A Converse to Lagrange's Theorem for Finite Abelian Groups

# A Converse to Lagrange's Theorem for Finite Abelian Groups

Recall from the Lagrange's Theorem page that if $G$ is a finite group and $H$ is a subgroup of $G$ then Lagrange's Theorem states that the order of $H$ must divide the order of $G$.

Thus for example, if $G$ is a group of order $12$ and $H$ is a subgroup of $G$ then $H$ must have order $1$, $2$, $3$, $4$, $6$, or $12$.

The next question to ask is if $G$ is a finite group, then for every positive divisor $d$ of $|G|$, does there exist a subgroup $H$ of $G$ with $|H| = d$? This is the converse of Lagrange's theorem, and unfortunately the answer is no. However, under certain circumstances it is true.

In the Theorem below we will see that the converse of Lagrange's theorem is true for finite abelian groups.

Later on the The Sylow Theorems page we will see that if $p$ is a prime and $|G| = p^km$ with $p \nmid m$ then indeed there is a subgroup of $H$ of $G$ with $|H| = p^k$.
 Theorem 1: Let $G$ be a finite group. If $G$ is abelian then for every positive divisor $d$ of $|G|$ there exists a subgroup $H$ of $G$ with $|H| = d$.

## Example 1

Let $G$ be a group of order $p$ where $p$ is a prime. Then $G$ must be isomorphic to the cyclic group $\mathbb{Z}_p$. Since every cyclic group is abelian, we see that $G$ satisfies the requirements for Theorem 1. So for every positive divisor $d$ of $|G| = p$ there is a subgroup of order $d$.

But $p$ is a prime number, so if $d > 0$ is such that $d \mid p$ then $d = 1$ or $d = p$. Indeed, the trivial group and whole group can be taken as the group $H$. So in this particular example, Theorem 1 doesn't really say much.

## Example 2

Let $G = K_4 = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Then $G$ is a group of order $4$ (this group is called the Klein Four-Group, and is the direct product of $\mathbb{Z}_2$ and $\mathbb{Z}_2$). Since $\mathbb{Z}_2$ is abelian, so is $K_4$.

So by Theorem 1, for $d = 2 \mid 4 = |K_4|$ there exists a subgroup $H$ of $K_4$ with $|H| = 2$. Indeed, $H = \mathbb{Z}/2\mathbb{Z}$ is a subgroup of $H$ with $|H| = 2$.