A group of functions and a subgroup $H$ in it Let $A$ be a non-empty set and consider $S(A)$ the set of all bijective functions from $A$ to itself, with composition as group operation. Fix $a_0 \in A$. Define $H = \{f \in S(A): f(a_0) = a_0\}$. Prove that $H$ is a subgroup of $S(A)$.
does this tell me $a_0$ is an identity or that $f(a_0)$ is an identity?
I am assuming I need to use something to do with subgroup theorems but I am having trouble understanding how to use these.

EDIT/ADDED
Ok that makes more sense since $f(a_0)$ = $a_0$ and $g(a_0)$ = $a_0$ we can conclude that $f^{-1}$ $f(a_0)$ = $f^{-1}$ $a_0$   so we know $f^{-1}$ is in the set?   So $f^{-1}$ $g(a_0) $= $a_0$  and the by the theorem we may conclude $H$ is a subgroup of G? 

As a side note would love to know WHERE you guys find the element signs and how to put subscripts on things if you could link me to a list of stuff you can put into here with some instructions. 
Someone has pointed me a direction a few times but i have never found something I understood. I am not a programer but if I can find a lost of things I am sure I can figure it out usually I just search until I find a post with what I want click edit and take the symbols I want but if theirs instructions somewhere I would very much appreciate that.
 A: $H = \{f \in S(A): f(a_0) = a_0\}$ is the set of bijective functions in $S(A)$ such that $f$ maps $a_0$ to $a_0$.
Here, $a_0$ is a fixed element which every bijective function in $S(A)$ that maps $a_0$ to itself is a member. $H$ includes the identity map/function, but not all $f \in H$ are the identity function.
To prove $H$ is a subgroup of $S(A)$: 


*

*Clearly, $H$ includes as its identity $f_I$, the identity map of $S(A)$, since $f_I(a_0)$ is clearly $a_0$.

*Now show that for any $f, g \in H$, $f\circ g \in H$: i.e., show that $H$ is closed under the function composition. 

*Show that for any $f \in H, f^{-1} \in H$. Hint: This must be the case, as every $f \in H$ is bijective, and if $f \in H$, then $f^{-1}$ exists, and since $f\in H$, $f(a_0) = a_0$ and $f^{-1}(f(a_0)) = f^{-1}(a_0) = f_I(a_0) = a_0$, since by definition, $f^{-1}\circ f = f_I$, so $f^{-1} \in H$


Then you will have shown $\,H\,$ is a subgroup of $\,S(A)$.

As for your second question (or aside), you might want to visit MathJax/LaTeX: a tutorial and "bookmark" it as a favorite!
A: Hint: To check something is a subgroup, it suffices to show that $H$ is nonempty (this clear since the identity function is an element [why?]), the composition of two functions is an element of the group (also easy to argue since every function fixes $a_0$), and that the inverse of a function is an element of the group (which follows because each element is a bijection).
I leave the details to you, but hopefully with what is given above, you can work through this mostly on your own.
A: The standard way to show that something is a subgroup is to show:


*

*It is nonempty

*If $f, g \in H$ then $fg^{-1} \in H$


I'll leave (1) to you and give you a hint for (2).  If $f, g \in H$ then you know $f(a_0) = a_0$ and $g(a_0) = a_0$.  Apply the function $fg^{-1}$ to $a_0$.  The result should be $a_0$ and once you've shown this you can conclude that $fg^{-1} \in H$.
