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How to just come quickly to the conclusion that $A_{5}$ is a simple group? Because it came in various competitive examinations, I understand that $A_{n}$ is generated by 3 cycles and the result will follow but I was thinking about some other approach by which I can show that $A_{5}$ is a simple group like by using any powerful theorem.

Also how about $GL(2,\mathbb{R})$ ?

Any help is great!

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    Show that it doesn't have a nontrivial normal subgroup. :P (But seriously, I don't know that there is a short simple proof. Here is an approachable one that is almost 4 pages long, with some details left to the reader: http://people.math.gatech.edu/~ecroot/A5.pdf ) – 2017-01-29
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    (Also, $\mathrm{GL}(2, \mathbb{R})$ is *not* simple) – 2017-01-29
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    As Morgan mentioned, $\mathbf{GL}(2, \mathbf R)$ is not simple. More generally, $\mathbf{GL}(n, \mathbf R)$ is not simple for all $n$. Hint: what's its center? – 2017-01-29
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    @AlexMacedo i think center is $\alpha . I$ , $\alpha \in \mathbb{Z}$ ? where $I$ is the identity matrix of order 2 . – 2017-01-29
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    @BAYMAX yes, if you mean $\alpha \in \mathbf R$ (instead of $\mathbf Z$). That's the way to go :-) – 2017-01-29
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    @AlexMacedo, yes yes i agree , then from center how to proceed ? – 2017-01-29
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    @BAYMAX the center is a normal subgroup – 2017-01-29
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    @AlexMacedo nice :) – 2017-01-29

3 Answers 3

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$A_n$ is a simple group for $n\ge 5$. The idea of the proof consists in showing that a normal subgroup of $A_n$ contains a $3$-cycle, hence it contains all $3$-cycles (by conjugation). As $A_n$ is generated by the $3$-cycles, the result follows.

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    For an actual proof, one can consider an element that is not a three cycle and has a maximal number of fixed points. Then one can show this can be conjugated to an element with more fixed points, hence the result. – 2017-01-29
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The following theorem is due to E. Galois and a proof can be found in Basic Algebra 1 by N. Jacobson:

Theorem. For all $n\geqslant 5$, the alternating group $A_n$ is simple.

Here are the main steps of the proof:

Proposition 1. For all $n\geqslant 3$, the alternating group $A_n$ is generated by the set of $3$-cycles.

Sketch of a proof. By induction on the number of fixed points, any element of $S_n$ is a product of transpositions. Moreover, since an element of $A_n$ has signature $1$ and a transposition has signature $-1$, the number of transpositions appearing in its decomposition must be even. To conclude, notice that any product of two distinct transpositions is a $3$-cycle. $\Box$

Proposition 2. For all $n\geqslant 5$, the $3$-cycles are conjugated in $S_n$.

Idea of a proof. Let $(a,b,c)$ and $(d,e,f)$ be two $3$-cycles. Since $n\geqslant 3$, there exists $\sigma\in S_n$ such that $\sigma(a)=d$, $\sigma(b)=e$, $\sigma(c)=f$ and using the conjugation formula, one has: $$\sigma(a,b,c)\sigma^{-1}=(d,e,f).$$ If $\sigma$ has signature $1$, we are done. Otherwise, since $n\geqslant 5$, there exists $i\neq j$ in $\{1,\cdots,n\}\setminus\{a,b,c\}$, therefore $\tau:=(i,j)\sigma$ is in $A_n$ and one has: $$\tau(a,b,c)\tau^{-1}=(d,e,f).$$ Whence the result. $\Box$

Proposition 3. Let $n\geqslant 5$ and $H\neq\{\textrm{id}\}$ be a normal subgroup of $A_n$, then there exists a $3$-cycle in $A_n$.

Sketch of a proof. The key point is that a $3$-cycle is an element of $A_n$ whose number of fixed points is maximal. $\Box$

Regarding $\textrm{GL}_2(\mathbb{R})$, $\textrm{SL}_2(\mathbb{R})$ is a nontrivial simple group of $\textrm{GL}_2(\mathbb{R})$, it is the kernel of the determinant.

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    What happens if $3-cycle$ is an element of $A_{n}$ , do we get any contradiction ? – 2017-01-29
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    You wrote "$A_n$ is not simple", but you meant to write that $A_n$ is simple (and hence not solvable). – 2017-01-29
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    The $3$-cycles are elements of $A_n$, they are even generators. This is the proposition 1. Notice that proposition $1$, $2$, $3$ give that a normal subgroup of $A_n$ which is not $\{\textrm{id}\}$ is $A_n$. – 2017-01-29
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    @AlexKruckman Thank you for the correction! This is now fixed! – 2017-01-29
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If you only care about $A_5$ and not all alternating groups, it's not too hard to see that it's simple if you happen to know its conjugacy class sizes: $1,\, 12,\, 12,\, 15,\, 20$.

If you don't remember their sizes (I didn't), conjugacy classes of $S_5$ are well-known and easy to work with:

\begin{array}{c|c|c|c|c|c|c} \text{representative} & 1 & (12) & (123) & (1234) & (12345) & (12)(34) & (12)(345) \\ \hline \text{size} & 1 & 10 & 20 & 30 & 24 & 15 & 20 \\ \text{parity} & \text{even} & \text{odd} &\text{even} & \text{odd} & \text{even} & \text{even} & \text{odd} \\ \text{splits} & \text{no} & & \text{no} & &\text{yes} &\text{no} & \end{array}

Here, we note the parity of elements in the conjugacy class, and whether this class splits in $A_n$

Recall that the conjugacy class $\operatorname{cl}_{S_n}(g)$ of an element $g$ in $S_n$ is not necessarily the same as it is in $A_n$: If $\operatorname{cl}_{S_n}(g) \neq \operatorname{cl}_{A_n}(g)$, the conjugacy class of $g$ in $A_n$ "splits" in half, one containing $g$, the other $g^{-1}$. If $g$ is made of cycles of distinct, odd lengths, then the conjugacy class splits; otherwise, it does not (see here, or here for a proof).

Now, $H \lhd G$ iff $H$ is a union of conjugacy classes of $G$. Just looking at the sizes of conjugacy classes of $A_5$, none of their unions (aside from the identity and whole group) could be a subgroup.

This method isn't really viable in general, but it is for $A_5$.