elementary Properties of cyclic subgroups

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

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.

Shadow · **Posted:** Thu, 8 Mar 2012 05:41:48 UTC

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.

Lagrange's theorem.

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

Great, the proof uses cosets and we haven't even discussed cosets! This assigned textbook is very backwards. But how does my interpretation hold since i don't really know what cosets are?

Shadow · **Posted:** Thu, 8 Mar 2012 06:00:38 UTC

DgrayMan wrote:

Great, the proof uses cosets and we haven't even discussed cosets! This assigned textbook is very backwards.

Have you not learned Lagrange yet? If so, then use the fact that b=a^k

for some k. Then you know that, if n is the order of a, then b^n=(a^k)^n=e

, so the order of b divides n.

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

No i havent heard it used in this course yet. The book we are using is " A book of absract algebra" 2nd edition by charles c. pinter. The chapter of cosets is two chapters away from where we are at.

Shadow · **Posted:** Thu, 8 Mar 2012 06:05:17 UTC

DgrayMan wrote:

No i havent heard it used in this course yet. The book we are using is " A book of absract algebra" 2nd edition by charles c. pinter. The chapter of cosets is two chapters away from where we are at.

Ah, then just use the other argument I said.

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

Do you recommend any books on abstract algebra that would be helpful?

Shadow · **Posted:** Thu, 8 Mar 2012 06:09:10 UTC

DgrayMan wrote:

Do you recommend any books on abstract algebra that would be helpful?

The book I learned out of was Herstein, but I can hardly recommend that one to you. I have seen good things from Rotman as well as Dummit and Foote though, maybe check those out, though from what I understand they're primarily geared at graduate level things, so they might not be as good either. My undergraduate location used Nicholson for the undergrads, so that's what I'd say you should check out. I haven't used it myself, but the university was always good to me, so I'll put my faith in it.

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

also, does this prove <a> is a subset of ?

Shadow · **Posted:** Thu, 8 Mar 2012 06:12:14 UTC

DgrayMan wrote:

also, does this prove <a> is a subset of ?

Hmm? No, that's not true is a subset of <a>, since is a subset of G which is equal to <a>.

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

ah i see, it would have to be a=b^k for <a> to be a subset of

Shadow · **Posted:** Thu, 8 Mar 2012 06:29:44 UTC

DgrayMan wrote:

ah i see, it would have to be a=b^k for <a> to be a subset of

yes.

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

Now by proving that <a> is a subset of and is a subset of <a>, does this mean <a>=?

Shadow · **Posted:** Thu, 8 Mar 2012 15:54:53 UTC

DgrayMan wrote:

Now by proving that <a> is a subset of and is a subset of <a>, does this mean <a>=?

If you can manage it, yes, but you realize that it probably isn't true, right?

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

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

...

S.O.S. Mathematics CyberBoard

elementary Properties of cyclic subgroups

Who is online