Is finite group isomorphic with the direct prouct of the Sylow $p$ subgroups?

Question

My question is the like the title

'($\forall$ finite group, G) $\simeq$ ($\Pi$ the Sylow $p$ subgroups of the G)'

I want to know this statement holds or not.(I'm not sure my proof is correct)

My proof) I'm not sure my trial is correct or not.

When the $\vert G\vert = p^n$, it is trivial that G is isomorphic with sylow p subgroup itself.

Let $\vert G \vert$ = $p_{1}^n p_{2}^m k$ for prime $p_1,p_2$ and $k\neq 1$ with $(p_1, k) = (p_2,k) =1$

Then, $\exists P_1$ and $P_2$ $s.t.$ $\vert P_1 \vert = p_1^n$ and $\vert P_2 \vert = p_2^m$ respectively.

(By sylow First thm, $P_1$ and $P_2$ are sylow subgroups.)

Also $P_1 \cap P_2 = \{ e \}$ ($e$ means identity of the $G$)

And clearly, $P_1, P_2 \leq G$ which means subset(subgroups)

Hence, $\vert P_1P_2 \vert$ = ${\vert P_1 \vert \vert P_2 \vert} \over \vert P_1 \cap P_2 \vert$ = $\vert P_1 \vert \vert P_2 \vert$

Therefore $k$ can be prime decomposed by the other primes, $p_i(\neq p_1, p_2)$

Same process Like the above, We can conclude

$\vert G \vert$ = $\Pi _1 ^n \vert P_i \vert \Rightarrow G \simeq \Pi _1 ^n \ P_i$ (Since each $P_i$ are subsets of the G )

Answer 1

Your statement is incorrect. The correct statement is:

A finite group is isomorphic to the direct product of it's Sylow subgroups if and only if all of the group's Sylow subgroups are normal.

See for instance Theorem 3.3 of this blurb of Keith Conrad.

Answer 2

This is false for $S_3$ for example. Your proof shows that the group is the internal product of its Sylow $p$-subgroups, which is true but much weaker than showing it is the direct product.

Answer 3

Others already explained that you don't get a direct product of groups for that needs the elements from distinct subgroups to commute (or both to be normal). You won't get even a semi-direct product for that would require one factor to be normal.

There is another fundamental problem in your approach. You won't even get a set of the prescribed size.

A building block in your argument seems to be the formula that if $H\le G, K\le G$, then we can form the set $$ HK=\{hk\mid h\in H, k\in K\}, $$ and it is an easy exercise to show that the size of this set is $$ |HK|=\frac{|H|\cdot|K|}{|H\cap K|}.\qquad(*) $$ But then you run into the following problem:

Formula $(*)$ is guaranteed to hold only if $H$ and $K$ are both subgroups.

You are still ok, when you form the product of two Sylow subgroups $P_1P_2$ (belonging to distinct primes $p_1\neq p_2$. But, $P_1P_2$ is not necessarily a subgroup, so when you add a third prime things may break down, and you don't necessarily have $|P_1P_2P_3|=|P_1|\cdot|P_2|\cdot|P_3|$.

As an example of this consider the group $A_5$ of order $60$. Let $$ \begin{aligned} P_1&=\langle (12)(34), (13)(24)\rangle,\ \text{a Sylow $2$-subgroup,}\\ P_2&=\langle (135)\rangle,\ \text{a Sylow $3$-subgroup,}\\ P_3&=\langle (14352)\rangle,\ \text{a Sylow $5$-subgroup.} \end{aligned} $$ When forming the products the following happens: $$ ((12)(34))(135)=(14352). $$ Meaning that some elements of $P_3$ already belong the product $P_1P_2$. It follows that there is more than one way of writing $1_G$ in the form $1_G=x_1x_2x_3$ with $x_i\in P_i, i=1,2,3$. Consequently the set $P_1P_2P_3$ has less than the expected $4\cdot3\cdot5$ elements, and you have failed to get all $60$.

• In my example it is possible to "fix" the approach by using $P_2'=\langle (123)\rangle$ instead of $P_2$. For then both $P_1$ and $P_2'$ would be subgroups of a copy of $A_4$, and the product $P_1P_2'$ would necessarily be all of $A_4$. In other words, a subgroup. Implying that $(P_1P_2')P_3$ would be all of $A_5$ (but still neither a direct or even a semi-direct product).
• But, it is impossible to similarly "fix" all the cases. For example, the order in which the prime factors are processed will matter. If you want to make a subgroup of order $20$ as a product of a Sylow $2$-subgroup and a Sylow $5$-subgroup you run into the obstacle that $A_5$ has no subgroups of order $20$. Proof left as an exercise (consider the left action of $A_5$ on the cosets of such a subgroup to get a homomorphism from $A_5$ to $S_3$ and use the fact that $A_5$ is simple).
• In general the set $HK$ is a subgroup if and only if the sets $HK$ and $KH$ are the same set. This holds if at least one of $H$, $K$ is a normal subgroup of $G$. In general we don't have any normal Sylow subgroups, so your rocket may not even lift off let alone proceed to the second stage.

Answer 4

Consider the symmetric group $S_3$

If this group was a direct product of two of its Sylow subgroups $P_2$ and $P_3$, we would have

$$S_3 \cong P_2 \times P_3 \cong C_2\times C_3\cong C_6$$

and then $S_3$ is an abelian group, contradiction.

Groups for which this property holds are called nilpotent groups. The following statements are equivalent for a finite group $G$:

(1) $G$ is nilpotent

(2) Every Sylow subgroup of $G$ is normal.

(3) Every maximal subgroup of $G$ is normal.

(4) Every proper subset of $G$ is properly contained in its normalizer.

(5) $G$ is the direct product of its Sylow subgroups.

In particular, $S_3$ is not nilpotent. It is solvable though.