Welcome to mathimatikoi.org forum; Enjoy your visit here.

## Does exists an epimorphism?

Groups, Rings, Domains, Modules, etc, Galois theory
Grigorios Kostakos
Founder
Articles: 0
Posts: 460
Joined: Mon Nov 09, 2015 1:36 am
Location: Ioannina, Greece

### Does exists an epimorphism?

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 ${\mathbb{Z}}_4$.
Grigorios Kostakos

Tags:
Demetres
Former Team Member
Articles: 0
Posts: 77
Joined: Mon Nov 09, 2015 11:52 am
Location: Limassol/Pyla Cyprus
Contact:

### Re: Does exists an epimorphism?

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 $\tau$.]
Grigorios Kostakos
Founder
Articles: 0
Posts: 460
Joined: Mon Nov 09, 2015 1:36 am
Location: Ioannina, Greece

### Re: Does exists an epimorphism?

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 ${\mathbb{Z}}_4$.
Grigorios Kostakos