Proving that the generator of $U$ is normal if $\forall u \in U, g\in G$ $gug^{-1} \in U$

Question

This is from Herstein.

$4.$ $\;a)$ Given a group $G$ and a subset $U$ denote by $\hat U$ the smallest subgroup of $G$ which contains $U$ (the subgroup generated by $U$). Prove there is a subgroup $\hat U$ of $G$.

$\;\;\;\;\;b)$ If $gug^{-1} \in U$ for all $g \in G$ and $u \in U$, prove that $\hat U$ is a normal subgroup of $G$.

My Work: Let $\mathscr A$ be the collection of subgroups of $G$ that contain $U$. This collection is non-empty since it includes $G$ itself. I defined the subgroup generated by $U$ to be, $$ \hat U = \bigcap_{A \in \mathscr A} A $$ Then we can prove that $\hat U$ is a subgroup of $G$, it contains $U$ and if $H$ is any subgroup of $G$ that contains $U$ then $\hat U$ is a subgroup of $H$.

But I am having trouble proving part $b)$. Any help is appreciated. A hint would be fabulous.

Thanks in advance.

Answer 1

Hint: Show that $$V:=\{v\in \hat{U}: gvg^{-1}\in \hat{U},\forall g\in G\}$$ is a subgroup containing $U$. It is clear that $V\subseteq\hat{U}$. Also by minimality of $\hat{U}$, it follows that $V\supseteq\hat{U}$. Hence $V=\hat{U}$, this means for all $v\in\hat{U}$ we have $gvg^{-1}\in\hat{U},\forall g\in G$, which means $\hat{U}$ is normal.

Answer 2

First show that for every $g\in G$ and $A\in \mathscr A$, $gAg^{=1}\in \mathscr A$. Indeed,

$$A\in \mathscr A\Rightarrow U\subset A\Rightarrow U=gUg^{-1}\subset gAg^{-1}.$$

Now

$$\hat U=\bigcap_{A\in\mathscr A}A\subset gAg^{-1}$$

for every $A\in \mathscr A$. Hence

$$\hat U\subset\bigcap_{A\in\mathscr A} gAg^{-1}=g\hat Ug^{-1}.$$

This can also be written as

$$g^{-1}\hat Ug\subset \hat U.$$

Replacing $g$ by $g^{-1}$ gives

$$g\hat Ug^{-1}\subset\hat U.$$

Since we have inclusions bothways, the identity is proved.

Answer 3

First we prove that $\forall \hat u \in \hat U$, $\exists u, v \in U$, s.t. $\hat u = uv$.

To do so we define set $\hat W = \{ uv | uv \in \hat U, u, v \in U \}$, and prove $\hat W$ is exactly $\hat U$.

Consider $\forall w_1, w_2 \in \hat W$.

By definition of $\hat W$, we have $w_1, w_2 \in \hat U$. Since $\hat U$ is a group, $w_1 w_2^{-1} \in \hat U$.

By definition of $\hat W$, we write $ w_1 = u_1 v_1$, $w_2 = u_2 v_2$. So $w_1 w_2^{-1} = (u_1 v_1) (u_2 v_2)^{-1} = (u_1 u_2) (v_2^{-1} v_1^{-1})$.

Since $U$ and $\hat U$ are group, $u_1 u_2, v_2^{-1} v_1^{-1} \in U$. So by definition of $\hat W$, $w_1 w_2^{-1} \in \hat W$.

So $\hat W$ is also a group.

But $U \subset \hat W$, and $\hat U$ is the smallest group that contains $U$, we have $\hat W = \hat U$.

With this we can proceed to

b) If $gug^{−1} \in U$ for all $g\in G$ and $u\in U$, prove that $\hat U$ is a normal subgroup of G.

We just proved above that $\forall \hat u \in \hat U$, there $\exists u, v \in U$ s.t. $\hat u = uv$.

So $g\hat u g^{-1} = g uv g^{-1} = (g u g^{-1}) (g v g^{-1})$, where $gug^{-1}, gvg^{-1} \in U$, so $g \hat u g^{-1} \in \hat U$.

So $\hat U$ is normal.