Constructing a surjective morphism $D_4 \to \Bbb Z_2$

Question

I'm given two groups:

• $D_{4, \circ}$, which is the dihedral group of order $8$, consisting of the elements $1, a, a^2, a^3, b, ab, a^2b, a^3b$; here, $a$ means a rotation and $b$ a reflection
• the group $\mathbb{Z}_{2, +} = \left\{[0]_2, [1]_2 \right\}$

I now need to construct a surjective morphism $$ g : D_{4, \circ} \rightarrow \mathbb{Z}_{2,+}. $$ I was wondering what is the general strategy for doing this. I have been doing this with trial and error, for example I tried to the mappings $$ g(1) = g(a^3) = g(a^2 b) = [0]_2 $$ and $$ g(a) = g(b) = g(ab) = [1]_2, $$ but then I see that $$ g(x \circ y ) \neq g(x) + g(y) $$ and hence $g$ is not a morphism under this mapping. So is there some better method to quickly identify this surjective morphism?

Answer 1

As Tobias hints, there's a general and important feature of dihedral groups that one should have in mind that leads quickly to an answer.

In case one doesn't see it immediately, here's a useful trick that applies more generally and only requires knowing a little about quotients. Given any group homomorphism $\phi: G \to H$ into an abelian group $H$, we have for any $a, b \in G$ that $$\phi(aba^{-1}a^{-1}) = \phi(a) \phi(b) \phi(a)^{-1} \phi(b)^{-1} = \phi(a) \phi(a)^{-1} \phi(b) \phi(b)^{-1} = e_H ,$$ so $\phi$ restricts to the identity on the commutator subgroup $[G, G] = \{a b a^{-1} b^{-1} : a, b \in G\}$, which one can show is normal. Thus, $\phi$ descends to a group homomorphism $\tilde \phi : G / [G, G] \to H$.

In our case, $[D_4, D_4] = \langle a^2 \rangle \cong \Bbb Z_2$ and so $D_4 / [D_4, D_4] = \langle \bar a \rangle \times \langle \bar b \rangle \cong \Bbb Z_2 \times \Bbb Z_2$, where $\bar a$ and $\bar b$ are the respective images of $a$ and $b$ under the quotient map $\pi : D_4 \to D_4 / [D_4, D_4]$. Now, we can pick any surjective homomorphism $\alpha : \Bbb Z_2 \times \Bbb Z_2 \to \Bbb Z_2$ (there are three of these), so that the map $\alpha \circ \pi$ is a surjective homomorphism $D_4 \to \Bbb Z_2$. (In fact, by construction, these three maps are all of the possible answers.)

One choice for $\alpha$ is defined by $\alpha(\bar a) = 0, \alpha(\bar b) = 1$, which corresponds to the homomorphism $\phi: D_4 \to \Bbb Z_2$ defined by $\phi(a) = 0, \phi(b) = 1$, and this is the map Tobias hinted at. If we recall that $D_n$ acts as the symmetry group of a regular $n$-gon, we can interpret this map geometrically: $a$ acts by some rotation of order $n$, and $b$ is a reflection, so that every element acts by a rotation or a reflection. In particular, our map $\phi: D_4 \mapsto \Bbb Z_2$ is the one that maps rotations to $0 \in \Bbb Z_2$ and reflections to $1 \in \Bbb Z_2$.

Answer 2

Hint: Define $g:D_4\rightarrow \Bbb{Z}_2$ by $g(x)=0$ if $x \in \{1,a,a^2,a^3\}$ and $g(x)=1$ otherwise. I'll leave it up to you to show that $g$ is the desired homomorphism.

Answer 3

Yes. Note that what $a$ generates in $D_4$ is a normal subgroup of $D_4$ (let's call it $B$), so $D_4/B \cong \mathbb{Z}_2$. Now we have a surjective homomorphism from $D_4$ to $D_4/B$ (the map that takes $x$ to the class of $x$ in $D_4/B$), so by composition it is not difficult to see what a surjective homomorphism from $D_4$ to $\mathbb{Z}_2$ should be.