Thread: 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.

Originally Posted by bex23

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,
So I think your identity should read , unless the problem mentioned that the group is commutative.

Originally Posted by bex23

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...




You can do likewise for a.

Originally Posted by bex23

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 . 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 ...Contradiction.
Similarly for x=q. Thus we are left with only pq.

Originally Posted by Isomorphism

Firstly observe that for a general group,
So I think your identity should read , unless the problem mentioned that the group is commutative.



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




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 . 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 ...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

Originally Posted by bex23

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:


Raise both sides to the power of p.So that:

Originally Posted by bex23

Again, from what info do you deduce that (ab)^pq=e.

Well we know that .



I have used the result .

Thanks for all your help. I finally see where you are coming from.

Bex