# Isomorphism of Direct product of groups

• Apr 15th 2013, 11:22 AM
MAX09
Isomorphism of Direct product of groups
find which of the following groups is isomorphic to S3 $\displaystyle \bigoplus$ Z2.

a) Z12 b) A4 c) D6 d) Z6 $\displaystyle \bigoplus$ Z2

I eliminate option a because Z12 is cyclic whereas S3 $\displaystyle \bigoplus$ 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 $\displaystyle \bigoplus$ Z2 isomorphic to Z6 $\displaystyle \bigoplus$ 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...

Thanks
• Apr 15th 2013, 11:38 PM
Gusbob
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.
• Apr 16th 2013, 11:20 PM
MAX09
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.
• Apr 17th 2013, 01:02 AM
Gusbob
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?
• Apr 17th 2013, 01:21 AM
xixi
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$\displaystyle D_{12}$.
• Apr 17th 2013, 11:06 PM
MAX09
Re: Isomorphism of Direct product of groups
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?
• Apr 18th 2013, 02:29 AM
xixi
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.