# Thread: order of x (Abstract Algebra)

1. ## order of x (Abstract Algebra)

Can someone help?

Prove: If x and g are elements of the group G, prove that |x| = |g^(-1)xg|. Deduce that |ab| = |ba| for every a, b in G.

|x| means the order of x. 1 is the identity element in this case.

For G a group and x in G, the order of x to be the smallest positive integer n such that x^n = 1 (that is |x| = n).

2. For the first part, let $|x| = n$ and start taking products of $gxg^{-1}$ until you see a pattern. Notice that $(gxg^{-1})^2 = (gxg^{-1})(gxg^{-1}) = gx(g^{-1}g)xg^{-1} = gx^2g^{-1}$. Similarly, $(gxg^{-1})^n = gx^ng^{-1}$. I think you can take it from here.

For the second part, let $|ab| = n$. Then, if e is the identity,

$b = be$

$\Longrightarrow b = b((ab)(ab)...(ab))$ for n copies of ab

$\Longrightarrow b = ((ba)(ba)...(ba))b$ by associativity (notice that you have n copies of ba)

$\Longrightarrow e = (ba)^n$ by cancellation

$\Longrightarrow |ba| \le n$

If $|ba| = k < n$, then $(ba)^k = e$.

So, $(ba)^kb = b$.

By associativity, $b(ab)^k = b$.

By cancellation, $(ab)^k = e$, which is a contradiction.

Thus, $|ba| = n$.

3. Originally Posted by dori1123

Deduce that |ab| = |ba| for every a, b in G.
we have: $ba=b(ab)b^{-1}.$ thus $|ab|=|ba|$ by the first part of the problem.

4. Thank you for helping me with the proof. But above only prove |x|=|g^(-1)xg| if the order of x is finite. What if the order of x is infinite? The definition I was given is that if the order of x is infinite, then we write |x| = 0. I don't know how to show |g^(-1)xg| = 0.

5. Originally Posted by dori1123
Thank you for helping me with the proof. But above only prove |x|=|g^(-1)xg| if the order of x is finite. What if the order of x is infinite? The definition I was given is that if the order of x is infinite, then we write |x| = 0. I don't know how to show |g^(-1)xg| = 0.
If $g^{-1}xg$ has finite order then so does $g(g^{-1}xg)g^{-1} = x$ by above.
Thus, they are both finite or infinite.
And if finite then they have the same order.