# Pronormality and subnormality imply normality

A subgroup $H$ of a group $G$ is said to be pronormal if for each $g\in G$, the subgroups $H$ and $gHg^{-1}$ are conjugate in $\langle H, gHg^{-1}\rangle$

Let $H$ be a pronormal subgroup of $G$ and suppose that $H$ is subnormal in $G$. Then I need to show that $H \unlhd G$.

Some facts to use: (a) if $H\leq K\leq G$, then $H$ pr $K$ (b) if $K\unlhd G$ then $G = N_G(H)K$

Since $H \triangleleft \triangleleft\; G$, there exists a chain of subgroups $$H=H_0 \unlhd H_1 \unlhd \dots \unlhd H_n = G$$

I will proceed by induction on $n$.

For $n=1$, clearly $H\unlhd H_1 = G$. I'm not sure how to proceed for the inductive step.

• You are almost there. By induction, $H \unlhd H_{n-1}$. Put $K=H_{n-1}$ and use (b) to get $G=N_G(H)K=N_G(H)$. – Derek Holt May 7 '16 at 10:08
• I'm not sure how by induction we get $H \unlhd H_{n-1}$. I always thought for the inductive step, we let $n=k$ so that $H \unlhd H_k$, and then we have to prove that $H\unlhd H_{k+1}$? – R Maharaj May 7 '16 at 10:31
• For $n \in {\mathbb Z}$, $n \ge 1$, let $P(n)$ be the statement that whwnever $H\ {\rm pr}\ G$ and $H$ is subnormal of depth at most $n$, then $H \unlhd G$. As you said yourself, this is clear for $n=1$. Now we assume that $P(n-1)$ is true. So that means that $H \unlhd H_{n-1}$ in your subnormal chain, and it follows using (b) that $H \unlhd H_n=G$, So we have shown that $P(n-1) \Rightarrow P(n)$. – Derek Holt May 7 '16 at 11:18