# Confusions in group theory

So I have two questions $1)$ on multiplicative inverses, and $2)$ on group action.

$1)$ Why is the group action $GL(n,\mathbb{R})$ on to a set $\mathbb{R}^n$ faithful but not transitive?

$2)$ Why is the symmetric group $S_n$ acting on $X=\{1,2,3,\dots,n\}$ faithful and transitive?

This chapter on group actions really knocking my head out, I dont get what's happening at all, can anyone help me clarify the above examples?

Next I want to ask about multiplicative inverses. What are multiplicative inverses in some set $R$ exactly? For instance is it any set of $x$ that satisfies the inequality $ax= 1\bmod n$ ,for $x \in R$?

• Yes that is what I mean, my mistake I will edit it. – Aurora Borealis Nov 28 '17 at 11:44

1) Because there is no $g\in GL(n,\mathbb{R})$ such that $g.(0,0,\ldots,0)=(1,0,\ldots,0)$.
2) If $p,q\in\{1,2,\ldots,n\}$ (with $p\neq q$), let $\tau$ be the element of $S_n$ such that $\tau(p)=q$, $\tau(q)=p$ and $\tau(o)=o$ if $o\in\{1,2,\ldots,n\}\setminus\{p,q\}$. Then $\tau(p)=q$. Therefore, the action is transitive. And it is faithful because, by the definition of $S_n$, if $\sigma\in S_n\setminus\{e\}$, then there is a $p\in\{1,2,\ldots,n\}$ such that $\sigma(p)\neq p$. Indeed (and by the same argument) when a group $G$ is the group of all bijections from a set $X$ into itself, then the natural action of $G$ on $X$ is faithful.