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Examine if there is an epimorphism from the dihedral group \[D_4=\bigl\langle{\sigma,\tau\;|\; \sigma^4=\tau^2={\rm{id}}\,,\; \tau\sigma\tau=\sigma^3}\bigr\rangle\] onto the additive group \({\mathbb{Z}}_4\).

No there is not. Suppose \( \phi: D_4 \to \mathbb{Z}_4\) is a homomorphism. Then
\[ 3\phi(s) = \phi(s^3) = \phi(tst) = \phi(t) + \phi(s) + \phi(t) \] and so \(2\phi(s) = 2\phi(t).\) But since \(2\phi(s) = 2\phi(t) = \phi(t^2) = 0\), then \(\phi(s),\phi(t) \in \{0,2\}\). So \(\phi(D_4) \subseteq \{0,2\}\) and so \(\phi\) is not onto.

Edit: I accidentally solved the problem with the relation \(sts = s^3\) instead of the correct \(tst = s^3\). I corrected it now. [In the solution I use \(s\) in place of \(\sigma\) and \(t\) in place of \(\tau\).]

A 2nd solution:

Suppose that there exists an epimorphism \(f: D_4\longrightarrow {\mathbb{Z}}_4\,,\) where \[D_4=\bigl\langle{\sigma,\tau\;|\; \sigma^4=\tau^2={\rm{id}}\,,\; \tau\sigma\tau=\sigma^3}\bigr\rangle\,.\] Then by 1st Isomorphism theorem we have that \(D_4/{\ker{f}}\cong{\mathbb{Z}}_4\) and \(\ker{f}\) must be a normal subgroup of \(D_4\) of order\[\circ(\ker{f})=\frac{\circ(D_4)}{\circ({\mathbb{Z}}_4)}=\frac{8}{4}=2\,.\] But the only normal subgroup of order \(2\) is \(\langle{\sigma^2}\rangle\). So \[D_4/{\ker{f}}=D_4/{\langle{\sigma^2}\rangle}=\bigl\{{\langle{\sigma^2}\rangle\,,\,\langle{\sigma^2}\rangle\sigma\,,\,\langle{\sigma^2}\rangle\tau\,,\,\langle{\sigma^2}\rangle\sigma\tau}\bigr\}\,.\] The order of every non-zero element of \(D_4/{\langle{\sigma^2}\rangle}\) is \(2\). On the other hand, \({\mathbb{Z}}_4\) has an element of order \(4\). So \(D_4/{\ker{f}}\ncong{\mathbb{Z}}_4\) and, therefore, does not exists an epimorphism from the dihedral group \(D_4\) onto the additive group \({\mathbb{Z}}_4\).