I am looking at Group Theory notes on Centralizer and Normalizer for next semester and come up with this question:

Let $H$ be a subgroup of $G$, and let $g$ be an element in $G$. Show that

$$(a)\ C_G(H^g) = C_G(H)^g$$

$$(b)\ N_G(H^g) = N_G(H)^g.$$


For (a), I sure the LHS is as follow but correct me if I am wrong:

$C_G(H^g) = \{x \in G : xg^{-1}hg = g^{-1}hgx = xghg^{-1} = ghg^{-1}x, \forall h^g \in H^g$ and where $ g, g^{-1} \in G\} $

But I am not sure for the RHS of (a). Does it literally mean $[C_G(H)]^g$?

For the LHS of (b), correct me if I am wrong here:

$$N_G(H^g) = \{x \in G : xH^g = H^gx\}$$ $$ = \{x \in G : xg^{-1}Hg = g^{-1}Hgx = xgHg^{-1} = gHg^{-1}x\}$$

Again for the RHS, does it mean literally $[N_G(H)]^g$?

Thank you very much for your time and help. Happy holidays to you all!

  • $\begingroup$ I've seen this notation other places, and absolutely hate it. It's so confusing. $\endgroup$ – user100463 Jan 2 '17 at 23:22

Yes the right hand side of both equations literally means what it says.

Here is a hint for both problems: write down precisely what it means that some element $x$ is a member of the left hand side. Then write down precisely what it means that some element $x$ is a member of the right hand side. Then manipulate both expressions so make them look more like each other. If you can manipulate them into the same expression then you're done and you can write down the proof nicely.

Here is something to get you started:

$$ \begin{aligned} x \in C_G(H^g) &\iff \dots \\ &\iff \dots \\ &\iff x^{g^{-1}}\in C_G(H) \\ &\iff x \in C_G(H)^g \end{aligned}$$

Spoiler: solution for (a)

$$ \begin{aligned} x \in C_G(H^g) &\iff (\forall h\in H)( (h^g)^x = h^g) \\ &\iff (\forall h\in H)(h^{gxg^{-1}}=h) \\ &\iff x^{g^{-1}}\in C_G(H) \\ &\iff x \in C_G(H)^g \end{aligned}$$

Spoiler: solution for (b)

$$ \begin{aligned} x \in N_G(H^g) &\iff (H^g)^x = H^g \\ &\iff H^{gxg^{-1}}=H \\ &\iff x^{g^{-1}}\in N_G(H) \\ &\iff x \in N_G(H)^g \end{aligned}$$

  • $\begingroup$ Awesome! How I love this forum! You have gave me hints more than I have bargained for, I think I can get it going. I have up-voted your response. One question though: How did you upload that Spoiler? Thanks again. $\endgroup$ – A.Magnus Dec 13 '14 at 17:37
  • $\begingroup$ If you type ">!" (without the quotation marks) in front of a sentence then it will turn into a spoiler. $\endgroup$ – Myself Dec 13 '14 at 17:42
  • $\begingroup$ One question again: From the way you manipulated the conjugate, it looks like conjugate is behaving similar to exponent. Is conjugate the generalization of exponent? Thanks again. $\endgroup$ – A.Magnus Dec 13 '14 at 20:45
  • $\begingroup$ No, not really. It's best to see them as entirely different things. It is true however that $(x^g)^h = x^{gh}$ both when the superscript denotes exponent (so $g$ and $h$ are natural numbers) and when it denotes conjugate, but this is more a notational coindidence of two facts with different meaning and different proofs. For conjugates the proof consists of repeatedly replacing every occurence of the superscript $a^b$ by $b^{-1}ab$ and working everything out until you see they are equal. $\endgroup$ – Myself Dec 13 '14 at 21:02

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