I have prove that $PSL(2, 5)$ is isomorphic to $A_5$.

I understand, that I should do it with Sylow subgroups. There are 5 2-subgroups in $PSL(2, 5)$, but I can't describe them.

After that I can use group action on $PSL(2, 5)$ (conjugation with Sylow subgroups) and get homomorphism to $S_5$. Image of this homomorphism is our $A_5$.

So I have two problems with this prove : I can't describe Sylow subgroups of $PSL(2, 5)$ and don't understand how show that image of homomorphism is $A_5$.

I can't use the fact that $PSL(2, 5)$ is simple.

  • $\begingroup$ Your Sylow 2-subgroups have what order? Are they cyclic? You can answer this by finding one subgroup, since they are all conjugate. $\endgroup$ – Morgan Rodgers Oct 8 '17 at 19:29
  • $\begingroup$ Yes, I understand that, but I have problems with finding this subgroup. $\endgroup$ – wuzzapcom Oct 8 '17 at 19:35
  • $\begingroup$ Maybe, this question and the answer I gave here can have some interest. $\endgroup$ – Jean Marie Oct 8 '17 at 19:50

For this specific question, an explicit description of the Sylow $2$-subgroups is not necessary, instead you can just use the fact that $\operatorname{PSL}(2,5)$ is simple.

But if you want to do this by calculation, the image of the following set in $\operatorname{PSL}(2,5)$ is a $2$-Sylow:

$$\left \{ \pmatrix{ 1 & 0 \\ 0 & 1}, \pmatrix{ 1 & 0 \\ 0 & -1},\pmatrix{0 & 1 \\ 1 & 0},\pmatrix{ 0 & 1 \\ -1 & 0} \right \} $$

With explicit calculations you can show that the action on the conjugates is faithful and that the image of the homomorphism lies in $A_5$; hence you have an isomorphism between $\operatorname{PSL}(2,5)$ and $A_5$.

  • $\begingroup$ Thank you for your answer, but I should use Sylow subgroups in this prove. $\endgroup$ – wuzzapcom Oct 8 '17 at 19:32
  • $\begingroup$ With simplicity, you can show that the homomorphism $\operatorname{PSL}(2,5) \rightarrow S_5$ given by the action on $2$-Sylows is injective. Then it is easy to conclude that the image is equal to $A_5$. $\endgroup$ – spin Oct 8 '17 at 19:36
  • $\begingroup$ Yes, but my lecturer forbade this prove. $\endgroup$ – wuzzapcom Oct 8 '17 at 19:39
  • $\begingroup$ Thank you again. Could you explain how you got this 2-Sylow subgroup? $\endgroup$ – wuzzapcom Oct 8 '17 at 19:48
  • $\begingroup$ You can deduce that the Sylow subgroup has to be isomorphic to $C_2 \times C_2$, in particular generated by two commuting elements of order $2$. So just find an element of order $2$ in $\operatorname{PSL}(2,5)$ and compute its centralizer. $\endgroup$ – spin Oct 9 '17 at 8:02

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