# Conjugacy class

• Oct 19th 2009, 12:03 AM
jackie
Conjugacy class
I'm stuck on this problem. Hope someone can give me help.
Let H be a normal subgroup of G, show that if $\displaystyle x \in H$ then $\displaystyle ord(x^H)=ord(x^G)$ or $\displaystyle ord(x^H)=\frac{1}{2}ord(x^G)$
I think I need to construct an isomorphism, but I have no clue how to do it here.
• Oct 19th 2009, 07:47 AM
tonio
Quote:

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 $\displaystyle x \in H$ then $\displaystyle ord(x^H)=ord(x^G)$ or $\displaystyle ord(x^H)=\frac{1}{2}ord(x^G)$
I think I need to construct an isomorphism, but I have no clue how to do it here.

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

Tonio
• Oct 19th 2009, 08:01 AM
jackie
Quote:

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

Tonio

G and H act on itself by conjugation
I think $\displaystyle x^H=\{gxg^{-1}:g \in H\}$={conjugates of x in H}
and similarly for $\displaystyle x^G$. Also, G is finite
This problem is from our homework handout
• Oct 19th 2009, 08:31 AM
tonio
Quote:

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

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

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

Tonio
• Oct 19th 2009, 02:09 PM
NonCommAlg
Quote:

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 $\displaystyle x \in H$ then $\displaystyle ord(x^H)=ord(x^G)$ or $\displaystyle ord(x^H)=\frac{1}{2}ord(x^G)$
I think I need to construct an isomorphism, but I have no clue how to do it here.

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

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

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

however, it can be shown that for any finite group $\displaystyle G,$ any normal subgroup $\displaystyle H$ of $\displaystyle G,$ and any $\displaystyle x \in H$ we have either $\displaystyle |x^G|=|x^H|$ or $\displaystyle |x^G| \geq 2|x^H|.$
• Oct 19th 2009, 07:44 PM
jackie
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 $\displaystyle [G:C_G(x)]=ord(x^G)$, similarly $\displaystyle [H:C_H(x)]=ord(x^H)$. I also used the second isomorphism in my proof.