup of $D_4$ and the other two cases are trivial.

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