Originally Posted by

**mpetnuch** Hello everyone. I received a take-home exam for my algebra class. I was doing the problems and I think I found a typo in the exam. The question is as follows:

Let $\displaystyle \phi : S_3 \to \mathbb{Z}_{3003}$ be a homomorphism and show $\displaystyle |\phi(S_3)| =$ 1, 2, or 3.

Now I though that $\displaystyle Im(\phi)$ is a subgroup of $\displaystyle \mathbb{Z}_{3003}$ and hence, by Lagrange theorem, must divide 3003.

Now $\displaystyle \phi(S_3)$ can have at most 6 elements (i.e. the order of $\displaystyle S_3$). But the only numbers less than or equal to 6 that divide 3003 are 1 and 3.

Also, if $\displaystyle |\phi(S_3)| = 3$ implies $\displaystyle \phi$ maps two elements to the identity of $\displaystyle \mathbb{Z}_{3003}$. This implies that the kernel has order 2. But the kernel of a homomorphism is a normal subgroup, and $\displaystyle S_3$ has no normal subgroup of order 2.

So $\displaystyle |\phi(S_3)| = 1$ is the only possibility???

Right?