I have just started going though "An Introduction to the theory of groups" by J.J Rotman. I have questions above the following two exercises:
" The identity function $1_X$ on a the set $X$ is a permutation. Show that $1_x \alpha = \alpha 1_x$ for all $\alpha \in S_X$
My Proposed solution :
The identity map $ 1_X : X \to X $ is defined as : $$1_X(x) = x: \forall x \in X$$
Now if we have some $\alpha \in S_X$ then the product: $$ \alpha 1_X ( x) = \alpha ( 1_X( x) ) $$
$$ = \alpha ( x) :\forall x \in X$$
and :
$$ 1_X \alpha (x) = 1_X ( \alpha(x)) $$ $$ = \alpha (x) : \forall x \in X $$
so $1_X \alpha = \alpha 1_X : \forall \alpha\in S_X $
Is this sufficient enough?
"For each $\alpha \in S_X$, prove that there is $ \beta \in S_X $ with $\alpha\beta =1 = \beta \alpha $"
Now obviously $\alpha $ is a bijection and thus there exist an inverse and this inverse is a element of $S_X$ if we let $\beta = \alpha^{-1} $ then the above is true. However I am unsure how to rigorously express this.
For every $y \in X $ there exists an unique $x \in X$ such that $\alpha(x) = y$ , so define $$\beta(y) = \beta(\alpha(x))= x $$ This is a good definition because $\alpha$ is bijective, and then you have to verify that $\beta\alpha = \alpha\beta = 1$