1. Find every homomorphism from $C_{10}$ group to $S_3$.
  2. Find every homomorphism from $S_3$ to $C_{6}$

I know that : $S_3$ is permutation group of $3!=6$ elements, and

$C_{10} = \{cos(\frac{2k\pi}{10}) + isin(\frac{2k\pi}{10}) : k = 0,1,....,n-1 \}$

$\forall_{x,y \in S_3} f(x \cdot y) = f(x) \odot f(y) $

I'have solved several problems like find homomorphism from $\mathbb{Z_n} \rightarrow \mathbb{Z_m}$.

Could you give me some tips?


A hint for #1:

Recall the isomorphism theorems. Suppose we have a homomorphism $\phi:C_{10} \rightarrow S_3$. Then:

$$C_{10}/\ker(\phi) \cong \operatorname{Im}(\phi)$$

In particular, the cardinality of the factor group on the left must match the cardinality of the image on the right. Further, it is a fact that $\operatorname{Im}(\phi)$ is a subgroup of $S_3$. Notice that Lagrange's theorem greatly restricts the number of possible homomorphisms!

A hint for #2:

Recall that, for any group homomorphism $\phi:G_1 \rightarrow G_2$, $\ker(\phi)$ is a normal subgroup of $G_1$. What are the normal subgroups of $S_3$?

  • $\begingroup$ I think that : $o(C_{10})=10$ and $o(S_3)$ = 6, co $LCM=(10,6)=2$ and Let $|\frac{C_{10}}{kerf}|=2$ and $C_{10}=\langle x\rangle$. Then one can see that $f_1(x)=(1,2)$, $f_2=(2,3)$ and $f_3=(1,3)$ are homomorphisms. ( because $(1,2),(2,3),(1,3)$ are elements of order = 2 in $S_3$) and $id$ ofc $\endgroup$ – MatJ Nov 18 '14 at 13:10
  • $\begingroup$ My mistake : $GCD$ I meant $\endgroup$ – MatJ Nov 18 '14 at 13:25

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