# Finite commutative p group

1. Feb 26, 2014

### jimmycricket

1. The problem statement, all variables and given/known data

Let $A = A(p)\times A'$ where $A(p)$ is a finite commutative p-group (i.e the group has order $p^a$ for $p$ prime and $a>0$) and $A'$ is a finite commutative group whose order is not divisible by $p$.
Prove that all elements of $A$ of orders $p^k, k\geq0$ belong to $A(p)$

3. The attempt at a solution
I dont know where to begin with this. Im quite sure that if the order of $A'$ is not divisible by $p$ then the order of any element of $A'$ is not divisible by $p^k$. Is this usefull or not?

2. Feb 26, 2014

### kduna

Let $e_1, e_2$ denote the identities of $A(p)$ and $A'$ respectively.

If $(a_1, a_2) \in A$, $(a_1, a_2) = (e_1, e_2)$ iff $a_1 = e_1$ and $a_2 = e_2$.

Suppose $(a_1, a_2)$ has order $p^k$ for some $k > 0$. Then it must be that

$a_1^{p^k} = e_1$ and $a_2^{p^k} = e_2$.

But then the order of $a_2$ divides $p^k$. So the order of $a_2$ is $p^m$ for some m. This is a problem if $a_2 \neq e_2$.

Note that in the statement of the problem, they regard $A(p)$ as a subgroup of $A$. But to be proper, $A(p)$ is isomorphic to the subgroup $A(p) \times \{e_2 \}$.

3. Feb 26, 2014

### kduna

Also, in group theory, people tend to stay away from using the word commutative. Instead you should use abelian. The reason for this is that it is nice to have separate words for commuting in multiplication and commuting in addition when you get to ring theory.