Conjugacy class

**jackie** · October 19th 2009, 12:03 AM

I'm stuck on this problem. Hope someone can give me help.
Let H be a normal subgroup of G, show that if $x \in H$ then $ord(x^H)=ord(x^G)$ or $ord(x^H)=\frac{1}{2}ord(x^G)$
I think I need to construct an isomorphism, but I have no clue how to do it here.

**tonio** · October 19th 2009, 07:47 AM

Originally Posted by jackie

I'm stuck on this problem. Hope someone can give me help.
Let H be a normal subgroup of G, show that if $x \in H$ then $ord(x^H)=ord(x^G)$ or $ord(x^H)=\frac{1}{2}ord(x^G)$
I think I need to construct an isomorphism, but I have no clue how to do it here.

What is $x^H$ ? Is this the normal closure of $x$ (of $<x>$ ) in H? Wouldn't $ord(x^H)=\frac{1}{2}ord(x^H)$ mean that $|H|$ has to be even?
Where did you take this problem from?

Tonio

**jackie** · October 19th 2009, 08:01 AM

Originally Posted by tonio

What is $x^H$ ? Is this the normal closure of $x$ (of $<x>$ ) in H? Wouldn't $ord(x^H)=\frac{1}{2}ord(x^H)$ mean that $|H|$ has to be even?
Where did you take this problem from?

Tonio

G and H act on itself by conjugation
I think $x^H=\{gxg^{-1}:g \in H\}$ ={conjugates of x in H}
and similarly for $x^G$ . Also, G is finite
This problem is from our homework handout

**tonio** · October 19th 2009, 08:31 AM

Originally Posted by jackie

G and H act on itself by conjugation
I think $x^H=\{gxg^{-1}:g \in H\}$ ={conjugates of x in H}
and similarly for $x^G$ . Also, G is finite
This problem is from our homework handout

I think it must be $x^H=<gxg^{-1}; g \in H>$ meaning: the subgroup generated by the conjugates of x by elements of H = the normal closure of <x> in H.

As $H \vartriangleleft G$ , both sbgps. $x^H \mbox { and } x^G$ are contained in H

Tonio

**NonCommAlg** · October 19th 2009, 02:09 PM

Originally Posted by jackie

I'm stuck on this problem. Hope someone can give me help.
Let H be a normal subgroup of G, show that if $x \in H$ then $ord(x^H)=ord(x^G)$ or $ord(x^H)=\frac{1}{2}ord(x^G)$
I think I need to construct an isomorphism, but I have no clue how to do it here.

this claim is false! (unless by $x^G$ and $x^H$ you mean something else!) here's a counter-example:

let $p \geq 5$ be a prime number and define $G=\left \{\begin{pmatrix}a & b \\ 0 & c \end{pmatrix}: \ \ a,b,c \in \mathbb{Z}/p\mathbb{Z}, \ ac \neq 0 \right \}$ and $H=\left \{\begin{pmatrix}1 & a \\ 0 & 1 \end{pmatrix}: \ \ a \in \mathbb{Z}/p\mathbb{Z} \right \}.$ also choose $x=\begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}.$

see that $H$ is a normal subgroup of $G.$ a little work shows that $|x^H|=|\{hxh^{-1}: \ h \in H \}|=1$ and $|x^G|=|\{gxg^{-1}: \ g \in G \}|=p-1.$

however, it can be shown that for any finite group $G,$ any normal subgroup $H$ of $G,$ and any $x \in H$ we have either $|x^G|=|x^H|$ or $|x^G| \geq 2|x^H|.$

**jackie** · October 19th 2009, 07:44 PM

Thank you very much for the help, tonio and NonCommAlg.
You are right NonCommAlg. The question was supposed to include that H is a normal subgroup of index 2 in finite G.
I used the result $[G:C_G(x)]=ord(x^G)$ , similarly $[H:C_H(x)]=ord(x^H)$ . I also used the second isomorphism in my proof.

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