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Given:

$\bullet$ A finite group $G$, an index 2 subgroup $H$, an element $a \in H$
$\bullet$ $[a]_H$ and $[a]_G$ are the conjugacy class in $H$ of $a$ and the conjugacy class in $G$ of $a$, respectively

To prove:
$[a]_H = [a]_G$ or $[a]_H$ is half the size of $[a]_G$, depending on whether or not the centralizer $Z_G(a)$ is contained in $H$.

Attempt:
$H$ being of index 2 means that $H$ is normal, which means that $H \cdot Z_G(a) $ is a subgroup of $G$. By the Second Isomorphism Theorem, $(H \cdot Z_G(a)) \, / \,H \cong Z_G(a) \, / \, (H \cap Z_G(a))$. If $Z_G(a)$ is contained in $H$, then $H \cap Z_G(a) = Z_G(a)$ and the group on the right-hand side of the isomorphism is trivial, and therefore the group on the left-hand side is trivial as well, and so $|H \cdot Z_G(a))| = |H|$.

Where do I go from here? Can I say that $H$ must be equal to $H \cdot Z_G(a))$? Even if I could, I don't know how to continue.

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  • 1
    $\begingroup$ I would recommend thinking about group actions. $G$ acts transitively on $[a]_G$ by conjugation. What is the kernel of this action? What happens when one restricts the action of $G$ to $H$? What does the Orbit-Stabilizer say about these various actions? $\endgroup$
    – Furlo Roth
    Commented Apr 14, 2019 at 21:37
  • $\begingroup$ @user655377 The kernel of the action of $G$ on $[a]_G$ is the centralizer of $a$ in $G$. If the action is restricted to $H$, then the action is the same only if $[a]_G \subset [a]_H$. The O.S. theorem says that $|G| = |[a]_G| \cdot |Z_G(a)| $ and $|H| = |[a]_H| \cdot |Z_H(a)|$. $\endgroup$
    – Junglemath
    Commented Apr 14, 2019 at 21:45
  • $\begingroup$ @user655377 Sorry, I don't see why $[a]_G = [a]_H$ iff $Z_G(a) = Z_H(a)$. Also, what does "it" refer to in your next sentence, $Z_H(a) or Z_G(a)$? $\endgroup$
    – Junglemath
    Commented Apr 15, 2019 at 1:03
  • $\begingroup$ Sorry, I reversed the cases in my comment which was confusing, so instead I gave most of the details in an answer. $\endgroup$
    – Furlo Roth
    Commented Apr 15, 2019 at 1:45

1 Answer 1

Reset to default
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Step 1. From the orbit stablizer theorem:

$$|G| = |[a]_G| \cdot |Z_G(a)|, \quad |H| = |[a]_H| \cdot |Z_H(a)|.$$

Hence

$$\frac{|[a]_G|}{|[a]_H|} = \frac{|G|}{|H|} \cdot \left(\frac{|Z_G(a)|}{|Z_H(a)|}\right)^{-1} = 2 \cdot \left(\frac{|Z_G(a)|}{|Z_H(a)|}\right)^{-1}$$

Step 2. We have $Z_H(a) = Z_G(a) \cap H$.

Step 3. There is a homomorphism:

$$\varphi: Z_G(a) \hookrightarrow G \rightarrow G/H.$$

The kernel is precisely $Z_G(a) \cap H = Z_H(a)$. Thus, by the first isomorphism theorem,

$$Z_G(a)/Z_H(a) \simeq \mathrm{im}(\varphi) \subset G/H,$$

Since $G/H$ has order $2$, it follows that $Z_G(a)/Z_H(a)$ is either trivial or a group of order 2.

Case 1. $Z_G(a) \subset H$. This means that $Z_H(a) = Z_G(a)$, and thus $Z_G(a)/Z_H(a)$ is trivial, and so

$$\frac{|[a]_G|}{|[a]_H|} = 2.$$

Case 2. $Z_G(a) \not\subset H$. This means that $Z_H(a) \ne Z_G(a)$, and thus $Z_G(a)/Z_H(a)$ is non-trivial, and so (by the above) of order $2$, and

$$\frac{|[a]_G|}{|[a]_H|} = \frac{2}{2} = 1.$$

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  • $\begingroup$ I think this works. Although, why did you write $\left(\frac{|Z_G(a)|}{|Z_H(a)|} \right)^{-1}$ instead of $ \frac{|Z_H(a)|}{|Z_G(a)|}$? Also, it was recommended to use the Second Isomorphism Theorem, but you did it alternately, using the First Isomorphism Theorem. Nice solution nevertheless. $\endgroup$
    – Junglemath
    Commented Apr 15, 2019 at 2:28
  • $\begingroup$ Writing in that way was to emphasize the fact that the $|[a]_G|/|[a]_H|$ was the ratio of the order of the group $G/H$ of order $2$ to the order of the group $Z_G(a)/Z_H(a) \subset G/H$. $\endgroup$
    – Furlo Roth
    Commented Apr 15, 2019 at 4:35
  • $\begingroup$ I think it's better to understand the first isomorphism theorem well and forget the other ones which are all just easy consequences of the first isomorphism theorem anyway. $\endgroup$
    – Furlo Roth
    Commented Apr 15, 2019 at 4:38
  • $\begingroup$ @FurloRoth They are all consequences, but I'm not sure they are easy consequences. $\endgroup$
    – user5826
    Commented Jun 27, 2019 at 23:54

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