Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is it true that the only non-trivial (subgroups other than the trivial group and the group itself) subgroups of $\mathbb{Z} \times \mathbb{Z}_2$ are all isomorphic to $\mathbb{Z}$? My intuition tells me this is true, but is there any formal way to see this?

share|improve this question
    
$n\mathbb{Z}\times\mathbb{Z}_2$ –  yoyo Mar 21 '13 at 22:43
    
More in particular , the non-trivial subgroup $\,\{0\}\times \Bbb Z_2\cong \Bbb Z_2\ncong \Bbb Z\,$ ... –  DonAntonio Mar 21 '13 at 22:50
    
I assume $\mathbb{Z}_2$ is again the wrong notation for $\mathbb{Z}/2\mathbb{Z}$? –  Martin Brandenburg Mar 22 '13 at 20:50
add comment

3 Answers

Well you have $\Bbb{Z}_{2}$, of course, and more generally all subgroups of the form $n \Bbb{Z} \times \Bbb{Z}_{2}$, which are definitely not isomorphic to $\Bbb{Z}$.

share|improve this answer
add comment

There is an well known fact proposed by Kaplansky that says:

If $G$ is an infinite group which is isomorphic to every proper subgroup, then $G\cong \mathbb Z$.

If your group has this property so it should be $\mathbb Z$, but as you see via @Andreas's answer, it is wrong.

share|improve this answer
    
+1 for adding an important theorem to know! +1 just because! ;-) –  amWhy Mar 23 '13 at 0:42
add comment

The direct product, by definition, has homomorphisms to both arguments, and by construction, the kernel of that homomorphism consists of the elements of the product of all other arguments, with the identity used in the remaining index.

share|improve this answer
1  
I really don't understand: if by the first "arguments" you mean "the direct factors", then what are those "all other arguments"? –  DonAntonio Mar 21 '13 at 22:43
1  
The homomorphism to the factor has kernel the product of the other factors, with an extra "blank" index added, i.e. always holding the identity of the factor you want a morphism to. –  Loki Clock Mar 21 '13 at 22:44
1  
perhaps you have the very wrong impression that I'm a beginning mathematician instead of a professional one, but I really have no many doubts in this subject. I am asking directly and sincerely because I really don't understand why you believe your answer addresses the OP's question. Of course, I might be missing something, but I can't see what... The OP's asking about subgroups of a group which is a direct product. How your answer helps him? –  DonAntonio Mar 21 '13 at 23:21
1  
My impression is that you're trying to make a point. But you aren't. You're underestimating the OP's ability to Google and to ask me about real problems with the answer. –  Loki Clock Mar 21 '13 at 23:29
1  
Oh, my! Well, I suppose that's your opinion. Discussion closed. –  DonAntonio Mar 21 '13 at 23:32
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.