📂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 $|G|$. Specifically, the following holds.

$$ |G : H| = |G| / |H| $$

Proof

If you think about it a little, it can only hold by common sense, and the proof is correspondingly simple.

All cosets have the same number of elements. Since $H$ is also one of the cosets of $G$, the cardinality of the cosets of $H$ is $|H|$. Since the cosets form a partition of $G$, adding up the cardinalities of all cosets gives $|G|$. Letting the number of cosets of $H$, $|G : H|$, be $r$, $$ |G| = |H| + \cdots |H| = r |H| $$ therefore $|H|$ is a divisor of $|G|$.

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Corollary

For a finite group $G$, if $|G|$ is a prime number, then $G$ is a cyclic group.

As an example, for a prime number $p$, $\mathbb{Z}_{p}$ is a cyclic group without any need to check.

Counterexample ²

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

If we assume that the converse of Lagrange’s theorem holds, then since $\displaystyle | A_{4} | = {{4!} \over {2}} = 12$, there would exist an $H \leqslant A_{4}$ satisfying $|H| = 6$. Since $A_{4}$ is specifically composed of three kinds, 12 cyclic groups as follows, $H$ consists of 6 of these cycles.

• The identity element of length $1$: $$e$$
• The cycles of length $3$, the $3$-cycles: $$(1,2,3) \\ (1,3,2) \\ (1,2,4) \\ (1,4,2) \\ (1,3,4) \\ (1,4,3) \\ (2,3,4) \\ (2,4,3)$$
• The products of transpositions of length $2$, the Klein four: $$(1,2)(3,4) \\ (1,3)(2,4) \\ (1,4)(2,3)$$ The set $V \leqslant A_{4}$ collecting only these together with the identity element $e$ is isomorphic to the Klein four-group. In that sense, in this post let us simply call these the Klein four.

Now, examining what relationship $H$ has with these elements…

1. Since $H$ is a group, it must have the identity element $e$, so in effect we only need to choose $5$ cycles.
2. Under the assumption that Lagrange’s theorem holds, there must also exist an $H_{2} \subset H$ of cardinality $2$, so $H$ must contain at least one Klein four.
3. Since there are only three Klein fours, $H$ must contain at least two $3$-cycles.
4. If $\alpha \in H$, then $\alpha^2 \in H$ must hold. But since $(1,i,j)^{2} = (1,j,i)$, if $H$ contains $3$-cycles of length, they must exist in an even number. Since it must contain the identity element, it cannot have $6$ of them.

In summary, $H$ must have at least one Klein four and at least two $3$-cycles, and in particular the $3$-cycles must be even in number, so the only possible cases are whether there are two or four $3$-cycles. Let us just split into cases and think about it.

• Case 1. Two $3$-cycles, three Klein fours
  $$(a,b,c) \circ (a,c)(b,d) = (b,d,c) \notin H$$ Whichever cycle $(a,b,c)$ $H$ contains, since it has all the Klein fours, we can obtain $(b,d,c)$ as above, and therefore $H$ 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 of $H$ be $(a,b,c)$ and the Klein four be $(a,b)(c,d)$. The point is that $d$, which is not contained in $(a,b,c)$, is in the Klein four. $$ (a,b,c) \circ (a,b)(c,d) = (c,d,a) = (a,c,d) $$ Thus $H$ must have exactly the $3$-cycles $(a,b,c), (a,c,b), (a,c,d), (a,d,c)$. However, $$ (a,d,c) \circ (a,b)(c,d) = (b, d, a) = (a,b,d) \notin H $$ so again $H$ is not closed under the product of cycles.

As a counterexample showing that the converse of Lagrange’s theorem does not hold, we have confirmed that $H$ cannot be a group.

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