The alternating group on five elements is simple

The al­ter­nat­ing group \(A_5\) on five el­e­ments is sim­ple.

Proof

Re­call that \(A_5\) has or­der \(60\), so La­grange’s the­o­rem states that any sub­group of \(A_5\) has or­der di­vid­ing \(60\).

Sup­pose \(H\) is a nor­mal sub­group of \(A_5\), which is not the triv­ial sub­group \(\{ e \}\). If \(H\) has or­der di­visi­ble by \(3\), then by Cauchy’s the­o­rem there is a \(3\)-cy­cle in \(H\) (be­cause the \(3\)-cy­cles are the only el­e­ments with or­der \(3\) in \(A_5\)). Be­cause \(H\) is a union of con­ju­gacy classes, and be­cause the \(3\)-cy­cles form a con­ju­gacy class in \(A_n\) for \(n > 4\), \(H\) would there­fore con­tain ev­ery \(3\)-cy­cle; but then it would be the en­tire al­ter­nat­ing group.

If in­stead \(H\) has or­der di­visi­ble by \(2\), then there is a dou­ble trans­po­si­tion such as \((12)(34)\) in \(H\), since these are the only el­e­ments of or­der \(2\) in \(A_5\). But then \(H\) con­tains the en­tire con­ju­gacy class so it con­tains ev­ery dou­ble trans­po­si­tion; in par­tic­u­lar, it con­tains \((12)(34)\) and \((15)(34)\), so it con­tains \((15)(34)(12)(34) = (125)\). Hence as be­fore \(H\) con­tains ev­ery \(3\)-cy­cle so is the en­tire al­ter­nat­ing group.

So \(H\) must have or­der ex­actly \(5\), by La­grange’s the­o­rem; so it con­tains an el­e­ment of or­der \(5\) since prime or­der groups are cyclic.

The only such el­e­ments of \(A_n\) are \(5\)-cy­cles; but the con­ju­gacy class of a \(5\)-cy­cle is of size \(12\), which is too big to fit in \(H\) which has size \(5\).

Children:

Parents:

  • Alternating group

    The al­ter­nat­ing group is the only nor­mal sub­group of the sym­met­ric group (on five or more gen­er­a­tors).