# Math Help - Homomorphisms of Symmetric Group of order 3

1. ## Homomorphisms of Symmetric Group of order 3

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

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

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

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

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

Right?

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

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

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

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

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

Right?
I want to help you but I am afraid to because this is a take home final exam. I am not sure if your professor would allow that even though you are not asking for answers. The best option is to sent your professor an e-mail and ask him.

I am planning to get exempt from abstract algebra course, so I do not want to help you out because I might end up taking the same exam you are taking!

3. Just so you know, it was a typo :-) So my analysis bases on the original problem was right.

4. Originally Posted by mpetnuch
Just so you know, it was a typo :-) So my analysis bases on the original problem was right.
Yes there was a mistake. I told my classmate in my other class that you were saying something about a mistake. And he said that there was and he showed me the same thing you were showing.

By the way: I cannot take the exam as a take home (like you) I need to take the real full exam.

Is you exam over? I want to show you an unbelievable theorem from group theory that was proven only in the 1980's. It is really beautiful.

5. I handed in my exam today. And although it was the "final" it is not the final exam. LOL. We have an in class exam on Tuesday just on Rings/Fields. The take home was comprehensive. BTW, if you are taking the same one I did, it is very easy.

6. Originally Posted by mpetnuch
We have an in class exam on Tuesday just on Rings/Fields.
But you never did anything on fields . Except the stuff on rings, factor rings, ideals, ideals of F[x]. Are you talking about that?

7. Um I don't know who told you we never did anything on fields but the exam on Tuesday is on chapters 5 and 6. I think you have the book, but anyway Chapter 5 is on rings, ideal, factor rings, etc... Chapter 6 is on fields, alegraic extensions. We did not do the construction section. However, although no homework was assigned from it, he basically covered most of the finite fields section as well (but left out and refused to mention the uniqness of field of same order up to isomorphism and that every field has order a product a primes - mostly because on the take home he had problems where you were supposed to figure out that there was only one field of order 4 and no fields of order 6 and he didn't want you to use these theorems).