# If f is a group homomorphism from S_3 into ZZ_6, then what is the order of f(S_3) ?

Sep 9, 2017

$1$ or $2$

#### Explanation:

Since ${\mathbb{Z}}_{6}$ contains an element of order $6$, but ${S}_{3}$ does not, there is no group isomorphism. So the image of ${S}_{3}$ in ${Z}_{6}$ is of order strictly dividing $6$.

Consider the image of the transposition $\left(\left(1 , 2\right)\right)$ under $f$. It must either be an element of order $1$ or of order $2$. It cannot have order $3$.

So we either have $f \left(\left(\left(1 , 2\right)\right)\right) = \hat{0}$ or $f \left(\left(\left(1 , 2\right)\right)\right) = \hat{3}$

Note that if $\left(\left(m , n\right)\right) \in {S}_{3}$ then there is some permutation $p$ in ${S}_{3}$ such that: ${p}^{- 1} \left(\left(1 , 2\right)\right) p = \left(\left(m , n\right)\right)$. Hence:

• If $f \left(\left(\left(1 , 2\right)\right)\right) = \hat{0}$ then $f \left(\left(\left(m , n\right)\right)\right) = \hat{0}$ for all transpositions in ${S}_{3}$ and hence $f \left({S}_{3}\right) = \left\{\hat{0}\right\}$

• If $f \left(\left(\left(1 , 2\right)\right)\right) = \hat{3}$ then $f \left(\left(\left(m , n\right)\right)\right) = \hat{3}$ for all transpositions in ${S}_{3}$ and hence $f \left({S}_{3}\right) = \left\{\hat{0} , \hat{3}\right\}$