$A_{5}$ is a simple group?

Question

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!

Answer 1

$A_n$ 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.

Answer 2

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.

Answer 3

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 $A_5$.