3

I am trying to figure out the following.

Suppose $G$ is a group and is finite; let $H$ be a proper subgroup. Show that the conjugates of $H$ do not cover $G$ (that is, there is some $g \in G$ which is not conjugate to an element of $H$).

As I understand it, we say that given $a,b \in G$, $a$ is conjugate with $b$ if $\exists g' \in G (a = g'bg'^{-1})$. Since $H$ is a proper subgroup of $G$, then there $\exists g \in G$ such that $g \notin H$. Since $g \notin H$, then $\forall h \in H (g \neq h)$. Suppose then that $\exists q \in G, \exists r \in H (g = qrq^{-1})$ $-$ i.e., that $g$ is conjugate with with some element of $H$. Then $q^{-1}gq = r$. However, I'm not sure where to go from here. I'm guessing that I have to somehow use the assumption that $G$ is finite however I'm not sure how to.

letsmakemuffinstogether
  • 1,834

4 Answers4

4

For $h,g\in G$ let $h^g=ghg^{-1}$; for a fixed $g\in G$, the map $h\mapsto h^g$ is an automorphism of $G$. Let $N=\{g\in G:H^g=H\}$, where $H^g=\{h^g:h\in H\}$; this is the normalizer of $H$, and it’s not hard to check that $N$ is a subgroup of $G$.

  • Show that $H\subseteq N$.
  • Show that the number of conjugates of $H$, including $H$ itself, is $|G/N|$.
  • Show that each conjugate of $H$ is a subgroup of $G$.

Since the identity element belongs to each conjugate of $H$, we have

$$\left|\bigcup_{g\in G}H^g\right|<|G/N|\cdot|H|=\frac{|G|}{|N|}\cdot|H|\le\frac{|G|}{|H|}\cdot|H|=|G|\;.$$

Brian M. Scott
  • 616,228
2

Consider the normalizer of $H$, $N(H)=\{g\in G|gH=Hg\}$. It is not hard to check that $N(H)$ is a subgroup of $G$ and $H$ is a normal subgroup of $N(H)$.

We shall show that the number of the conjugate subgroups of $H$ equals to the number of cosets of $N(H)$ in $G$.

Method I: Consider the map $\phi:gHg^{-1}\to gN(H)$. We shall show that the map $\phi$ is bijective.

If $g_1Hg_1^{-1}=g_2Hg_2^{-1}$ then $(g_2^{-1}g_1)H=H(g_2^{-1}g_1)$. Thus $g_2^{-1}g_1\in N(H)$, this is, $g_1N(H)=g_2N(H)$. Hence the map $\phi$ is well-defined.

If $g_1N(H)=g_2N(H)$ then $g_2^{-1}g_1\in N(H)$. Thus $(g_2^{-1}g_1)H=H(g_2^{-1}g_1)$, this is, $g_1Hg_1^{-1}=g_2Hg_2^{-1}$. Hence the map $\phi$ is injective.

Obviously, the map $\phi$ is surjective.

Method II: Lemma: Let $G$ act transitively on a set $S$ and $K=Stab(x)$ for some $x\in S$. Then the action of $G$ on $S$ is equivalent to the action of $G$ on the coset space $G/K$.

Let $S=\{gHg^{-1}|g\in G\}$. Consider the action of G on S defined by $x\circ gHg^{-1}=x(gHg^{-1})x^{-1}$ for $x\in G$ and $gHg^{-1}\in S$. It is easy to check that $G$ acts transitively on S.

Since $H\in S$, let $K=Stab(H)=\{g\in G|g\circ H=H\}=\{g\in G|gHg^{-1}=H\}$ $=\{g\in G|gH=Hg\}=N(H)$.

Hence, by Lemma, we have $|S|=|G/N(H)|$.

wzstrong
  • 553
0

Hints:

  1. What is the maximum number of possible conjugates of $H$? (Recall that $G$ acts transitively on the set of conjugates of $H$.)
  2. What is the maximum number of elements these conjugates could cover?
Ted
  • 33,788
0

Consider the normalizer $N_G(H)$.The number of conjugate groups of H is $|G/N_G(H)|$.Now calculate the cardinality of all conjugate groups of H:
$|\cup_{g\in G}H^g|\leqslant |G/N_G(H)|\cdot (|H|-1)+1\leqslant |G|-|G/N_G(H)|+1<|G| $

user12580
  • 857