Homomorphism Group From $S_3$ to $\Bbb Z_5 \setminus \{0\}$

Question

Let $A=\{1,2,3\}$. Given that $$S_3=\{f:A \to A \mid \text{$f$ is a bijective function}\}$$ is a group under composition of function. Let $\Bbb Z_5^*=\{\bar{x} \in \Bbb Z \mid \bar{x} \ne \bar{0}\}$ is a group under multiplication modulo $5$. Define a map $P:S_3 \to \Bbb Z_5^*$ by $P(a)=\overline{a(1)}$ for all $a \in S_3$. Determine whether $P$ is a group homomorphism.

Attempt:

EDIT:

Consider $$a= \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix}, \qquad b= \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} \in S_3.$$ We obtain $$P(a \circ b)=\overline{(a \circ b)(1)}=\overline{a(b(1))}=\overline{a(2)}=\overline{1}\ne \overline{4}=\overline{2} \cdot \overline{2}=\overline{a(1)} \cdot \overline{b(1)}=P(a) \cdot P(b).$$ Hence, $P$ is not a group homomorphism.

Is it correct? Any ideas? Thanks in advanced.

Answer 1

Pretty good.

Or, what would the kernel be? $\rm ker(P)=\{a\in S_3\mid P(a)=1\}=\{a\in S_3\mid a(1)=1\}=\{e,(23)\}\cong S_2.$

But then $\lvert \rm im(P)\rvert =3\not\mid 4=\lvert \Bbb Z_5^*\rvert\rightarrow\leftarrow.$

Or, $S_2$ isn't normal in $S_3.$

Answer 2

Yes, that's correct.

Another idea. If $P$ is a homomorphism, then $o(P(a))\mid o(a)$ for every $a\in S_3$. Take $a=(12)$; then, $P(a)=\bar 2$; but $o(\bar 2)=4\nmid 2=o(a)$. So, $P$ is not a homomorphism from $S_3$ to $\Bbb Z_5^*$.