Every Sylow subgroup is normal, then $G$ has a subgroup of order $m$ for every division $m$ of $|G|$

Some trouble working out an algebra problem.

Suppose that every Sylow subgroup of a finite group $$G$$ is normal. Prove that $$G$$ has a subgroup of order $$m$$ for every divisor $$m$$ of $$|G|$$.

• Title and question are two different things. – Randall Dec 9 '18 at 5:00

Let $$|G| = n \in \mathbb{N}$$ and let $$n = p_1^{k_1} \cdot p_2^{k_2} \cdots p_r^{k_r}$$ be the prime factor decomposition of $$n$$, where $$r \in \mathbb{N}, k_i \in \mathbb{N}$$ and $$p_i$$ is prime $$\forall i \in \{1,2,3,\cdots,r \}$$.
We know that if for every $$i \in \{1,2,3,\cdots,r \}$$ the $$p_i$$-Sylow subgroups of $$G$$ are normal, then there is a unique $$p_i$$-Sylow subgroup $$S_i$$ (because the $$p_i$$-Sylow subgroups are conjugate). This implies that there also exists an isomorphism $$f: S_1 \times S_2 \times \cdots \times S_r \to G,$$ with $$f( (x_1, x_2, \cdots, x_r)) = x_1x_2\cdots x_r.$$
Now, we also know that every $$p$$-group of order $$p^t, t \in \mathbb{N}$$ has a subgroup of order $$p^k$$ for each $$k \in \{0,1,2,3,\cdots,t \}$$ (A $$p$$-group of order $$p^n$$ has a normal subgroup of order $$p^k$$ for each $$0\le k \le n$$).
Now let $$m$$ be a divisior of $$n$$. Then, $$m = p_1^{i_1} \cdot p_2^{i_2} \cdots p_r^{i_r}$$, where $$i_s \in \{0,1,2,\cdots, k_s \}, \forall s \in \{1,2,3,\cdots,r \}$$.
Now, we have that the $$p_1$$-Sylow subgroup has a subgroup of order $$p_1^{i_1}$$, the $$p_2$$-Sylow subgroup has a subgroup of order $$p_2^{i_2}$$ and so on, every $$p_u$$-Sylow subgroup of $$G, u \in \{1,2,3,\cdots,r \}$$ has a subgroup of order $$p_u^{i_u}$$. The direct product from the isomorphism transfers this subgroup of $$S_1 \times S_2 \times \cdots \times S_r$$ to a subgroup of $$G$$.