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This is part of practice midterm that I have been given (our prof doesn't post any solutions to it) I'd like to know whats right before I write the midterm on Monday this was actually a 4 part question, I'm posting just 1 piece as a question in its own cause it was much too long.

Let $G$ be an abelian group.

Suppose that $a$ is in $G$ and has order $m$ (such that $m$ is finite) and that the positive integer $k$ divides $m$.

(ii) Let $a$ be in $G$ and $l$ be a positive integer. State the theorem which says $\langle a^l\rangle=\langle a^k\rangle$ for some $k$ which divides $m$.

I have a theorem in my textbook that says as follows.

Let G be a finite cyclic group of order n with $a \in G$ as a generator for any integer m, the subgroup generated by $a^{m}$ is the same as the subgroup generated by $ a^{d}$ where $d=(m,n)$.
if this is the correct theorem can someone please show me why? thanks

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3 Answers 3

up vote 1 down vote accepted

I'll prove a more general of fact about cyclic groups, which would be good to know. Every subgroup of a cyclic group $\langle a\rangle$ of order $m$ is cyclic of order $l$ dividing $m$. For each $l$ dividing $m$, there is exactly one subgroup of order $l$: it is $\langle a^{m/l}\rangle$. The question ii) follows immediately from that.

Since $\langle a\rangle$ is isomorphic to $\mathbb{Z}_m$ via $n\longmapsto a^n$, it suffices to work in $\mathbb{Z}_m$.

Let $H$ be a subgroup of $\mathbb{Z}_m$. We can assume $H\neq\{0\}$, and we consider its smallest positive element $k$ in $\{0,\ldots,n-1\}$.

By Euclidean division, every $h\in H$ can be written $h=qk+r$ with $0\leq r<k$. Now note that $qk$ belongs to $H$, so $r=h-qh$ lies in $H$. By minimality of $k$, we must have $r=0$. So $h=qk$. This proves that $H\subseteq \langle k\rangle $. The converse is obvious.

So $H$ is cyclic, generated by $k$, its smallest positive element. If $l$ denotes the order of $H$, i.e. the order of $k$, we must have $$H=\{0,k,2k,\ldots,(l-1)k\}\qquad\mbox{and}\qquad kl=m.$$

It is now clear that for each $l$ dividing $m$, there is exactly one subgroup of order $l$. And it is cyclic. This finishes to prove the claims above.

Finally, let us prove the claim you stated in this context: $H=\langle j\rangle$ and consider $k=\gcd(j,m)$. Since $k$ divides $j$, we clearly have $H=\langle j\rangle\subseteq \langle k\rangle$. Now assume $kn$ belongs to the latter. By Bezout, there exist $u,v$ such that $k=uj+vm$. So $kn=ujn+vmn\equiv ujn [m]$. Hence $kn$ belongs to $H=\langle j\rangle$. So we do have $H=\langle j\rangle=\langle k\rangle=\langle \gcd(j,m) \rangle$.

Applying the above to some $H=\langle a^j\rangle$ yields two ways of getting $k$ dividing $m$ such that $H=\langle a^k\rangle$. The second is more explicit since we take $k=\gcd(j,m)$.

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For the theorem in your textbook, note that $(a^d)^{m/d}=a^m$ and $(a^m)^{b}=a^d$ where $b$ is the integer that $bm+cn=d$, which you can find using Euclidean Algorithm.

This means they can generate each other. Therefore, the groups they generate are the same.

You can use this to prove the question.

The question says $\forall l>0, \exists k|m,\langle a^l\rangle=\langle a^k\rangle$. Using the theorem, we know that $\langle a^l\rangle=\langle a^d\rangle$ where $d=gcd(l, n)$. Since $a$ is of order $m$, $d$ must divides $m$. Therefore, pick $d$ as your $k$.

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Well i feel like a bit of a fool i just couldn't read what the question was even asking thats why it made no sense tyvm. –  Faust7 Mar 17 '13 at 5:27

Yes you can apply the theorem you cite immediately with $(\langle a\rangle,m,l)$ repectively in the roles of $(G,n,m)$ (here $\langle a\rangle$ is the cyclic subgroup generated by $a$); the desired value for $k$ is the value $d$ of the theorem, which is, in view of the name changes, $k=\gcd(m,l)$.

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