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Math Help - Mathematical Induction

  1. #1
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    Mathematical Induction

    I have a couple of questions that I a unsure on. The first part have done but hopefully you can confirm I have done this right the rest I am not sure on. Can you point me in the right direction.

    Suppose G is a group with elements a and b or orders p and q resectively and p and q are distinct primes

    a) Show that (ab)^k=a(ab)^(k-1) b

    This is what I have. The only way I could figure to do this was to start from the RHS and prove that it equals the LHS.

    a(ab)^(k-1) b=a*a^(k-1) *b(k-1) *b
    =a*a^k *a^-1 *b^k*b^-1 *b
    =a*a^k *a^-1 *b^k e (as b^-1 b =e)
    =a*a^k *a^-1 *b^k

    Now since multiplication of integers is associative, aa^ka^-1=aa^-1a^k. Thus

    =a*a^-1*a^k *b^k
    =e*a^k *b^k (as aa^-1=e)
    =a^k*b^k
    =(ab)^k as required

    Have I done this correctly?

    b)Show that (ab)^k=e if and only if a^k=b^k=e.

    Now I know that I need to do mathematical induction here but I always have trouble with this. Can you give me a starting point or a similar example so I can follow what you are saying as you do it.

    c)Show that the order of ab is pq.

    I think I must be having a stupid day as to me if a and b have orders p and q respectively then obviously ab=pq and I am not sure how to use mathematical induction to get the answer.
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  2. #2
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    Quote Originally Posted by bex23 View Post
    I have a couple of questions that I a unsure on. The first part have done but hopefully you can confirm I have done this right the rest I am not sure on. Can you point me in the right direction.

    Suppose G is a group with elements a and b or orders p and q resectively and p and q are distinct primes

    a) Show that (ab)^k=a(ab)^(k-1) b
    Firstly observe that for a general group, (ab)^k = b^k a^k
    So I think your identity should read (ab)^k=b(ab)^{k-1}a, unless the problem mentioned that the group is commutative.

    Quote Originally Posted by bex23 View Post
    b)Show that (ab)^k=e if and only if a^k=b^k=e.

    Now I know that I need to do mathematical induction here but I always have trouble with this. Can you give me a starting point or a similar example so I can follow what you are saying as you do it.
    Have they asked you to use Induction? You can do without it...

    (ab)^{kp} = e \Rightarrow b^{kp} = e \Rightarrow q|kp
    \text{ But since p is a prime we have, } q|k \Rightarrow b^k = e

    You can do likewise for a.

    Quote Originally Posted by bex23 View Post
    c)Show that the order of ab is pq.

    I think I must be having a stupid day as to me if a and b have orders p and q respectively then obviously ab=pq and I am not sure how to use mathematical induction to get the answer.
    Let the order of (ab) be x. Now, since (ab)^{pq} = e, x|pq. But p and q are primes, the only divisors of pq are 1,p,q and pq. Its easy to see that if x=p, then e =(ab)^x = (ab)^p = b^pa^p = b^p \Rightarrow p|q...Contradiction.
    Similarly for x=q. Thus we are left with only pq.
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  3. #3
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    Quote Originally Posted by Isomorphism View Post
    Firstly observe that for a general group, (ab)^k = b^k a^k
    So I think your identity should read (ab)^k=b(ab)^{k-1}a, unless the problem mentioned that the group is commutative.



    Have they asked you to use Induction? You can do without it...

    (ab)^{kp} = e \Rightarrow b^{kp} = e \Rightarrow q|kp
    \text{ But since p is a prime we have, } q|k \Rightarrow b^k = e

    You can do likewise for a.
    How do you get (ab)^kp. I know that p is the order of a but how does it come to be in the power?

    Let the order of (ab) be x. Now, since (ab)^{pq} = e, x|pq. But p and q are primes, the only divisors of pq are 1,p,q and pq. Its easy to see that if x=p, then e =(ab)^x = (ab)^p = b^pa^p = b^p \Rightarrow p|q...Contradiction.
    Similarly for x=q. Thus we are left with only pq.
    Again, from what info do you deduce that (ab)^pq=e.

    Hopefully you can clear this up for me

    Thanx
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by bex23 View Post
    How do you get (ab)^kp. I know that p is the order of a but how does it come to be in the power?
    Starting point was:
    (ab)^k = e

    Raise both sides to the power of p.So that:
    (ab)^{kp} = e^p = e

    Quote Originally Posted by bex23 View Post
    Again, from what info do you deduce that (ab)^pq=e.
    Well we know that a^p = b^q = e.

    (ab)^{pq} = b^{pq}a^{pq} = (b^q)^p (a^p)^q = e^p e^q = e

    I have used the result (ab)^n = b^na^n.
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  5. #5
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    Thanks for all your help. I finally see where you are coming from.

    Bex
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