Group Theory: Ord(x) = Ord(g^-1xg)

[basil32]

Homework Statement

If x and g are elements of the group G, prove that |x|=|g^-1xg|. Deduce that |ab| =|ba| for all a,b $\in$ G

Homework Equations

The Attempt at a Solution

I've written a proof but does not seem quite right:
if (g^-1xg)ⁿ= x^m then n must be equal m
g^-nxⁿgⁿ = x^m then
xⁿgⁿ =gⁿx^m
xⁿ =gⁿx^mg^-n
x^m =g^-nxⁿgⁿ
xⁿ = gⁿ(g^-nx^mgⁿ)g^-n
xⁿ = x^m
n = m

 

[Dick]

There's more than one thing wrong with that. The problems start with (g^(-1)xg)^n=g^(-n)x^ng^n. That's only true if the group is commutative. I'll give you a big hint. (g^(-1)xg)^2=(g^(-1)xg)(g^(-1)xg). What's that?

 

[basil32]

There's more than one thing wrong with that. The problems start with (g^(-1)xg)^n=g^(-n)x^ng^n. That's only true if the group is commutative. I'll give you a big hint. (g^(-1)xg)^2=(g^(-1)xg)(g^(-1)xg). What's that?

(g^-1xg)² = (g^-1xg)(g^-1xg) => g^-1x²g
so by induction (g^-1xg)ⁿ = g^-1xⁿg
and?

 

[Dick]

(g^-1xg)² = (g^-1xg)(g^-1xg) => g^-1x²g
so by induction (g^-1xg)ⁿ = g^-1xⁿg
and?

Now it's your turn. Think about what order of x and order of g^(-1)xg mean. Show they are the same number.

 

[basil32]

Now it's your turn. Think about what order of x and order of g^(-1)xg mean. Show they are the same number.

g^-1xⁿg = e
xⁿg=ge
xⁿ = g.g^-1
xⁿ = e

Thanks.

 

[Deveno]

g^-1xⁿg = e
xⁿg=ge
xⁿ = g.g^-1
xⁿ = e

Thanks.

just because g^-1xⁿg = e, and xⁿ = e, does NOT mean that |g^-1xg| = |x|.

at best, it only shows that |g^-1xg| divides |x| (assuming by n, you mean |x|).

you need to show that no smaller positive integer will work (this isn't that hard, try proof by contradiction).

then there's still part 2...

 

[HallsofIvy]

It is also, by the way, not true in a general group that is $x^m= x^n$, m= n.

 

[Deveno]

right. an obvious example is in the multiplicative group of non-zero real numbers.

(-1)² = 1² = 1, but -1 ≠ 1.

 

[basil32]

A bit of a blow. I thought I finished but apparently it requires much more careful thinking.

you need to show that no smaller positive integer will work (this isn't that hard, try proof by contradiction).

The choice of n is the smallest positive integer that xⁿ=e. So any m < n would contradict this. What's to prove?

 

[gauss^2]

The choice of n is the smallest positive integer that xⁿ=e. So any m < n would contradict this. What's to prove?

Prove that $(g^{-1} x g)^m = e$ cannot happen for a positive integer $m < n$. Use the fact mentioned earlier that $(g^{-1} x g)^m = g^{-1} x^m g$ for any integer $m$ to derive a contradiction based on the assumption that $1 \leq m < n$ and $(g^{-1} x g)^m = e$ both hold.

 

[Deveno]

A bit of a blow. I thought I finished but apparently it requires much more careful thinking.


The choice of n is the smallest positive integer that xⁿ=e. So any m < n would contradict this. What's to prove?

that n is, in fact, the order of g^-1xg. we know no smaller m works for x, but g^-1xg, isn't x, it's some other element of G, so you need to prove no smaller m will work for g^-1xg.

suppose (as an imaginary example) that g^-1xg = x², and |x| = 4.

then g^-1xg would have order 2, and so would still be e if raised to the 4th power since e² = e.

do not confuse the statements xⁿ = e, and |x| = n. this can get you into trouble. the second implies the first, but NOT the other way around.

you have to get used to this sort of thing, because often you will consider functions (homomorphisms) from one group to another f:G→H. if f is not 1-1 (and you will encounter such functions), than f will map several elements to the identity of H.

due to the nature of homomorphisms, it will be the case that if |x| = n, (f(x))ⁿ will be e_H. but this does not mean |f(x)| = n.

so the map G→G given by x→g^-1xg is special, it doesn't display this kind of behavior.