Thread: Induction Proofs

Prove by induction:

1. If g and h are commuting elements of G, prove g commutes with every positive integer power of h: ghⁿ= hⁿg for all positive integers n.Then prove g^mhⁿ = hⁿg^m for all positive integers m and n. Is this true when m and n are arbitrary integers?

2. If g and h are commuting elements of G, prove (gh)ⁿ = gⁿhⁿ for all positive integers n.


Here's my attempt at solving 1, I'm just not sure if it's correct, or if more needs to be proved.

ghⁿ = hⁿg

Base step: let n = 1
gh = hg

Assume gh^k = h^kg

gh^k+1= h^k+1g
gh^kh = h^khg
gh^k = h^kg

Is that done correctly, and if so would it end the proof? And can the others be proved in the same way?

The first line after your "Assume g(h^k)=(h^k)g" was unjustified. You hadn't yet given a reason for asserting that g(h^(k+1))=(h^(k+1))g. In fact, proving that statement is the goal of the induction.
Prop1: Let g,h in group G such that gh = hg. Then for all integers n>=1, g(h^n)=(h^n)g.
Proof: By induction. Define Statement(n) = "g(h^n)=(h^n)g"
Statement(1) is true, since gh=hg is given.
Assume Statement(k) is true for some k>=1. Then g(h^k)=(h^k)g. Multiply on right byh: (g(h^k))h=((h^k)g)h.
Now associative property of group operation gives: g((h^k)h)=(h^k)(gh). But (h^k)h=h^(k+1) by def of exponent, and gh = hg is given.
Thus g(h^(k+1))=(h^k)(hg). Apply associative again to the RHS, getting g(h^(k+1))=((h^k)h)g and then again note that (h^k)h=h^(k+1).
Thus g(h^(k+1))=(h^(k+1))g. Thus Statement(k+1) is true (since that's exactly what Statement(k+1) asserts).
Thus have proven that Statement(1) is true, and also: if Statement(k) is true for any k>=1, then Statement(k+1) must also be true.
The line above means that we've proven by induction that Statement(n) is true for all n>=1.
Thus g(h^n)=(h^n)g for all n>=1. QED

Originally Posted by Patriotx12

Prove by induction:

1. If g and h are commuting elements of G, prove g commutes with every positive integer power of h: ghⁿ= hⁿg for all positive integers n.Then prove g^mhⁿ = hⁿg^m for all positive integers m and n. Is this true when m and n are arbitrary integers?

2. If g and h are commuting elements of G, prove (gh)ⁿ = gⁿhⁿ for all positive integers n.


Here's my attempt at solving 1, I'm just not sure if it's correct, or if more needs to be proved.

ghⁿ = hⁿg

Base step: let n = 1
gh = hg

True but it would be better to state "it is given that g and h commute".

Assume gh^k = h^kg

gh^k+1= h^k+1g
gh^kh = h^khg

How do you go from this line to the next? That is, how do you justify cancelling those "h"s?

gh^k = h^kg

Is that done correctly, and if so would it end the proof? And can the others be proved in the same way?

Also, you have gone from what you want to get to what you are assuming in the "induction hypothesis"- that's okay (and is called "synthetic proof") as long as every step is reversible but I think a direct proof is always preferable.
From gh^k= h^kg, the induction hypothesis, multiply both sides of the equation, on the [b]right[b], by h:
gh[sup]k[sup]h= h^kgh. The left side is clearly gh^k+1 but how do you get h^kgh to h^k+1g? (Remember that our basic hypothesis is that g and h commute.)

Now try the same thing with the next problem.