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In the tradition of this question,

why is $\operatorname{PGL}(2,4)\cong A_5$?

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(As Jyrki correctly points out) that's an easy question because there is a natural action of $PSL_2(\mathbb F_q)$ on a (q+1)-point set. Less trivial one is why $PSL_2(\mathbb F_5)$ is also $A_5$. – Grigory M Aug 17 '11 at 7:34
That's also easy, since ${\rm PSL}(2,5)$ has $5$ Sylow $2$-subgroups (it can't have $3$, otherwise there would be a normal subgroup of index at most $6$, and then would have only one Sylow $5$-subgroup, which is not the case). – Geoff Robinson Aug 17 '11 at 8:15
Can we interpret $PGL(2,4)$ geometrically as the full group of rotational symmetries of the Icosahedron/Dodecahedron? – AnonymousCoward Aug 17 '11 at 8:45
@Geoff Sure, sure. But that's a proof that proves everything but explains nothing, in my opinion. (The real question is, more or less, what are 5 geometric objects PSL(2,5) acts upon?) – Grigory M Aug 17 '11 at 9:13
@Grigory:opinion is opinion. Not sure why a geometric proof "explains" and an algebraic proof does not. Things are as they are:we can try to understand them, but we can't always explain "why" they are as they are. – Geoff Robinson Aug 17 '11 at 11:10
up vote 2 down vote accepted

A minor comment: questions such as these maybe should be phrased slightly differently, in my view. In this case, the question is perhaps "what is a good explanation of this isomorphism, or an easy way to see it?"

An algebraic way to see the isomorphism is something like this. Note that ${\rm GL}(2,4)$ has a center of order $3$, consisting of scalar matrices. The group of scalar matrices of order dividing $3$ is a normal subgroup $N$ which intersects ${\rm SL}(2,4)$ in the identity, because no non-identity element of $N$ has determinant $1$. Hence we see that ${\rm PGL}(2,4)$ is isomorphic to ${\rm SL}(2,4)$. Also, it has order $60$. A Sylow $5$ subgroup $S$ of ${\rm SL}(2,4)$ permutes the Sylow $2$-subgroups of ${\rm SL}(2,4)$ by conjugation. None of these Sylow $2$-subgroups is stabilised by $S$, since they each have order $4$, and admit no automorphism of order $5$. Hence there are at least $5$ Sylow $2$-subgroups of ${\rm SL}(2,4).$ By Sylow's theorem, there are no more than that. ${\rm SL}(2,4)$ has no subgroup $M$ of index $2$, for such a subgroup would be normal, and would either have one or six Sylow $5$-subgroups. If there were one, it would be normal in ${\rm SL}(2,4)$, which is impossible (for example, an element of order $5$ in ${\rm SL}(2,4)$ has centralizer of order $5$, so the normalizer of a Sylow $5$-subgroup has order at most $20$). If there were $6$ subgroups of order $5$ in $M$, then there would only be $6$ elements in $M$ not of order $5$, and then $M$ would have a normal Sylow $3$-subgroup. But that would be normal in ${\rm SL}(2,4)$, while an element of order $3$ in ${\rm SL}(2,4)$ has centralizer of order $3$, so the normalizer of a Sylow $3$-subgroup in ${\rm SL}(2,4)$ has order at most $6$.

The permutation action of ${\rm SL}(2,4)$ on its Sylow $2$-subgroup yields a homomorphism from ${\rm SL}(2,4)$ into $A_5$ (since ${\rm SL}(2,4)$ has no subgroup of index $2$). We have already noted that no Sylow $2$-subgroup of ${\rm SL}(2,4)$ is normalized by an element of order $5$. ${\rm SL}(2,4)$ has no non-trivial normal $2$-subgroup (such a subgroup could not be a Sylow $2$-subgroup, as we have seen. On the other hand, if it had order $2$, it would be a central subgroup, and ${\rm SL}(2,4)$ has trivial center). We have also seen that a Sylow $3$-subgroup of ${\rm SL}(2,4)$ is not normal. Thus no non-identity element of ${\rm SL}(2,4)$ normalizes every Sylow $2$-subgroup. Hence ${\rm SL}(2,4)$ is isomorphic to a subgroup of $A_5$ and must be all of $A_5$ by order considerations.

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A starting point is that a 2-dimensional vector space over the field of 4 elements has 5 1-dimensional subspaces (=the projective line over $F_4$). The group $PGL(2,4)$ then acts on the set of those lines (well, $GL(2,4)$ acts naturally, but the scalar matrices keep the lines fixed setwise). Thus we have a homomorphism $f:PGL(2,4)\rightarrow S_5$. Moreover, if a linear mapping keeps all those 1-dimensional subspaces fixed (setwise), then that linear mapping has to be a scalar multiplication. So $f$ is injective. Thus the image of $f$ is of order 60, and hence must be equal to $A_5$.

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