Let $f: D_4 \rightarrow C_{24}$ be a homomorphism. Show that for all $a \in D_4$, the following is true $f(a)^2 = e$. Let $f: D_4 \rightarrow C_{24}$ be a homomorphism. Show that for all $a \in D_4$, the following is true $f(a)^2 = e$.


What I thought we could do, was write out $D_4$, as it has just 8 elements.
$D_4 = \{1,\rho,\rho^2,\rho^3,\sigma,\sigma\rho,\sigma\rho^2,\sigma\rho^3 \}$ with $\sigma$ being the reflection, having order 2 and $\rho$ being the rotation, having order 4. Then we could just show for every element here the condition $f(a)^2 = e$ holds. As we know that $f(a)^2 = f(a^2)$, we know that the elements with an order that divides 2 out of $D_4$, work. That are $1$, $\rho^2$, $\sigma$ and $\sigma\rho^2$. We still need to show that the condition holds for the other elements of $D_4$. 

That is where I don't know what to do anymore. Is what I have done correct? If so, how should I continue? Are there better ways of tackling such problems? 

Thanks for reading,

K.
Let $\varphi:D_4\to C_{24}$ be an arbitrary homomorphism. Let $a,b\in C_{24}$ be such that $\rho\mapsto a$ and $\sigma\mapsto b$. Then we have two relations: $bab=a^{-1}$ and $b^2=0$. Since $C_{24}$ is commutative, these relations implies that $a=a^{-1}$, or that $a^2=0$. Hence, we have that $a^2=b^2=0$, so $\varphi(x)^2=0$ for any possible $x\in D_4$. 

My original, incorrect answer. Thanks to Zoe H for pointing out where I went wrong.


  I don't believe this is correct. Define a homomorphism $\phi: D_4\to\mathbb{Z}/24\mathbb{Z}$ given by $\rho\mapsto 6$ and $\sigma\mapsto 0$. Then $\phi(\rho)^2 = 12\ne 0$.
Then we could just show for every element here the condition $f(a)^2 = e$ holds. As we know that $f(a)^2 = f(a^2)$, we know that the elements with an order that divides 2 out of $D_4$, work. That are $1$, $\rho^2$, $\sigma$ and $\sigma\rho^2$.


That's true, and a nice idea! But, it turns out that the only elements whose orders don't divide $2$ are $\rho$ and $\rho^3$: products of a single rotation and a single reflection (like  $\sigma \rho$) are reflections, and have order $2$.

Thus, to continue with this approach, you can focus on showing that $a^2 \in \ker f$ for $a \in \{\rho, \rho^3\}$ and any homomorphism $f\colon D_4 \to C_{24}$. You can see that $(\rho)^2 = \rho^2 = (\rho^3)^2$, so you just need to figure out why $\rho^2 \in \ker f$ for any $f$.

One approach to do that: It turns out that the commutator subgroup of $D_4$ is $\langle \rho^2 \rangle$ (although it suffices here to know only that $\rho^2$ is in the commutator subgroup). Assuming you've learned the theorem that "$G/\ker f$ is abelian if and only if the commutator subgroup of $G$ is contained in $\ker f$", look at the possibilities for $G/\ker f$. The only possible sizes for $G/\ker f$ here are $1, 2$, and $4$ (why not $8$?), and all groups of these orders are abelian.



An alternative approach is simply to look at the possible sizes for normal subgroups of $D_4$ (i.e., possibilities for $\ker f$) and use the first isomorphism theorem to think about $f(D_4) \cong D_4 / \ker f$.

As above, the only possibilities for $|f(D_4)| = |D_4 / \ker f|$ are $1, 2,$ or $4$.

Certainly there's nothing to worry about if $|f(D_4)| \in \{1,2\}$.

The only normal subgroup of $D_4$ having order $2$ is $\langle \rho^2 \rangle$ (the only other subgroups of order $2$ are those generated by a single reflection, and these are never normal), which happens to be the center $\Bbb Z(D_4)$ of $D_4$. Probably you've learned that $G / \Bbb Z(G)$ can never be cyclic, implying that $D_4 / \langle \rho^2 \rangle$, having order $4$ and not being cyclic, must be isomorphic to $C_2 \times C_2$, where every element has order dividing $2$.