Abstract Nonsense

Crushing one theorem at a time

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.

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April 19, 2011 - Posted by | Group Theory, Uncategorized | , , ,

2 Comments »

  1. […] lemma that if is the Sylow -subgroup of and the Sylow -subgroup of then . We thus have by a previous theorem that but since and (and are primes) that and so that […]

    Pingback by Groups of Order pq (pt. I) « Abstract Nonsense | April 19, 2011 | Reply

  2. […] that and so is cyclic as claimed. Now since all the Sylow -subgroups of are normal we know from basic Sylow theory that where is the unique element of . But since each is cyclic and for we have from basic […]

    Pingback by Unit Group of a Finite Field is Cyclic « Abstract Nonsense | September 21, 2011 | Reply


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