I had come across following term 'subgroup generated by subset' in Herstein Excercise.
Given any group G and subset U ,$\bar{U}$ be smallest subgroup containg that.
I thought that can be given by $\bar{U}$=$\cap_{U\in G_i}G_i$ where $G_i$ is subgroup in G .Is there in other defination exist for $\bar{U}$.
Next question says to prove that $\bar{U}$ is normal if $gug^{-1}\in U $ $\forall g\in G$
To prove that this I have to show that $g\bar{U}g^{-1}=\bar{U}$
By defination above $g\cap_{U\in G_i}G_i g^{-1}$=$\cap_{U\in G_i}gG_ig^{-1}$=$\cap_{U\in G_i}G_i$
I had not concrete argument to write last thing .$G_i$ is subgroup containg U. Please help me to find correct argument to write that .
The definition $\bar{U} = \cap_{U \in G_i} G_i$ is the standard definition for the subgroup generated by the subset $U$. Another way to define it is to construct it explicitly. You can definite $\bar{U}$ as the subset of all finite products of the type $u_1^{k_1}u_2^{k_2}\cdots u_r^{k_r}$, where $u_i \in U, k_i \in \mathbb{Z}$. However you have to show that this is a subgroup, which shouldn't be hard using the standard method.
You can use the explicit construction for the last part. For any $\bar{u} \in \bar{U}$ we have that $\bar{u} = u_1^{k_1}u_2^{k_2}\cdots u_r^{k_r}$ for some. $u_i \in U, k_i \in \mathbb{Z}$. Then we have:
$$g\bar{u}g^{-1} = gu_1^{k_1}u_2^{k_2}\cdots u_r^{k_r}g^{-1} = (gu_1g^{-1})^{k_1}(gu_2g^{-1})^{k_2}\cdots (gu_rg^{-1})^{k_r} = v_1^{k_1}v_2^{k_2}\cdots v_r^{k_r} \in \bar{U}$$
Hence the proof that $\bar{U}$ is normal if $gug^{-1} \in U$ for every $g \in G, u \in U$