Converse of Lagrange’s theorem does not hold

Lagrange’s theorem, states that for any finite group \(G\), the order (number of elements) of every subgroup \(H\) of \(G\) divides the order of \(G\) (denoted by \(\vert G \vert\)).

Lagrange’s theorem raises the converse question as to whether every divisor \(d\) of the order of a group is the order of some subgroup. According to Cauchy’s theorem this is true when \(d\) is a prime.

However, this does not hold in general: given a finite group \(G\) and a divisor \(d\) of \(\vert G \vert\), there does not necessarily exist a subgroup of \(G\) with order \(d\). The alternating group \(G = A_4\), which has \(12\) elements has no subgroup of order \(6\). We prove it below.

\(G\) consists of:

  • The identity or neutral element \(e\).
  • The three elements that are product of disjoint transpositions. Those \(3\) elements with \(e\) make up a subgroup \(V \subset H\) (\(V\) is isomorphic to the Klein four-group)
  • The eight 3-cycles.

Suppose that \(H \subset G\) is a subgroup of order \(6\) and that \(H^\prime\) denotes the intersection \(H \cap V\). \(H^\prime\) is a subgroup of \(H\) and \(V\). By Lagrange’s theorem, \(H^\prime\) order divides \(4\) and \(6\). So \(\vert H^\prime \vert\) is equal to \(1\) or to \(2\). If \(\vert H^\prime \vert=1\), the map \((h,v) \mapsto h \cdot v\) defined from \(H \times V\) to \(G\) is one-to-one. Which doesn’t make sense as \(G\) would have at least \(24\) elements. Therefore \(\vert H^\prime \vert=2\) and \(H\) is made up the identity \(e\), an element \(v\) which is product of two disjoint transpositions and six 3-cycles.

Also the index \(\vert G : H\vert\) is equal to \(2\) and consequently \(H\) is a normal subgroup of \(G\). We recall the argument. For \(a \in G \setminus H\) the left cosets \(H, aH\) form a partition of \(G\). Similarly, the right cosets \(H, Ha\) form a partition of \(G\). As \(aH \neq H\), we have \(aH =Ha\) which allows to conclude.

We denote \(v=(i,j)(k,l)\) with \(\{i,j,k,l\} = \{1,2,3,4\}\) and \(t\) the 3-cycle \((i,j,k)\). We have \(tvt^{-1}=(j,k)(i,l) \neq (i,j)(k,l)\) and \(tvt^{-1} \in H\) as \(H \triangleleft G\). In contradiction with the cardinality of \(\vert H^\prime \vert = \vert H \cap V \vert= 2\). We have finally proven that \(A_4\) has no subgroup of order \(6\).

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