Question: Any simple group of order 60 is isomorphic to $A\_5$

Let $G$ be a simple group of order $60$ 

.(**Assumption**) 
$G$ has a subgroup of order 12 say $H$ .

Then by Extended Cayley's Theorem $\exists f:G\rightarrow S\_5$ such that $ker f\subset H$. Since $G$ is simple $ker f=\{e\}$ ;hence $G$ is isomorphic to a subgroup $T$ of $S\_5$

I think if I can show that $T=A\_5$ I am done .But how to show that?any help

Answer: 
From the [Extreme Value Theorem](http://en.wikipedia.org/wiki/Extreme_value_theorem):

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> For any continuous $f$ defined on $[a,b]$, there exists $c\_1,c\_2\in[a,b]$ such that for all $x\in[a,b]$ $f(c\_1)\leq f(x)\leq f(c\_2)$.
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We get a continuous version of Rolle:

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> If $f$ is continuous on $[a,b]$ and $f(a)=f(b)$ then there exists an $c\_0\in(a,b)$ such that $f(c\_0)$ is either the minimum or the maximum value of $f$ on $[a,b]$.
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> 
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**Proof**: If $f(x)$ is constant on $[a,b]$ then you can pick any $x\_0\in(a,b)$.

Otherwise, assume $f(c)>f(a)$ for some $c\in(a,b)$. (The second case, $f(c)<f(a)$, is the same.) By the Extreme Value Theorem, there exists $c\_0$ such that $f(c\_0)\geq f(x)$ for all $x\in[a,b]$. Since $f(c\_0)\geq f(c)>f(a)=f(b)$, we know that $c\_0\neq a,b$, so $c\_0\in(a,b)$.

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Rolle's theorem follows from this by the proof that, if $f$ is differentiable on $(a,b)$ then a maximum or minimum on $(a,b)$ must have derivative zero.
There is a theroem that states something pretty close to what you want :

Let $a, b \in \mathbb{R}, a<b$ and $f:[a,b]\to\mathbb{R}$ continous over $[a,b]$. 

Then, $\exists m, M \in [a, b]$ such that $f([a, b])=[m, M]$ and hence :
$$\forall x \in [a, b], m \leq f(x) \leq M$$

However, from the moment $f$ is not continuous, this is not necessarily true. e.g. $f(x) = \frac{1}{x^2}, f(-2) = f(2)$ but it has no maximum on $[-2, 2]$.

Note also that Rolle's theorem is different, it does not say that there is a maximum but more specifically that there is a value for which the derivative is $0$ (which is why it needs the function do be differentiable). 

If you allow $±\infty$ to be maximum/minimum values, this won't change anything. They can be infimum and supremum but not maximum and minimum because your function will never reach it.

Another example is for instance the function $1-e^{-x}$. It will never reach one, so one is not a maximum.