# Mathematical Induction

• May 12th 2008, 07:46 AM
bex23
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.
• May 12th 2008, 08:15 AM
Isomorphism
Quote:

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, $\displaystyle (ab)^k = b^k a^k$
So I think your identity should read $\displaystyle (ab)^k=b(ab)^{k-1}a$, unless the problem mentioned that the group is commutative.

Quote:

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...

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

You can do likewise for a.

Quote:

Originally Posted by bex23
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 $\displaystyle (ab)^{pq} = e, x|pq$. 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 $\displaystyle e =(ab)^x = (ab)^p = b^pa^p = b^p \Rightarrow p|q$...Contradiction.
Similarly for x=q. Thus we are left with only pq.
• May 13th 2008, 12:37 AM
bex23
Quote:

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

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

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

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?

Quote:

Let the order of (ab) be x. Now, since $\displaystyle (ab)^{pq} = e, x|pq$. 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 $\displaystyle e =(ab)^x = (ab)^p = b^pa^p = b^p \Rightarrow p|q$...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
• May 13th 2008, 01:31 AM
Isomorphism
Quote:

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:
$\displaystyle (ab)^k = e$

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

Quote:

Originally Posted by bex23
Again, from what info do you deduce that (ab)^pq=e.

Well we know that $\displaystyle a^p = b^q = e$.

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

I have used the result $\displaystyle (ab)^n = b^na^n$.
• May 14th 2008, 06:39 AM
bex23
Thanks for all your help. I finally see where you are coming from. (Clapping)

Bex