# The intersection of the conjugates of $H\leq G$ is *exactly* the union of the conjugacy classes of $G$ that are fully in $H$

There are several posts discussing the intersection of the conjugates of a subgroup, but I think this provides at least something new in this forum. I'd like to know if the reasoning and conclusion are valid, and whether or not this follows trivially from some other conclusion made elsewhere.

Assume the following:

• $$H\leq G$$ is a subgroup.
• $$N:=\bigcap_{g\in G}gHg^{-1}$$ is the group intersection of all conjugates in $$G$$ of $$H$$.
• $$C:=\bigcup_{i\in I}C_i$$ is the set union of all conjugacy classes $$C_i$$ of $$G$$ where $$C_i\subseteq H$$.

The goal is to show that $$N=_{\mathbf{Set}}C$$.

Show that $$C\subseteq N$$:

• Let $$Q\subseteq H$$ be any set invariant under conjugation in $$G$$. This means that $$gQg^{-1}=Q$$ for all $$g\in G$$. Therefore, $$Q$$ is in all conjugates of $$H$$, and is thus also in $$N$$. This can be applied to any $$C_i$$ in the union $$C$$.

Show that $$N\subseteq C$$:

• Let $$x\in N$$. If $$g^{-1}xg\in H$$ for all $$g\in G$$, then the conjugacy class of $$x$$ is also in $$H$$. Equivalently, we can show that $$x\in gHg^{-1}$$ for all $$g\in G$$. By definition of $$N$$, this is true for any choice of $$g$$; therefore, the conjugacy class of $$x$$ is contained in $$H$$, thereby also including that class in $$C$$.

Since $$C\subseteq N$$ and $$N\subseteq C$$, we can conclude that $$N=_{\mathbf{Set}}C$$. $$\square$$

At the very least, this shows that if $$H$$ contains a nontrivial conjugacy class then $$N$$ is also nontrivial. I'm fine with that result because it puts a lower bound on what $$N$$ can be. On the other hand, it implies that in many cases $$H$$ might already be normal if it contains a nontrivial conjugacy class. (Perhaps this is common for permutation groups? Seems plausible.)

• This is true (and you don't need to assume that $H$ is a subgroup, it's true for $H$ an arbitrary subset) but it doesn't seem particularly easy to show that $H$ contains a nontrivial conjugacy class so I don't know what use it might have. To my mind a much more useful characterization of $N$, which can often be used to calculate it immediately, is that it's the kernel of the action of $G$ on $G/H$. In particular $G$ acts faithfully on $G/H$ iff $N$ is trivial. Oct 22, 2020 at 9:39
• What is the meaning of $G/H$ when $H$ isn't normal? Is it just an arbitrary choice of left or right cosets? Oct 22, 2020 at 10:05
• $G/H$ is the set of left cosets $\{gH \mid g \in G\}$. The action of $G$ on $G/H$ is defined by $g \cdot (g' H) = (gg') H$ (and now argue that this action is well-defined). Oct 22, 2020 at 11:52

First, $$N$$ is a normal subgroup of $$G$$ since: $$gNg^{-1}=g \bigg(\bigcap_{x \in G} xHx^{-1} \bigg) g^{-1} = \bigcap_{x \in G} (gx)H(gx)^{-1}=\bigcap_{x \in G} xHx^{-1}=N$$ We also have by definition that $$N \leqslant H$$. Moreover, for any $$K \unlhd G$$ and $$K \leqslant H$$, we have: $$K = \bigcap_{g \in G} gKg^{-1} \leqslant \bigcap_{g \in G} gHg^{-1}=N \implies K \leqslant N$$
Clearly, all the conjugacy classes of $$G$$ that $$N$$ is a union of, must be subsets of $$H$$, and thus, $$N \subseteq C$$. To show that $$N=C$$, it suffices to show that for every conjugacy class $$C'$$ that is a subset of $$H$$, there exists $$K \unlhd G$$ such that $$C' \subseteq K$$. This would force $$K \leqslant N$$ and in turn, $$C' \subseteq N$$, as required.
Define $$K$$ as: $$K=\bigcap_{g \in G} g \langle C' \rangle g^{-1}$$ As $$K$$ is the intersection of groups, we must have $$K$$ to be a group. Since $$K$$ is a subgroup of $$\langle C' \rangle$$, we have $$K \leqslant H$$. Moreover, $$K$$ is the intersection of conjugate groups, and thus must be normal. Finally, we can see that $$C' \subseteq K$$ since: $$C' = \bigcap_{g \in G} gC'g^{-1} \subseteq \bigcap_{g \in G} g \langle C' \rangle g^{-1} = K$$
• You glossed over what makes $C$ a group, i.e., how you know it's closed. Oct 22, 2020 at 11:14