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.

Let $G=(\mathbb{Z}_n,+)\,$ be a group where $\,n \geq 2$ and $d \neq 0$.

Find all possible values for $d$ and $n$ so that $\{0,d\}$ is a subgroup of $G$.

I know that for subset of $G$ to be a group there needs to be a closure under the addition operation.

So: $\,d+d = 2d\,$ belongs to $\,\{0,d\}$.

So $2d=0$.

But the problem is I'm not sure how to generalize this and to find all possible values for $\,d\,$ and $\,n\,$.

share|improve this question
    
Title is not well descriptive. –  B. S. Dec 30 '12 at 15:46
    
@BabakSorouh - For a first post I would say that its a good title, its very hard to give titles to questions at first –  Belgi Dec 30 '12 at 16:08
    
@Anna yes I learned Theorem of Lagrange –  Anna Dec 30 '12 at 16:13
add comment

3 Answers 3

up vote 5 down vote accepted

$(1)$ $\mathbb{Z}_{n}$ is cyclic and has a subgroup of order $m$ only if $m$ divides $n$ (by the theorem of Lagrange).

$(2)$ Any subgroup of a cyclic group is cyclic.

$(3)$ In your problem, any subgroup of the form $D\le G = \{0, d\}$ $d\ne 0$, $d \in \mathbb{Z}_n$ clearly has order $2$ and must be cyclic, so $D = \langle d \rangle$.

So yes, $d+d = 0$, and hence $d^{-1} = -d = d$, otherwise $|\langle d\rangle| = \{0, d\}\ne 2$.

By $(1)$, $|\langle d \rangle| = 2$ must divide $n.\;$ So $\;n = 2k\;$ where k is some positive integer. Hence $n$ must be even.

Now, what can you say about the possible element $d$ such that $d$ generates a subgroup of order $2$, given that $|\mathbb{Z}_{n}|$ is even $n = 2k$?

  • $|G| = 2:\;\;$ If $G = \mathbb{Z}_2$, $D = \{0, 1\} = G$.

  • $|G| > 2: \;\;k = d \;\; \implies\quad n = 2k = 2d \implies d = \dfrac{n}{2}.$

Try specific examples to get clear on what's going on here: What element in $\mathbb{Z}_4$, generates a subgroup of order $2$? Likewise, for $Z_{12}$ what is the element that generates a subgroup of order $2$?

share|improve this answer
    
Why aren't you using the notation with $+$ ? the group is abelian –  Belgi Dec 31 '12 at 12:38
    
@amWhy no thank you very much! –  Anna Dec 31 '12 at 17:57
    
You're welcome Anna! –  amWhy Dec 31 '12 at 18:01
add comment

$\mathbb{Z}_{n}$ is cyclic and have a subgroup of order $l$ $\iff l\mid n$. Moreover, this subgroup is unique.

In your case you want $l=2$ hence such a subgroup exist if and only if $n$ is even.

You also want $H=\langle d\rangle$ hence the order of $d$ is $?$ hence $d=?$

share|improve this answer
    
@amWhy - True. I think that Brian's answer also uses Lagrange when he deduced $n$ is even. –  Belgi Dec 30 '12 at 15:40
    
@amWhy - but now you got me curious on how to do this without Lagrange ;) –  Belgi Dec 30 '12 at 15:44
    
My guess is that this question is a precursor to introducing Lagrange...and hence, in that context, is more of a challenge. –  amWhy Dec 30 '12 at 15:49
    
@amWhy where Brian is using Lagrange? + still not sure about d... –  Anna Dec 30 '12 at 16:24
    
@Anna - How do you deduce $n$ is even without it ? –  Belgi Dec 30 '12 at 16:27
show 7 more comments

HINT: You know that $2d\equiv 0\pmod n$, so $n$ has to be even, and $d$ has to be ... what?

share|improve this answer
add comment

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.