# Prove that locally finite nilpotent group is direct product of its normal maximal $p$- subgroups

Definition- A group $G$ is said to be locally finite if every finite subset of $G$ generates a finite subgroup.

Now I have to prove the following proposition.

Proposition- Let $G$ be a locally finite nilpotent group. Then $G=\times\ O_p$ where $O_p$ is the normal maximal $p$ - subgroup of $G$ and direct product is taken over all primes $p$.

I have encountered locally finite groups for first time and any help on this proposition is appreciated.

I know that for finite nilpotent groups, $G$ can be expressed as direct product of nontrivial sylow subgroups, but here in locally finite nilpotent, sylow subgroups (maximal p-groups) can be infinite too.

If you can give a reference of this proof, it will be a huge help. Thanks.

• By the way, "locally finite nilpotent" is ambiguous. It could mean either locally finite and nilpotent, or locally finite and locally nilpotent. – Derek Holt Feb 1 '16 at 9:08

Since it is locally finite nilpotent, for any prime $p$, any finite collection of $p$-elements generate a $p$-group. But that immediately implies that the set of all $p$-elements in $G$ generate a $p$-group $O_p$, which is clearly a maximal $p$-group and normal in $G$.
$G$ local finite nilpotent also implies that, for distinct primes $p$ and $q$, all $p$-elements commute with all $q$-elements, so $O_p$ and $O_q$ commute. Also every element of $G$ can be written uniquely as a product of elements from the different $O_p$, so $G$ is equal to their direct product.
• Sorry, but this is not immediately clear to me that why "But that immediately implies that the set of all $p$-elements in $G$ generate a $p$-group $O_p$, which is clearly a maximal $p$-group and normal in $G$." – Bhaskar Vashishth Mar 15 '16 at 1:18
• set of all $p$-elements in $G$ generate a $p$-group $O_p$ – Bhaskar Vashishth Mar 28 '16 at 20:57
• The fact that $G$ is locally finite nilpotent implies that, if $x_1,\ldots,x_n$ are $p$-elements then so is their product $x_1x_2 \cdots x_n$. So the set of all $p$-elements generates a $p$-group. – Derek Holt Mar 28 '16 at 22:06