Exercise 3.3.10 (Hall subgroups)

Proof.

(a)

By hypothesis, $$|G:H|\wedge |H|=1.$$

Since $N⊴G$, $\mathit{𝐻𝑁}$ is a subgroup of $G$. By Proposition 13 of Section 3.2,

$$|\mathit{𝐻𝑁}|=\frac{|H||N|}{|H\cap N|},$$

thus

$$\begin{array}{lll}\hfill |N:H\cap N|=|\mathit{𝐻𝑁}:H|& & \hfill \text{(1)}\end{array}$$

(even if $H$ is not normal in $G$).

Since $|G:H|=|G:\mathit{𝐻𝑁}|\cdot |\mathit{𝐻𝑁}:N|$, then $|\mathit{𝐻𝑁}:N|$ divides $|G:H|$, where $|G:H|\wedge |H|=1$, so we obtain $|\mathit{𝐻𝑁}:N|\wedge |H|=1$. Using (1), this gives

$$\begin{array}{lll}\hfill |N:H\cap N|\wedge |H|=1.& & \hfill \text{(2)}\end{array}$$

By Lagrange Theorem, $|H\cap N|$ divides $|H|$, so (2) implies

$$|N:H\cap N|\wedge |H\cap N|=1.$$

This shows that $H\cap N$ is a Hall subgroup of $N$.

(b)

By hypothesis, $|G:H|\wedge |H|=1$. Since $|G:\mathit{𝐻𝑁}|$ divides $|G:H|$, we obtain

$$|G:\mathit{𝐻𝑁}|\wedge |H|=1,$$

and since $|H:H\cap N|$ divides $|H|$, this gives

$$|G:\mathit{𝐻𝑁}|\wedge |H:H\cap N|=1.$$

Moreover, $|H:H\cap N|=|\mathit{𝐻𝑁}:N|$, therefore

$$\begin{array}{lll}\hfill |G:\mathit{𝐻𝑁}|\wedge |\mathit{𝐻𝑁}:N|=1.& & \hfill \text{(3)}\end{array}$$

$\mathit{𝐻𝑁}∕N$ is a subgroup of $G∕N$, with index

$$|G∕N:\mathit{𝐻𝑁}∕N|=(|G|∕|N|)∕(|\mathit{𝐻𝑁}|∕|N|)=|G|∕|\mathit{𝐻𝑁}|=|G:\mathit{𝐻𝑁}|.$$

(We don’t use the Third Isomorphism Theorem, because $\mathit{𝐻𝑁}$ is not necessarily normal in $G$.)

Then (3) becomes

$$|G∕N:\mathit{𝐻𝑁}∕N|\wedge |\mathit{𝐻𝑁}∕N|=1,$$

which shows that $\mathit{𝐻𝑁}∕N$ is a Hall subgroup of $G∕N$.