📂Abstract Algebra

Proof of Lagrange's Theorem

Theorem ¹

If $H$ is a subgroup of a finite group $G$, then $|H|$ is a divisor of $|G|$.

Proof

Upon slight consideration, it’s logically inevitable and the proof is accordingly simple.

Each coset has the same number of elements. Since $H$ is also one of the cosets of $G$, the cardinality of the cosets of $H$ is $|H|$. As the cosets partition $G$, summing the cardinalities of all cosets equals $|G|$. Let the number of cosets of $H$ be $( G : H )$, and denote this as $r$. $$|G| = |H| + \cdots |H| = r |H|$$ Thus, $|H|$ is a divisor of $|G|$.

■

Corollary

If $G$ is a finite group and $|G|$ is a prime number, then $G$ is a cyclic group.

For example, for the prime number $p$, $\mathbb{Z}_{p}$ is unmistakably a cyclic group.

Counterexample ²

The alternating group $A_{4}$ serves as a counterexample to the converse of Lagrange’s theorem.

Assuming the converse of Lagrange’s theorem holds, then for $\displaystyle | A_{4} | = {{4!} \over {2}} = 12$, there must exist a $|H| = 6$ satisfying $H \leqslant A_{4}$. Specifically, $A_{4}$ consists of three types, 12 kinds, of cyclic groups, hence, $H$ is formed of six of these cycles:

• The identity of length $1$: $$e$$
• Cycles of length $3$, and $3$-cycles: $$(1,2,3) \\ (1,3,2) \\ (1,2,4) \\ (1,4,2) \\ (1,3,4) \\ (1,4,3) \\ (2,3,4) \\ (2,4,3)$$
• Products of transpositions of length $2$, Klein four-group: $$(1,2)(3,4) \\ (1,3)(2,4) \\ (1,4)(2,3)$$ These along with only the identity $e$ gathered form $V \leqslant A_{4}$, which is isomorphic to the Klein four-group. In that sense, in this post, let’s just call them Klein four.

Looking into what relation $H$ has with these elements…

1. As $H$ is a group, it must include the identity element $e$, hence, technically one needs to choose $5$ cycles.
2. Assuming Lagrange’s theorem holds, there must exist a cycle whose cardinality is $2$, therefore, $H$ must include at least one Klein four.
3. Since there are only three Klein fours, $H$ must include at least two $3$-cycles.
4. If $\alpha \in H$, then it must be $\alpha^2 \in H$. However, since $(1,i,j)^{2} = (1,j,i)$, if $H$ includes $3$-cycles, it must have them in even numbers. It can’t have $6$ cycles given it must include the identity.

In summary, $H$ must have at least one Klein four and at least two $3$-cycles, notably, the $3$-cycles must be in even numbers, so the possible scenario is only whether the $3$ cycles are two or four. Let’s just consider the cases.

• Case 1: Two $3$-cycles, three Klein fours
  $$(a,b,c) \circ (a,c)(b,d) = (b,d,c) \notin H$$ Regardless of which cycle $(a,b,c)$ is included, since $H$ has all Klein fours, we obtain $(b,d,c)$, and hence, $H$ is not closed under the product of cycles.

• Case 2: Four $3$-cycles, one Klein four
  Without loss of generality, let one of the $3$-cycles in $H$ be $(a,b,c)$, and the Klein four be $(a,b)(c,d)$. The point is that $d$, not included in $(a,b,c)$, exists in the Klein four. $$(a,b,c) \circ (a,b)(c,d) = (c,d,a) = (a,c,d)$$ Therefore, $H$ must precisely have the $3$-cycles $(a,b,c), (a,c,b), (a,c,d), (a,d,c)$. However, $$(a,d,c) \circ (a,b)(c,d) = (b, d, a) = (a,b,d) \notin H$$ thus, $H$ is not closed under the product of cycles as well.

We have presented $H$ as a counterexample demonstrating that the converse of Lagrange’s theorem does not hold, confirming that $H$ cannot be a group.

■