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

Statement
Let $$G$$ be a fact about::finite solvable group. Then, there exists a finite group $$H$$ that is a fact about::group having subgroups of all orders dividing the group order, and containing a subgroup isomorphic to $$G$$.

Facts used

 * 1) uses::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.