Proving $G_p$ is a Subgroup in Advanced Modern Algebra

[Artusartos]

Homework Statement

From Advanced Modern Algebra (Rotman):

Definition Let p be a prime. An abelian group G is p-primary if, for each $a \in G$, there is $n \geq 1$ with $p^na=0$. If we do not want to specify the prime p, we merely say that $G$ is primary.

If $G$ is any abelian group, then its p-primary component is $G_p = \{a \in G : p^na=0 for some n\geq 1\}$.

I was trying to prove that $G_p$ is a subgroup.

Homework Equations

The Attempt at a Solution

If $a$ and $b$ are in $G_p$, with $p^ka=0$ and $p^na=0$, then $p^{k+n}ab \in G_p$.

But now I need to show that if $a \in G_p$ then $a^{-1} \in G_p$. If the operation is addition, then of course it is true, since $\underbrace{a+...+a}_{p^k\text{ times}} = 0 \implies (-1)(\underbrace{a+...+a}_{p^k\text{ times}})=0$. But what if its any arbitrary operation?

 

[micromass]

I'm a bit confused. Since you're working in an abelian group, and since you denote the identity by $0$, it seems to me that you denote the group operation by $+$.

But here:

If $a$ and $b$ are in $G_p$, with $p^ka=0$ and $p^na=0$, then $p^{k+n}ab \in G_p$.

your group operation is multiplication. I don't think this can be right. I think that you need to prove that if $a,b\in G_p$, then there is some $n$ such that $p^n (a+b) = 0$.

 

[Artusartos]

I'm a bit confused. Since you're working in an abelian group, and since you denote the identity by $0$, it seems to me that you denote the group operation by $+$.

But here:



your group operation is multiplication. I don't think this can be right. I think that you need to prove that if $a,b\in G_p$, then there is some $n$ such that $p^n (a+b) = 0$.

Thanks. I was thinking that we had to prove this for any arbitrary group, regardless of operation. Usually, people tend use multiplication as the operation if the operation is unknown, right? That was the reason why I proved the first part that way, not necessarily because I had the operation of multiplication in my mind.

However, I got confused when I got to the second part...and as I said it is easy to prove it when the operation is addition, but what if it is not?

In any case, I'm assuming that the definition actually did mean addition...because (as you said), they write 0 for the identity element, right?

So if $a$ and $b$ are in $G_p$, then $p^{k+n}(a+b) = 0$, so $a+b \in G_P$, right?

 

[micromass]

Thanks. I was thinking that we had to prove this for any arbitrary group, regardless of operation. Usually, people tend use multiplication as the operation if the operation is unknown, right? That was the reason why I proved the first part that way, not necessarily because I had the operation of multiplication in my mind.

However, I got confused when I got to the second part...and as I said it is easy to prove it when the operation is addition, but what if it is not?

In any case, I'm assuming that the definition actually did mean addition...because (as you said), they write 0 for the identity element, right?

The general groups are always written using multiplication. But here we are dealing with an abelian group and they are usually written with addition notation. It's just a notation anyway.

So if $a$ and $b$ are in $G_p$, then $p^{k+n}(a+b) = 0$, so $a+b \in G_P$, right?

Seems right.