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I have tried to solve the Exercise 8.12, page 227, in Joseph J. Rotman, An Introduction to the Theory of Groups, Fourth Edition, Graduate texts in Mathematics, Springer-Verlag, New York (1995).

The exercise is: Prove that $$PSL(2,9)\cong A_6.$$

Let $G:=PSL(2,9)$, I can see that $|G|=360=2^3.3^2.5$ and it is enough to show that $n_5\ne 36$ (where $n_5$ be the number of Sylow $5-$subgroups) but I stuck here.

Thanks.

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Thank you very much. –  Q.TL Nov 20 '13 at 4:32
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But note that the earlier SE question is about proving that there is a unique simple group of order 360. A lot of that proof was about finding the number of Sylow $p$-subgroups, whether or not they are cyclic, etc. You don't need to do any of that here, because you are given two specific groups. –  Derek Holt Nov 20 '13 at 8:55

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All we need to do is to construct a subgroup of index 6 in $PSL_2(\mathbb F_9)$ (indeed, such a subgroup gives a non-trivial homomorphism to $S_6$; since $PSL_2(\mathbb F_9)$ and $A_6$ are both simple of the same order, this is an isomorphism).

Now there is a beautiful geometric construction of a (non-trivial) map $A_5\hookrightarrow PSL_2(\mathbb F_9)$: $A_5$ is the group of rotations of the dodecahedron so we have a map $A_5\hookrightarrow SO(3)\cong PSU_2\subset PSL_2(\mathbb C)$; this map is defined over $\mathbb Z[i,\frac12,\phi]$ (where $\phi$ is the golden ratio), and reduction mod 3 gives a map $A_5\hookrightarrow PSL_2(\mathbb F_3[i]\cong\mathbb F_9)$. Voilà.

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(Cf. math.stackexchange.com/a/119373 which uses reduction mod 5 instead to prove that $PSL_2(\mathbb F_5)\cong A_5$.) –  Grigory M Nov 26 '13 at 15:15

An easy argument is given in p.52, R.A.Wilson, "The finite simple groups", Graduate texts in mathematics, Springer-Verlag.

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