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.
find which of the following groups is isomorphic to S_{3} Z_{2}.
a) Z_{12} b) A_{4} c) D_{6} d) Z_{6} Z_{2}
I eliminate option a because Z12 is cyclic whereas S_{3} 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} Z_{2 }isomorphic to Z_{6} 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
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.
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.
is a subgroup of . Can you realise as a subgroup of either of your two remaining options?
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?