Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $p$ be a prime number, let $A$ be an abelian group, define $A(p)$ to be the subgroup of all elements that have power of $p$ order, i.e. $x \in A(p) \iff x\in A \ \wedge \ p^ex = 0$, in additive group notation. Note that for some $e$, $p^ex = 0 \iff p^{ord(x)}x = 0$, so that statement is correct. Now I want to prove that if $A(p)$ is finite then it's a $p$-group in the most general setting, so I was thinking: if $A$ is any finte group then there is an element of order $|A|$, but that's not true since not every finite group is cyclic. I have no idea what makes $A(p)$ special so that it's a $p$-group when finite.

For each $x \in A(p)$, $ \ x \mathbb{Z}$ is a finite cyclic of order $p^e$ where $e = ord(x)$ in $A(p)$. We can write $A(p) = x_1\mathbb{Z} + \dots + x_n\mathbb{Z}$, where $x_i$ runs through the nonzero elements of $A(p)$. Let's try to create a basis. If $k_1 x_1 + \dots + k_n x_n = 0$, then $k_n x_n = $ sum of rest of terms. If $x_n$ has order $p^e$, then without loss of generality, let $k_n \lt p^e$, since if not, then $k_n$ can be reduced to that since $(mp^e + k_n')x_n = 0 + k_n' x_n$ where $k_n'\lt p^e$. Almost there, still working on it...

Edit

Thanks, TBrendle. There are two definitions of $p$-group and they're equivalent. A finite group is a $p$-group iff all of its elements have $p$-power order. One direction is obvious. To show the converse, if $|A| = qm$ for any prime $q$ other than $p$ then it has an element of order $q$, by another theorem. Thus $|A| = p^e$ for some $e$.

share|improve this question
    
Do you know the definition of "$p$-group"? –  Zev Chonoles Sep 28 '13 at 21:49
    
Finite group with $p$ power size. –  Enjoys Math Sep 28 '13 at 21:51
    
I guess $A(p)$ has a basis too since it's finite? –  Enjoys Math Sep 28 '13 at 21:55
    
Let me try to prove this for a sec before answering. –  Enjoys Math Sep 28 '13 at 21:55
    
It seems like you're still confused about bases. As we just discussed on your earlier question, any torsion abelian group (other than the trivial group) does not have a basis. Therefore, unless $A(p)$ is trivial, it does not have a basis. This is regardless of it being finite or infinite. –  Zev Chonoles Sep 28 '13 at 23:21

1 Answer 1

up vote 4 down vote accepted

A $p$-group is usually defined as a group in which all elements have $p$-power order, so this is true by definition.

If you are working with an alternative definition where $p$-group means a finite group whose order is a power of $p$: If $A(p)$ is not a $p$-group then by Cauchy's theorem there is some element in $A(p)$ whose order is a prime $q$ with $q \neq p$, but this is a contradiction to the definition of $A(p)$.

If you are having trouble with this problem, you may not understand the definitions involved.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.