# Every finite solvable group is a subgroup of a finite group having subgroups of all orders dividing the group order

This article gives the statement, and possibly proof, of an embeddability theorem: a result that states that any group of a certain kind can be embedded in a group of a more restricted kind.
View a complete list of embeddability theorems

## Statement

Let $G$ be a Finite solvable group (?). Then, there exists a finite group $H$ that is a Group having subgroups of all orders dividing the group order (?), and containing a subgroup isomorphic to $G$.

## Facts used

1. ECD condition for pi-subgroups in solvable groups: This is an extended version of Sylow's theorem in finite solvable groups, stating that Hall subgroups of all permissible orders exist.
2. A cyclic group has subgroups of all orders dividing its order.

## Proof

Given: A finite solvable group $G$ of order $p_1^{k_1}p_2^{k_2} \dots p_r^{k_r}$, with $p_i$ prime and $k_i \ge 1$.

To prove: The direct product $H = G \times C_m$ has subgroups of all orders dividing its order, where $m = p_1^{k_1 - 1}p_2^{k_2 - 1} \dots p_r^{k_r - 1}$. Note that this direct product contains $G \times 1$, isomorphic to $G$, so this is sufficient.

Proof: $H$ has order:

$p_1^{2k_1 - 1}p_2^{2k_2 - 1} \dots p_r^{2k_r - 1}$.

Now, consider any divisor $d$ of the order of $H$, say:

$p_1^{j_1}p_2^{j_2} \dots p_r^{j_r}$.

We construct a subgroup of $H$ of this order. First, define $l_i$ as $j_i$ if $j_i < k_i$ and $j_i - k_i$ if $j_i \ge k_i$. Then, find a subgroup $L$ of $C_m$ of order:

$p_1^{l_1}p_2^{l_2} \dots p_r^{l_r}$.

Now, $j_i - l_i \in \{ 0, k_i \}$, so consider:

$p_1^{j_1 - l_1}p_2^{j_2 - l_2} \dots p_r^{j_r - l_r}$.

By fact (1), we conclude that $G$ has a normal subgroup $N$ of this order. Then, $N \times L$ is a subgroup of $H$ of order $d$, as required.