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I need to prove the following: Let $X$ be a set and $S_x$ the set of all bijections $f: X$ to $X$. Is $S_x$ a group under composition of functions?

I know that to prove something is a group I need to prove
$1)$ associativity,
$2)$ the existence of the identity,
$3)$ the existence of the inverse. I have already proved that the composition of functions is associative in a previous problem, though those functions were not bijective. I'm really stuck on the other two parts.

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    Are you comfortable with what the set of all bijections $X \to X$ is? For example, could you write them all down if the set is, say, $X = \{1,2,3\}$? – 2017-01-16
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    Yes, that doesn't seem to be my issue. I've never been perfect with that concept, but I think understand the basic meaning of bijections. – 2017-01-16
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    The "set of all bijections that maps a function from $X$ to $X$" --- that doesn't make sense. Do you mean "the set of all bijections from $X$ to $X$"? Clearing up linguistic misconceptions like this is a good first step for problems like this. – 2017-01-16
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    @JohnHughes I just edited it to be the problem to the best of my math typing ability. – 2017-01-16
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    Start with the easy ones. That's somewhat subjective, but most people will find that naming a bijection that works as identity is not hard. – 2017-01-16
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    OK. That's better. Now go ahead and as @Fabio suggests, propose a bijection that might serve as the identity element for "composition". – 2017-01-16
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    In previous problems using composition of functions, the identity has commonly been $x$. That's my gut reaction for this problem, as any of function composed with this one would result in said function. – 2017-01-16
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    Yes, you'd say that $f(x)=x$ is the identity. Now, are bijections invertible? – 2017-01-16
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    @FabioSomenzi Well shoot, chalk this up to working on little to no sleep while massively overcomplicating the problem. Thanks – 2017-01-16

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