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 Post subject: On group theory 4Posted: Mon Nov 09, 2015 3:38 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Find all the functions $\displaystyle{f:\left(\mathbb{Z}_{6},+\right)\longrightarrow \left(S_{3},\circ\right)}$ having the property $\displaystyle{f(x+y)=f(x)\circ f(y)\,,\forall\,x\,,y\in\mathbb{Z}_{6}\,\,\,\,\,\,}$ (homomorphism) .

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 Post subject: Re: On group theory 4Posted: Sun Jun 26, 2016 4:33 pm
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
Every element of symmetric group ${\cal{S}}_{3}=\big\langle{\rho,\,\tau \; \big| \ \;\rho^{3}=\tau^2={\rm{id}}, \ \tau\,\rho=\rho^2\,\tau}\big\rangle$ has order either $1$, or $2$, or $3$. Let $f:{\mathbb{Z}}_{6}\longrightarrow{\cal{S}}_{3}$ be a group homomorphism. Then must hold
\begin{align*}
&\left[\begin{array}{r}
f([1]_6)={\rm{id}}\; \veebar \;f([1]_6)=\rho\; \veebar \;f([1]_6)=\rho^2\; \veebar\\
f([1]_6)=\tau\; \veebar \;f([1]_6)=\rho\,\tau\; \veebar \;f([1]_6)=\rho^2\,\tau
\end{array}\right]\,.
\end{align*} In each case, demanding for every $[k]_6\in\mathbb{Z}_6\,,\; k\in\{0,1,\ldots,5\}$ that $f\big([k]_6\big)=k\,f([1]_6)$ we have six different homomorphisms, of which none is an epimorphism nor monomorphism. These homomorphisms are all the homomorphisms from ${\mathbb{Z}}_{6}$ to ${\cal{S}}_{3}$.

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Grigorios Kostakos

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