elementary Properties of cyclic subgroups

Shadow · **Posted:** Thu, 8 Mar 2012 21:01:53 UTC

daveyinaz wrote:

DgrayMan wrote:

If G=<a> and b is in G, prove the order of b is a factor of the order of a.

This is what ive tried coming up with.

I know if G={a0,a1,a2...a(n-1)}
then ord(a)=n

so for some k integer b=a^k

so ord(b)=n^k which is a factor of the order of a.

I understand how this is true but again its difficult for me trying to prove for all elements in a group.

I hope you didn't really mean to say the order of b = n^k

because this is not a factor of the order of

...

Yeah, he got his exponent rules messed up and forgot that order is the minimal thing.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

hmm. why wouldn't it be true? Since its not, how is it proven that <a>=?

Shadow · **Posted:** Fri, 9 Mar 2012 00:25:38 UTC

DgrayMan wrote:

hmm. why wouldn't it be true? Since its not, how is it proven that <a>=?

Well for one, $(a^k)^n=a^{nk}$ , and for another, I keep telling you there is no reason to believe <a>= for any choice of b, I mean what if b=e?

daveyinaz · **Joined:** Sat, 16 Aug 2008 04:47:19 UTC **Posts:** 208

DgrayMan wrote:

hmm. why wouldn't it be true? Since its not, how is it proven that <a>=?

You could always create an isomorphism between $\langle a \rangle$ and $\langle b \rangle$ .

Shadow · **Posted:** Fri, 9 Mar 2012 01:06:46 UTC

daveyinaz wrote:

DgrayMan wrote:

hmm. why wouldn't it be true? Since its not, how is it proven that <a>=?

You could always create an isomorphism between $\langle a \rangle$ and $\langle b \rangle$ .

If the order of b is not the order of a, then none such exists.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

Just wondering, cause my teacher said if they are subsets of each other then they are equal.

Shadow · **Posted:** Fri, 9 Mar 2012 03:59:26 UTC

DgrayMan wrote:

Just wondering, cause my teacher said if they are subsets of each other then they are equal.

Yes, that's true, but there's a difference between saying that and saying for an arbitrary b this is true. There are some b for which it is true, and some for which it is not.

It's of course true that if you can show $\langle a\rangle\subseteq\langla b\rangle$ then they're equal, but there was nothing in the original thing about that, you just had an arbitrary $b\in\langle a\rangle$ , and you were asking about the order of b, which is only marginally related to whether or not b is another generator of the group.

S.O.S. Mathematics CyberBoard

elementary Properties of cyclic subgroups

Who is online