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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 $\,n\,$.

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Title is not well descriptive. – Babak 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
up vote 4 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 $2$?

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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

$\mathbb{Z}_{n}$ 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 $d=?$

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@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

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

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