# A group of order 6 that has no element of order 6 is isomorphic to S_3

#### Samuelb88

1. The problem statement, all variables and given/known data
Suppose $G$ contains an element of order 3, but none of order 6. Show $G$ is isomorphic to $S_3$.

2. Relevant equations
I am not allowed to use Sylow's theorems, or quotient groups.

3. The attempt at a solution
I've established that $G$ contains a subgroup $H$ of order 3, and three other elements of order 2. I know that $H$ is normal in $G$, while the subgroups generated by elements of order 2 are not. I also know that $G$ permutes the elements of order 2 by conjugation, i.e. if $y \in G$ is of order 2, and $g \in G$, then $gyg^{-1}$ will always have order 2.

I want to claim that that $G$ permutes the elements of order 3 by conjugation as well, but I am not sure if this is true, and if it is, how to prove it.

My idea is that I can somehow establish that if $G$ permutes the elements of order 3 by conjugation as well, then I can begin to construct an isomorphism. Unfortunately, this is just guess work, and at my point in my algebra career, I don't see how to do this and I don't have any idea how to even construct an isomorphism other than write out a multiplication table and brute force it.

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#### Samuelb88

Can an element of G which is not the identity create an identity permutation by acting on it?
Sorry, I don't understand your question. What are you referring to it? Are you asking me if a nontrivial element of G can create a identity permutation by conjugation by that element of an element in H? #### Dick

Science Advisor
Homework Helper
Sorry, I don't understand your question. What are you referring to it? Are you asking me if a nontrivial element of G can create a identity permutation by conjugation by that element of an element in H? Sorry about my initial confusing posts. I'm asking you if an element g of G having order 2 can create an identity permutation on H by conjugation. If it doesn't, which permutation must it create? G can't create a representation of S3 by permuting H by conjugation. geg^(-1)=e. I'd stick with your original program of permuting the elements of order 2. If H={e,h,h^2} then the elements of order 2 are {g,gh,gh^2}, right? This is probably awfully close to simply creating a multiplication table.

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#### Samuelb88

If H={e,h,h^2} then the elements of order 2 are {g,gh,gh^2}, right?
Right! I've gotten that far, tho by different means. I guess I'm hung up on how to show the two groups are isomorphic. I suppose I could just map elements to each other in such a way such that the homomorphism property is satisfied. I think I could construct a homomorphism that would clearly be bijective and thus get my isomorphism.

#### Dick

Science Advisor
Homework Helper
Right! I've gotten that far, tho by different means. I guess I'm hung up on how to show the two groups are isomorphic. I suppose I could just map elements to each other in such a way such that the homomorphism property is satisfied. I think I could construct a homomorphism that would clearly be bijective and thus get my isomorphism.
I liked your idea of showing the group acting on the order 2 elements O={g,gh,gh^(-1)} by conjugation generates all permutations of them. Two transpositions can generate all of S3. Just show you get two such transpositions. Which elements of the group are likely to create transpositions of elements of O?

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