# Direct product of Sylow subgroups

Proposition II.7.5 of Hungerford's Algebra goes

Proposition 7.5. A finite group is nilpotent if and only if it is the direct product of its Sylow subgroups.

Let $G = (\mathbb{Z}_6, +)$. $G$ is abelian, so it is nilpotent and thus by the proposition is the direct product of its Sylow subgroups.

The Sylow subgroups of $G$ are $H = (\mathbb{Z}_3, +)$ and $K = (\mathbb{Z}_2, +)$, so the proposition says $G = H \times K$. In particular, $0 \in H$ and $0 \in K$, so $(0,0) \in H \times K = G$... but that's not true, since $G = \{0,1,2,3,4,5\}$.

Where is/are my misunderstanding(s)?

• Can you explain why you think it isn't true? Is it because you think the notation (0,0) does not literally appear in the list 0,1,2,3,4,5? – anon Nov 7 '13 at 5:46
• Yes, that's why. – Rainbow Nov 7 '13 at 5:52
• When it is written "it is the direct product" it really means "it is isomorphic to the direct product", and in your case the isomorphism is given by sending $(0,1)$ to $1$. – zarathustra Nov 7 '13 at 6:04
• @Antoine Actually the isomorphism $\Bbb Z/3\Bbb Z\times\Bbb Z/2\Bbb Z\cong\Bbb Z/6\Bbb Z$ sends $(0,1)$ to $3$ and $(1,1)$ to $1$. – anon Nov 7 '13 at 6:16
• @anon: of course, I should respect the policy to not post comments before 7am. Thank you for reporting :) – zarathustra Nov 7 '13 at 11:57

When we say a group "is a direct product," it suffices to simply interpret this as saying the group is isomorphic to said direct product. However it is important to understand the difference between thinking about direct products internally as opposed to externally.

There is a psychological difference between an internal and an external direct product.

• In an external direct product $H\times K$, the elements look like tuples $(h,k)$ with $h\in H$ and $k\in K$. Thus one may say that anything that is not a tuple of this form is not in $H\times K$.
• An internal direct product is structurally a direct product, but the elements may or may not literally be in the form of tuples. More on this now.

Let's look at how the internal direct product is motivated. We see that there is a "copy" of $H$ sitting inside the direct product $H\times K$ as $H\times 1=\{(h,1):h\in H\}$. Similarly there is a "copy" of $K$ sitting inside as $1\times K$. (If the groups are not written multiplicatively then the numeral "$1$" might be misleading.) Everything in $H\times K$ can be written as a product of something in $H\times1$ times something in $1\times K$ (in particular $(h,k)=(h,1)(1,k)$). The two subgroups $H\times1$ and $1\times K$ intersect trivially as $(1,1)$. And everything in $H\times1$ commutes with everything in $K\times1$.

And thus we have our definition. We say $G$ is a direct product of subgroups $H$ and $K$ if:

• $G=HK$, i.e. every element $g\in G$ can be written as $g=hk$ with $h\in H$ and $k\in K$;
• $H\cap K=1$, i.e. $H$ and $K$ intersect trivially;
• $[H,K]=1$, i.e. everything in $H$ commutes with everything in $K$.

Fact: If $H$ and $K$ are two groups, then the external direct product $H\times K$ is an internal direct product of its subgroups $H\times1$ and $1\times K$ (which are canonincal copies of $H$ and $K$ sitting inside $H\times K$). Conversely, if $G$ is an internal direct product of subgroups $H$ and $K$, then there is an isomorphism $H\times K\cong G$ given by $(h,k)\mapsto hk$ between the external direct product $H\times K$ and the original group $G$ (which, remember, is an internal direct product of $H$ and $K$).

This tells us that internal and external direct products are actually the same (the are isomorphic), but the elements inside might look different. In my opinion the best way to think about the difference is as follows: a direct product is external if it was constructed that way to begin with, and it is internal if it was only discovered that way after analyzing it. If I were to, say, merely relabel the elements of an external direct product while keeping the whole group structure the exact same, the concept of internal direct product allows us to still recognize the group as a direct product, even if the elements no longer look like tuples.

In particular, $\Bbb Z/6\Bbb Z$ is an internal direct product of its subgroups $2\Bbb Z/6\Bbb Z$ and $3\Bbb Z/6\Bbb Z$ (which are isomorphic to $\Bbb Z/3\Bbb Z$ and $\Bbb Z/2\Bbb Z$ respectively). More generally we have an isomorphism

$$\frac{\Bbb Z}{p_1^{e_1}\cdots p_r^{e_r}\Bbb Z}\cong\frac{\Bbb Z}{p_1^{e_1}\Bbb Z}\times\cdots\times\frac{\Bbb Z}{p_1^{e_1}\Bbb Z}.$$

This is the content of the abstract version of Sun-Ze (better known as the Chinese remainder theorem). It is recommended that students become familiar with the structure theory of abelian groups and cyclic groups in particular as well as the basic construction of direct products before dwelling on more advanced topics like subgroup series, Sylow subgroups and nilpotence.

• This was a great explanation. Thank you. – Rainbow Nov 14 '13 at 1:16