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 S3 Z2.
a) Z12 b) A4 c) D6 d) Z6 Z2
I eliminate option a because Z12 is cyclic whereas S3 Z2 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 S3 Z2 isomorphic to Z6 Z2 then we have S3 isomorphic to Z6, which is again a contradiction as Z6 is cyclic whereas S3 is not.
Is my argument right?
Also it would be great if I can get a head start with the other options too...
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 A4 should n't have an element of order 6, i realized that the order of any element of A4 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 D6, eh?