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removed a "clearly"
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DavidLHarden
DavidLHarden
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The most direct answer I know for this is the following:

Any maximal subgroup of a finite solvable group has index equal to a power of a prime. The unsolvability of $A_{5}$ is provable, therefore, by exhibiting a maximal subgroup of index 6 or 10 in $A_{5}$. These are both easy to see in terms of the geometry of the icosahedron.
The icosahedron has 12 vertices, so they come in 6 opposite pairs. The stabilizer of one of them has index 6, and it's obvious that this subgroup of index 6 is maximal because (considering a rotation of order 5 about one opposite pair of vertices) the action of $A_{5}$ on these 6 opposite pairs is clearly doubly transitive (and therefore primitive).
Likewise, the icosahedron has 20 faces, which come in 10 opposite pairs. To prove this action is primitive, note that the point stabilizer has elements of order 3 which fix only the pair of opposite faces we have chosen. Using only this, we conclude that the point stabilizer has suborbit lengths equal to 1 and some multiples of 3 (in fact, we get 10=1+3+6 from the suborbit lengths). Since 10 has no proper divisors congruent to 1 mod 3, this must be a primitive action.

The most direct answer I know for this is the following:

Any maximal subgroup of a finite solvable group has index equal to a power of a prime. The unsolvability of $A_{5}$ is provable, therefore, by exhibiting a maximal subgroup of index 6 or 10 in $A_{5}$. These are both easy to see in terms of the geometry of the icosahedron.
The icosahedron has 12 vertices, so they come in 6 opposite pairs. The stabilizer of one of them has index 6, and it's obvious that this subgroup of index 6 is maximal because (considering a rotation of order 5 about one opposite pair of vertices) the action of $A_{5}$ on these 6 opposite pairs is clearly doubly transitive (and therefore primitive).
Likewise, the icosahedron has 20 faces, which come in 10 opposite pairs. To prove this action is primitive, note that the point stabilizer has elements of order 3 which fix only the pair of opposite faces we have chosen. Using only this, we conclude that the point stabilizer has suborbit lengths equal to 1 and some multiples of 3 (in fact, we get 10=1+3+6 from the suborbit lengths). Since 10 has no proper divisors congruent to 1 mod 3, this must be a primitive action.

The most direct answer I know for this is the following:

Any maximal subgroup of a finite solvable group has index equal to a power of a prime. The unsolvability of $A_{5}$ is provable, therefore, by exhibiting a maximal subgroup of index 6 or 10 in $A_{5}$. These are both easy to see in terms of the geometry of the icosahedron.
The icosahedron has 12 vertices, so they come in 6 opposite pairs. The stabilizer of one of them has index 6, and it's obvious that this subgroup of index 6 is maximal because (considering a rotation of order 5 about one opposite pair of vertices) the action of $A_{5}$ on these 6 opposite pairs is doubly transitive (and therefore primitive).
Likewise, the icosahedron has 20 faces, which come in 10 opposite pairs. To prove this action is primitive, note that the point stabilizer has elements of order 3 which fix only the pair of opposite faces we have chosen. Using only this, we conclude that the point stabilizer has suborbit lengths equal to 1 and some multiples of 3 (in fact, we get 10=1+3+6 from the suborbit lengths). Since 10 has no proper divisors congruent to 1 mod 3, this must be a primitive action.

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DavidLHarden
DavidLHarden
  • 3.7k
  • 22
  • 28

The most direct answer I know for this is the following:

Any maximal subgroup of a finite solvable group has index equal to a power of a prime. The unsolvability of $A_{5}$ is provable, therefore, by exhibiting a maximal subgroup of index 6 or 10 in $A_{5}$. These are both easy to see in terms of the geometry of the icosahedron.
The icosahedron has 12 vertices, so they come in 6 opposite pairs. The stabilizer of one of them has index 6, and it's obvious that this subgroup of index 6 is maximal because (considering a rotation of order 5 about one opposite pair of vertices) the action of $A_{5}$ on these 6 opposite pairs is clearly doubly transitive (and therefore primitive).
Likewise, the icosahedron has 20 faces, which come in 10 opposite pairs. To prove this action is primitive, note that the point stabilizer has elements of order 3 which fix only the pair of opposite faces we have chosen. Using only this, we conclude that the point stabilizer has suborbit lengths equal to 1 and some multiples of 3 (in fact, we get 10=1+3+6 from the suborbit lengths). Since 10 has no proper divisors congruent to 1 mod 3, this must be a primitive action.