## 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.

*Motivation*

Often it happens, finagling around with the conditions on the number of Sylow -subgroups of some group that one finds that for every prime that divides . We shall show that this is a very fortuitous thing because then will be isomorphic to the direct product of those Sylow subgroups.

*A Lemma About Direct Product of Groups*

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

**Lemma: ***Let be a group and be subgroups of such that*

*Then, .*

**Proof: **By a previous theorem it suffices to prove that the elements of and commute and every element of can be written uniquely as the product of something in and something in . To see that the elements of and commute we merely note that for any and any by assumption that we have that and so . But, with equal validity since we have that and so . Thus, and so by assumption and so .

Now, to see that every element of can written uniquely as the product of something in and something in we first note that by it suffices to prove uniqueness. But, this is clear since if then and so evidently and so that by assumption or . Similarly we see that . The theorem then follows by previous discussion.

**The Theorem**

We now proceed to prove the aforementioned theorem. Namely:

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

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

Conversely, suppose that (the unique decomposition of into the product of distinct primes) and let be the Sylow -subgroup of . We note first that

and so . We note then that since we have that , and since , and we have by our lemma that . Similarly, one has that (recalling that since one has that )

and so . Thus, evidently using the same reasoning as the previous step , and so

Moreover (obviously since the above isomorphism holds) it’s true that . Continuing the logic we see that and . But, since

we may conclude that and so the conclusion follows.

**References:**

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

[…] 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 […]

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[…] 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 […]

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