# Thread:

1. ## |x|=|gxg^{-1}|

Let G be a group and let G act on itself by conjugation, so each g in G maps G to G by $x\mapsto gxg^{-1}$. For each g in G, prove that conjugation by g is an isomorphism from G onto itself. Deduce that x and $gxg^{-1}$ have the same order for all x in G and that for any subset A of G, $|A|=|gAg^{-1}|$.

I have shown this an isomorphism. I am not sure how to show x and $gxg^{-1}$ have the same order.

2. ## Re: |x|=|gxg^{-1}|

Originally Posted by dwsmith
Let G be a group and let G act on itself by conjugation, so each g in G maps G to G by $x\mapsto gxg^{-1}$. For each g in G, prove that conjugation by g is an isomorphism from G onto itself. Deduce that x and $gxg^{-1}$ have the same order for all x in G and that for any subset A of G, $|A|=|gAg^{-1}|$.

I have shown this an isomorphism. I am not sure how to show x and $gxg^{-1}$ have the same order.
If $\phi$ is an isomorphism then $\phi(x)^{|x|}=\phi(x^{|x|})=\phi(e)=e$ and $e=\phi^{-1}(e)=\phi^{-1}(\phi(x)^{|\phi(x)|})=x^{|\phi(x)|}$ imply $|\phi(x)|\mid |x|$ and $|x|\mid |\phi(x)|$ respectively....so.

3. ## Re: |x|=|gxg^{-1}|

there are different ways to do this:

1). suppose that x has order n. then x^n = e, so (gxg^-1)^n = (gxg^-1)(gxg^-1).....(gxg^-1) (n times)

= gx(g^-1g)x(g^-1g).....xg^-1 = (gx)(x)(x)....(x)g^-1 = g(x^n)g^-1 = geg^-1 = gg^-1 = e.

(ok, technically using induction on n would be better, but you get the idea). thus |gxg^-1| divides n.

now suppose that |gxg^-1| = k < n. then (gxg^-1)^k = g(x^k)g^-1 = e, so x^k = g^-1g = e, a contradiction.

hence |gxg^-1| = |x|.

2). since x-->gxg^-1 is an isomorphism, the order of gxg^-1 must be the order of x. why? suppose not.

case 2a) |x| < |gxg^-1|. in this case we can find a k with x^k = e, but (gxg^-1)^k ≠ e.

using our isomorphism e = geg^-1 = g(x^k)g^-1 = (gxg^-1)^k, contradicting our choice of k.

case 2b) |gxg^-1| < |x|. then we have (gxg^-1)^k = e, but x^k ≠ e, for some k.

since e = (gxg^-1)^k = g(x^k)g^-1, we have that x^k is in the kernel of our isomorphism.

but since an isomorphism is injective, the kernel is {e}, contradiction.

thus |gxg^-1| = |x|.

now, since x-->gxg^-1 is an isomorphism, it is bijective, so |A| and |gAg^-1| have to be equal.

4. ## Re: |x|=|gxg^{-1}|

Originally Posted by Drexel28
$\phi^{-1}(\phi(x)^{|\phi(x)|})$
Why is this phi of x to the order of phi of x?
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9171209335327148, "perplexity_flag": "middle"}