1

This was a challenge question posted by my professor in algebra course last semester, and I've been trying to solve it. I think there is probably a solution online but I couldn't find much information about this challenge.

Let $G$ be a non-trivial finite group, and let $H\le G$ be s.t. $|\text{Core}_G(H)|=1$, where $\text{Core}_G(H):=\underset{g\in G}{\cap}gHg^{-1}$.

  1. Prove that $\Delta_H(G):=G-\underset{g\in G}{\cup}gHg^{-1}$ is nonempty;
  2. Can $\Delta_H(G)$ contain an element of prime power order?

My thought:

I was discussing by case. Case 1: $Z(G)$ is non-trivial; Case 2: otherwise.

In case 1, say we have $x\in Z(G)$ with $x\ne e$. Then $x\notin H$, otherwise, we would have $x\in\text{Core}_G(H)$ which makes $|\text{Core}_G(H)|>1$. Thus $x\in\Delta_H(G)$, as $x\notin gHg^{-1}$ for any $g\in G$. Therefore $\Delta_H(G)$ is nonempty. Moreover, it turns out that $Z(G)\subset\Delta_H(G)$, so there must be an element of prime power order.

I have no clues on how to deal with case 2. I tried to think in terms of conjugacy classes (so $\text{Core}_G(H)$ is the intersection of all conjugacy classes of $h\in H$ and $\Delta_H(G)$ is the set of elements which are not in any conjugacy classes of $h\in H$), but it seems to not lead to a solution.

Shaun
  • 44,997
Yuqiao Huang
  • 728

2 Answers2

1

It is well known that no group can be the union of conjugates of a proper subgroup of finite index. In particular, no finite group can be the union of the conjugates of a proper subgroup. Since we are given that the core of $H$ is trivial, $H\neq G$ so part (1) follows from a counting argument.

For part (2), the answer is "yes, it's possible". To see this, let $p$ be an odd prime, and let $G$ be the nonabelian group of order $p^3$ and exponent $p$, with generators $x$ and $y$ ($G$ is also sometimes called the Heisenberg group of order $p^3$). Then $H=\langle x\rangle$ is a proper subgroup of order $p$, but is not normal. Since it is not normal, it's core must be a proper subgroup of $H$. But $H$ is of prime order, so the core of $H$ is trivial. The union of its conjugates cannot be all of $G$, as discussed above, yet every element of $\Delta_H(G)$ is an element of a $p$-group, hence has order that is a power of $p$ (in fact, equal to $p$).

For $p=2$, take $G=D_4$, the dihedral group of order $8$, $G=\langle r,s\mid r^4=s^2=1, sr=r^3s\rangle$, and let $H=\{e,s\}$. Again, $H$ is not normal but of prime order, hence its core is trivial. Any element in $\Delta_H(G)$, which cannot be trivial, is of order a power of $2$.

Arturo Magidin
  • 398,050
1

With respect to your question 1. the classical approach is a counting argument as pointed out by Arturo. But if you know some character theory of groups, notably induced characters and Frobenius Reciprocity, you can proceed as follows. Let $H \lt G$, with $|G:H|$ finite and assume $\bigcup_{g \in G}g^{-1}Hg=G$. Induce the trivial character $1_H$ of $H$: then $(1_H)^G(x)=\frac{1}{|H|}\sum_{g \in G}1_H^{\circ}(g^{-1}xg)=|G:H|$. (here $1_H^{\circ}(t)=1_H(t)=1$ if $t \in H$ and $1_H^{\circ}(t)=0$ if $t \notin H$). Hence $|G:H|=[(1_H)^G,1_G]_G=[1_H,(1_G)_H]_H=[1_H,1_H]_H=1$ by Frobenius Reciprocity. So, $G=H$, a contradiction.

Nicky Hekster
  • 49,281