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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 $|G|$.

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    $\begingroup$ Title and question are two different things. $\endgroup$ – Randall Dec 9 '18 at 5:00
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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 $G$.

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