# Size of conjugacy class in subgroup compared to size of conjugacy class in group

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.

• 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? Commented Apr 14, 2019 at 21:37
• @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 $|H| = |[a]_H| \cdot |Z_H(a)|$. Commented Apr 14, 2019 at 21:45
• @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, $Z_H(a) or Z_G(a)$? Commented Apr 15, 2019 at 1:03
• Sorry, I reversed the cases in my comment which was confusing, so instead I gave most of the details in an answer. Commented Apr 15, 2019 at 1:45

## 1 Answer

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

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

• 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. Commented Apr 15, 2019 at 2:28
• 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 $Z_G(a)/Z_H(a) \subset G/H$. Commented Apr 15, 2019 at 4:35
• 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. Commented Apr 15, 2019 at 4:38
• @FurloRoth They are all consequences, but I'm not sure they are easy consequences. Commented Jun 27, 2019 at 23:54