Isomorphism of Direct product of groups

find which of the following groups is isomorphic to S_{3} $\displaystyle \bigoplus $ Z_{2}.

a) Z_{12} b) A_{4} c) D_{6} d) Z_{6} $\displaystyle \bigoplus $ Z_{2}

I eliminate option a because Z12 is cyclic whereas S_{3} $\displaystyle \bigoplus $ Z_{2} is not because we know that the External direct product of G and H is cyclic if and only if the orders of G and H are relatively prime. Here it's not the case.

Here's my question. Can I eliminate option d using the following argument?

If S_{3} $\displaystyle \bigoplus $ Z_{2 }isomorphic to Z_{6} $\displaystyle \bigoplus $ Z_{2} then we have S_{3} isomorphic to Z_{6}, which is again a contradiction as Z_{6} is cyclic whereas S_{3} is not.

Is my argument right?

Also it would be great if I can get a head start with the other options too...

Thanks

Re: Isomorphism of Direct product of groups

An easier argument for d) is that it is abelian, whereas your original group is not. I don't think your original argument is justified.

Re: Isomorphism of Direct product of groups

Hi Gusbob,

I have found justification for my claim, yet I agree with you that your argument that the property of "being abelian" is a more convincing and more elegant solution. Any ideas about the other two options ... I just need to eliminate one more to arrive at the answer.

Re: Isomorphism of Direct product of groups

The most obvious hint is a giveaway, but I can't think of anything else at an elementary level short of writing an explicit isomorphism to the correct answer.

$\displaystyle S^3$ is a subgroup of $\displaystyle S^3\times Z_2$. Can you realise $\displaystyle S^3$ as a subgroup of either of your two remaining options?

Re: Isomorphism of Direct product of groups

$\displaystyle S_{3}\oplus Z_{2}$ is not isomorphic to $\displaystyle A_{4}$ because the element ((123),1) has order 6 while $\displaystyle A_{4}$ doesn't have any element of order 6. Actually $\displaystyle S_{3}\oplus Z_{2}$ is isomorphic to the dihedral group$\displaystyle D_{12}$.

Re: Isomorphism of Direct product of groups

xixi, your justification was very elegant. it took a while to strike me as to why A_{4} should n't have an element of order 6, i realized that the order of any element of A_{4 }is got to be the lcm of the cycles into which it can be split, which can never exceed 4 cos splitting 4 letters can only be done with at most 4 parts or lesser.

still i guess you mean to say that the answer is D_{6}, eh?

Re: Isomorphism of Direct product of groups

Yes, $\displaystyle S_{3} \oplus Z_{2}$ is isomorphic to $\displaystyle D_{12}$ (It is the dihedral group of order twelve) which though denoted $\displaystyle D_{6}$ in an alternate convention.In other words, it is the dihedral group of degree six.