Present and explain an example refuting the converse to Lagrange’s Theorem.
Solution
The group \(A_4\) has order \(12\), but it has no subgroup of order \(6\), even though \(6\) divides \(12\).
1. The group \(A_4\) has order \(12\) and contains eightdistinct elements of order \(3\).
The sign homomorphism \(S_4\to\{1,-1\}\) is surjective because a transposition has sign \(-1\). Its kernel \(A_4\) therefore has order \(4!/2=12\) [@DF04]. For a \(3\)-cycle, choose the three letters in four ways and then one of the two cyclic orientations. This gives exactly \(4\cdot2=8\) distinct \(3\)-cycles. Each is even, since \((a\,b\,c)=(a\,c)(a\,b)\), so all eight lie in \(A_4\). They all have order \(3\).
2. The group \(A_4\) has no subgroup of order \(6\).
If \(H\leq A_4\) had order \(6\), its index would be \(2\). Every index-two subgroup is normal: for an element outside \(H\), both its left coset and its right coset are the complement of \(H\), while an element of \(H\) has both cosets equal to \(H\). Thus the quotient map \(\pi:A_4\to A_4/H\) would be defined, with target of order \(2\).
For a \(3\)-cycle \(c\), the order of \(\pi(c)\) divides both \(3\) and \(2\), so \(\pi(c)=1\). Hence all eight distinct \(3\)-cycles from step 1 would belong to \(H\). Together with the identity they would give at least nine elements in a group of order six, a contradiction.
Lagrange’s theorem asserts that a subgroup order divides the group order. Here that divisibility holds for \(6\mid12\) but the requested subgroup does not exist, disproving the converse.