This is a question I have when I read proof of "group of order $15$ is cyclic". So let $H$ and $K$ be Sylow $p$-group of $G$ with order $3$ and $5$ respectively. I understand that they are normal subgroups and $H\cap K = \{e\}$ since $\gcd(3,5)=1$. But I just don't see why $HK = G$? The best I can do is to conclude that $HK$ is a subgroup of $G$.
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I don't get what you mean by juxtaposition of $H$ and $K$. But if you mean the order of the two groups equals the order of $G$, then you should see Lagrange theorem about the order of groups. Proofs of it are abundant, and essentially your question. Perhaps you have trouble with a specific are of proving it? There are most likely questions on it. – marshal craft Mar 19 '17 at 03:16
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We know that $H K$ is a subgroup of $G$ that contains $H$ and $K$. By Lagrange's theorem, the order of $H K$ is divisible by both $3$ and $5$. Therefore, $H K$ is of order $15$ so $H K = G$.
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Denote $H=\{h_1,h_2,h_3\}$ and $K=\{k_1,\dots,k_5\}$. Since $H\cap K = \{e\}$, if $h_{i_1}k_{j_1} = h_{i_2}k_{j_2}$, $\underbrace{h_{i_2}^{-1} h_{i_1}}_{\in H} = \underbrace{k_{i_2} k_{i_1}^{-1}}_{\in K} \in H\cap K = \{e\}$, so $h_{i_1}=h_{i_2}$ and $k_{j_1}=k_{j_2}$. In other words, an element in $HK$ is uniquely determined by $h_i$ and $k_j$. Since $|HK| = |H||K| = 3 \times 5 = 15 = |G|$ and $HK$ is a subgroup of $G$, we have $HK = G$.
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