# Abstract Nonsense

## Relation Between Sylow’s Theorems and Direct Product

Point of Post: In this post we discuss an interesting relationship between Sylow’s Theorems and the direct product of groups. Namely, we’ll show that all the Sylow subgroups of a group are normal if and only if the group is isomorphic to the direct product of those Sylow Subgroups.

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Motivation

Often it happens, finagling around with the conditions on the number of Sylow $p$-subgroups of some group $G$ that one finds that $\#\left(\text{Syl}_p(G)\right)=n_p=1$ for every prime $p$ that divides $G$. We shall show that this is a very fortuitous thing because then $G$ will be isomorphic to the direct product of those Sylow subgroups.

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A Lemma About Direct Product of Groups

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We shall now, for our convenience, prove a lemma similar to a previous one which characterizes the direct product of groups. Indeed:

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Lemma: Let $G$ be a group and $H,K$ be subgroups of $G$ such that

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\begin{aligned} &\mathbf{(1)}\quad H,K\unlhd G\\ &\mathbf{(2)}\quad HK=G\\ &\mathbf{(3)}\quad H\cap K=\{e\}\end{aligned}

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Then, $G\cong H\times K$.

Proof: By a previous theorem it suffices to prove that the elements of $H$ and $K$ commute and every element of $G$ can be written uniquely as the product of something in $H$ and something in $K$. To see that the elements of $H$ and $K$ commute we merely note that for any $h\in H$ and any $k\in K$ by assumption that $H\unlhd G$ we have that $khk^{-1}\in H$ and so $khk^{-1}h^{-1}\in H$. But, with equal validity since $K\unlhd G$ we have that $hk^{-1}h^{-1}\in K$ and so $khk^{-1}h^{-1}\in K$. Thus, $khk^{-1}h^{-1}\in H\cap K$ and so by assumption $khk^{-1}h^{-1}=e$ and so $kh=hk$.

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Now, to see that every element of $G$ can written uniquely as the product of something in $H$ and something in $K$ we first note that by $\mathbf{(2)}$ it suffices to prove uniqueness. But, this is clear since if $hk=h_1k_1$ then $h_1^{-1}h=k_1k^{-1}$ and so evidently $h_1^{-1}h\in H$ and $h_1h^{-1}=k_1k^{-1}\in K$ so that by assumption $h_1h^{-1}=e$ or $h_1=h$. Similarly we see that $k_1=k$. The theorem then follows by previous discussion. $\blacksquare$

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The Theorem

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We now proceed to prove the aforementioned theorem. Namely:

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Theorem: A finite group $G$ is isomorphic to the direct sum of its Sylow subgroups if and only if every Sylow subgroup is normal in $G$.

Proof: Clearly if $G$ is isomorphic to the direct product of its Sylow subgroups then they are evidently normal in $G$.

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Conversely, suppose that $|G|=p_1^{\alpha_1}\cdots p_n^{\alpha_n}$ (the unique decomposition of $|G|$ into the product of distinct primes) and let $P_j$ be the Sylow $p_j$-subgroup of $G$. We note first that

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$\left|P_1\cap P_2\right|\mid \left(|P_1|,|P_2|\right)=\left(p_1^{\alpha_1},p_2^{\alpha_2}\right)=1$

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and so $P_1\cap P_2=\{e\}$. We note then that since $P_1,P_2\unlhd G$ we have that $P_1P_2\leqslant G$, and since $P\cap P_2=\{e\}$, and $P_1,P_2\unlhd G$ we  have by our lemma that $P_1P_2\cong P_1\times P_2$. Similarly, one has that (recalling that since $P_1\cap P_2=\{e\}$ one has that $|P_1P_2|=|P_1||P_2|$)

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$\left|P_3\cap P_1P_2\right|\mid \left(|P_3|,|P_1P_2|\right)=\left(p_3^{\alpha_3},p_1^{\alpha_1}p_2^{\alpha_2}\right)=1$

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and so $P_1\cap P_2P_3=\{e\}$. Thus, evidently using the same reasoning as the previous step $P_1P_2P_3\leqslant G$, $P_1,P_2P_3\unlhd G$ and so

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$P_1P_2P_3\cong P_3\times(P_1P_2)\cong P_3\times (P_1\times P_2)\cong P_1\times P_2\times P_3$

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Moreover (obviously since the above isomorphism holds) it’s true that $|P_1P_2P_3|=|P_1||P_2||P_3|=p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3}$. Continuing the logic we see that $P_1\cdots P_n\leqslant G$ and $P_1\times P_n\cong P_1\times\cdots\times P_n$. But, since

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$\left|P_1\cdots P_n\right|=|P_1|\cdots|P_n|=p_1^{\alpha_1}\cdots p_n^{\alpha_n}=|G|$

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we may conclude that $P_1\cdots P_n=G$ and so the conclusion follows. $\blacksquare$

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References:

1. Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.