What is $pA$ when $p$ is a prime number and $A$ an abelian group?

Question

Let $A$ be a finite abelian $p$-group. I want to prove that $pA$ is also an abelian finite $p$-group, of order strictly less than the order of $A$.

The problem is that I don't even know what does the notation $pA$ mean. My guess is $$pA=\left\{pa\ |\ a\in A\right\}$$ If this is correct, then $pA$ is certainly a finite abelian group, and each element $pa$ of $pA$ has period $p^{r-1}$, if $p^r$ is the period of $a$. But how can I deduce that the order of $pA$ is strictly less than the order of $A$?

Answer 1

Your definition for $pA$ is correct (at least it makes a lot of sense, considering what we want to prove)

Here is a short proof for all the problem.

Notice $f:A\rightarrow A$ defined by $f(a)=pa$ is a homomorphism ( we need commutativity for this), the image is $pA$, so it is automatically an abelian subgroup. Now notice that by Cauchy's theorem there is $x$ so that $x$ has order $p$. So $f(x)=e$ and $f$ is not injective, since the group is finite $f$ is not surjective, and so $pA\neq A$ as desired.

Answer 2

Your guess is correct. To show that $pA$ is smaller than $A$, consider the map $f:A\to pA$ given by $f(a)=pa$. Is this map injective? Is it surjective?