$p$-component of finite abelian groups On Wikipedia it is stated that for finite abelian groups, p-component is used  to mean the Sylow p-subgroup, so the abelian group is the product of its p-components for primes p. Since every group can have several Sylow $p$-subgroups I am wondering whether the product will contain multiple factors of the same $p$-subgroup. By that I mean if $G$ has two Sylow $p$-subgroups and two Sylow $q$-sugroups will 
$$ G = H_p \times H'_p \times H_q \times H'_q$$
or will it be
$$ G = H_p \times H_q $$
The latter seems more plausible but I'm just not sure how to work it out and how to think about it. Any help will be greatly appreciated. 
 A: In an abelian group, we know that $ab=ba$ for every $a,b\in G$, and in particular, this means that $a^{-1}ba=b$ for every $a,b\in G$.  Thus for any subgroup $S$ and any $x\in G$, we have $x^{-1}Sx=S$.
The second Sylow theorem tells us that all Sylow $p$-subgroups are conjugate, that is, if $\operatorname{Syl}_p(G)$ is the set of Sylow $p$-subgroups of $G$, then for each pair $S,T\in\operatorname{Syl}_p(G)$ there exists an $x\in G$ such that $x^{-1}Sx=T$.
But since we know $x^{-1}Sx=S$ always, this means that $T=S$.  In other words, in an abelian group, there can be only one Sylow $p$-subgroup of every $p$ dividing $|G|$.
So you might ask why "$p$-component" is a term reserved for abelian groups: the answer is that, in fact, this is not strictly true.  $p$-components can also be defined for nilpotent groups.  There are several equivalent definitions for a nilpotent group, but the relevant one to this context is any group which has exactly one Sylow $p$-subgroup for every $p$ dividing $|G|$.
A: If $G$ is an Abelian group and $m\in\mathbb{N}$, $n>0$, define
$$
G[m]=\{\,x\in G: x^m=1\,\}.
$$
which is easily seen to be a subgroup of $G$. Note that, if $m>1$ divides $|G|$, then $G[m]$ is not trivial, because there is a prime $p$ dividing $m$ and $G$ has an element of order $p$ for each prime $p$ dividing $|G|$ (this is the only part of the Sylow theorems, known as Cauchy's theorem, we need).
If $|G|=mn$ with $\gcd(m,n)=1$, then $G=G[m]\times G[n]$. Indeed, if $am+bn=1$ for integers $a$ and $b$, you can write, for any $x\in G$,
$$
x=x^1=x^{an}x^{bm}
$$
where $x^{an}\in G[m]$ and $x^{bm}\in G[n]$. Moreover $G[m]\cap G[n]=\{1\}$ because, if $x\in G[m]\cap G[n]$, you have
$$
x=x^1=(x^m)^a(x^n)^b=1^a1^b=1.
$$
Now we can proceed by induction to show that $G$ is the product of its $p$-components
$$
G_p=\{\,x\in G:x^{p^k}=1\text{ for some }k>0\,\}
=\bigcup_{k>0}G[p^k]
$$
($p$ a prime).
The case $|G|=1$ is obvious. So we can assume $|G|>1$ and that the result is true for all groups having less elements than $G$. If $|G|$ is a prime power, there's nothing to prove. Otherwise $|G|=mn$, with $\gcd(m,n)=1$, $m>1$, $n>1$. So we can apply the induction hypothesis to $G[m]$ and $G[n]$, noting that the set of primes dividing $m$ and $n$ are disjoint.
