Recent content by xcr

Gotcha. And I could show that abk=b2ka by doing aba-1=b2 and then raising each side to the k gives abka-1=b2k. Move the a over and get abk=b2ka. Thank you so much Deveno and I like Serena. Hope ya'll both have a great holiday season.

Yea that was a typo, it should have been 5. So does that prove that the order of b is 31 or is there more that needs to be done?

Ok, so far I have that aba-1=b2, rewritten as b-1ab=ba. Raising both sides to the 5th power gives us e=(ba)5, therefore the order of ba is 5 or m where m|5. so m=1,2,3,4. m can't be 2,3,4 because they don't divide m. 1|5 but then we have ba=e, where ba=b-1ab. So b-1ab=e, giving us ab=b, which...

If you multiply both sides by a4 on the right, you get b=a4, so I see now that there is a possibility that ba=e where a and b do not equal e. But if you do b=am, then b5=(am)5=a5m=(a5)m=em=e. Not sure if this is correct but I'd say that this is a contradiction because 5 does not divide 31 and we...

Ok, so ba≠e because then a must equal e and b must equal e, which cannot be true because of our assumptions

I thought that since 2,3, and 4 did not divide 5 that the could not be the order?

Well if (ba)=e, then (ba)2=(ba)(ba)=ee=e, which is not true. Therefore ba≠e

I would just need to show that (ab)m, where m=1,2,3,4 does not equal e. So just show that (ab)1≠e, (ab)2=a(ba)b≠e, (ab)3=a(ba)2b≠e, (ab)4=a(ba)3b≠e.

ok so I took e=(ba)5=b(ab)4a=b(ab)(ab)(ab)(ab)a. I then took ab=b2a and substituted it in. After doing this REPEATEDLY I moved all the b's to the left and all the a's to the right. I ended up with e=b31a5, but a5=e, so e=b31e, or e=b31. Therefore the order of b is 31. Felt like I was doing the...

ab=b2a b-1ab=ba (b-1ab)5=(ba)5 (b-1a5b)=(ba)5 b-1eb=(ba)5 b-1b=(ba)5 e=(ba)5 Therefore |a|=|ba| As for proving that (ba)m≠e for m<5, do I just show that when m=1,2,3,4, you get b-1amb=(ba)m, which ≠e? Then do I just show that (aba-1)n=(b2)n, which can be rearranged as b-nabn=bna...

Homework Statement Let f:R→S be a homomorphism of rings. If J is an ideal in S and I={r∈R/f(r)∈J}, prove that I is an ideal in R that contains the kernal of f. Homework Equations The Attempt at a Solution I feel like I have the problem right, but would like to have someone look...

I understand how you got to this point: abn = b2na. The steps to get there go like this: aba-1=b2 so raising both sides to the nth power would look like (aba-1)n=(b2)n. On the left side we would have n terms of aba-1, but regrouping them would cause then the aa-1 terms to cancel. yielding...

Homework Statement Let G be a group with identity e. Let a and b be elements of G with a≠e, b≠e, (a^5)=e, and (aba^-1)=b^2. If b≠e, find the order of b. Homework Equations Maybe the statement if |a|=n and (a^m)=e, then n|m. Other ways of writing (aba^-1)=b^2: ab=(b^2)a...

Ok, I think I got it now. I got the quotient is (4x^2)-3x and the remainder is 2x+1. I multiplied the quotient and (2x^2)+x+1 and added the remainder and it came to the original polynomial. Thanks for checking over my work.

Homework Statement Apply the division algorithm for polynomials to find the quotient and remainder when (x^4)-(2x^3)+(x^2)-x+1 is divided by (2x^2)+x+1 in Z7. Homework Equations The Attempt at a Solution I worked the problem and got that the quotient was (4x^2)-3x-1 and the...