# Is there a surjective homomorphism from $D_{10}$ to $\mathbb Z _2\times \mathbb Z_2$

Is there a homomorphism $f\colon D_{10} \to \mathbb Z_2\times \mathbb Z_2$ that is onto?

Attempt: $D_{10} = \{e,s,r,...,r^9,sr,...,sr^9\}$, $\mathbb Z_2\times \mathbb Z_2 = \{(0,0),(1,0),(0,1),(1,1)\}$. Since $(1,0) + (0,1) = (1,1)$ let $f\colon D_{10} \to \mathbb Z_2\times \mathbb Z_2$ be defined by $f(e) = (0,0)$, $f(r) = (1,0)$, $f(s) = (0,1)$ such that the rest of $D_{10}$ follows. Is there a way to check if this (or any other map) is a homomorphism other than trying every two combinations in $D_{10}$?

Usually the first isomorphism theorem can be helpful in such cases. It says if $\phi: G \longrightarrow H$ is an onto homomorphism then $G/\text{Ker}(\phi) \cong H$. Furthermore $\left|G/\text{Ker}(\phi)\right|=|G|/|\text{Ker}(\phi)|=|H|$.
In your question if there is an onto homomorphism from $G=D_{10}$ to $H=\mathbb{Z}_2 \times \mathbb{Z}_2$. Then The size of the kernel will be $|\text{Ker}(\phi)|=5$. Because of the prime order this must be cyclic. So you should look for cyclic subgroups of $D_{10}$ which are of order $5$. For example, $\langle r^2\rangle$. So now you know that $\phi(r^2)=(0,0)$. Hopefully you will be able to take it from here.
• Fun fact: one can show that a group admits a surjection to $\mathbb Z/2\mathbb Z \times\mathbb Z/2\mathbb Z$ if and only if it is a union of three proper subgroups. (In this case $D_{10}$ is a union of $C_{10}$ and two copies of $D_5$.) – Myself Nov 25 '15 at 20:02