1

For a p-primary group $G$, if $\langle y\rangle$ is a cyclic group of order $p^i$ for any natural integer $i$, then $ng \in \langle y\rangle$ implies $g \in \langle y\rangle$ for any $g \in G$?

Also, in the terms of group theory, what does the symbol $G^p$ means for a group $G$?

I wanted to ask this because I was referring to this post:Example of Intersection of Pure Subgroup which is not Pure

Note that $G$ is abelian. Thanks in advance.

Community
  • 1
Seoral
  • 961

1 Answers1

3

The answer to your first question is "no". Take $G=C_p\times C_{p^2}$, a product of a cyclic group of order $p$ and a cyclic group of order $p^2$, and let $y=(1,1)$, which has order $p^2$. Then let $g=(0,1)$. Then $pg = (0,p) = p(1,1) = py$, so $pg\in \langle y\rangle$, but $g\notin \langle y\rangle$.

Added for completeness: As mentioned in the comment below, if you also have that $\gcd(n,p)=1$, however, then the conclusion does follow: since $G$ is a $p$-group, there exists $k$ such that $p^kg = 0$, and from $\gcd(n,p)=1$ we know there exist integers $r,s$ such that $1 = rn + sp^k$. Then, since $ng\in\langle y\rangle$, we have: $$g = 1g = (rn+sp^k)g = (rn)g + (sp^k)g = r(ng) + s(p^kg) = r(ng)+s0 = r(ng)\in\langle y\rangle.$$

Note that the above works in any group: if $H$ is a subgroup of $G$, $g\in G$ is of order $k$, and $g^n\in H$ with $\gcd(n,k)=1$, then $g\in H$, essentially because $\langle g\rangle = \langle g^n\rangle$ if $\gcd(n,k)=1$, so $$g\in\langle g\rangle = \langle g^n\rangle \subseteq H.$$

For your second question: for a group $G$, $G^p$ is the subgroup of $G$ generated by all elements of the form $x^p$ with $x\in G$; that is, the subgroup generated by the $p$th powers of all elements of $G$.

If $G$ is abelian, then this is simply the subgroup that consists of all $p$-powers, since $a^pb^p = (ab)^p$; for nonabelian groups, it may in general be larger.

Your first sentence is not very well-written, by the way: "any" is incorrect there, you mean "some" (otherwise, you are saying that for all natural numbers $i$, the subgroup $\langle y\rangle$ has order $p^i$; this is impossible, since the subgroup can only have one order).

Arturo Magidin
  • 398,050
  • @Arturo Magidin: Thanks for the answer, clarification and the nice counter example to my question above. – Seoral Feb 17 '11 at 09:02
  • @Seoral: Note, however, that if $n$ is relatively prime to $p$, then the conclusion does follow. Because $g$ is order $p^k$ for some $k$, and $\gcd(n,p^k)=1$ implies we can write $1=rn+sp^k$. So $g = (rn+sp^k)g = r(ng) + s(p^kg) = r(ng)+0 = r(ng)\in \langle y\rangle$, as $ng\in\langle y\rangle$. – Arturo Magidin Feb 17 '11 at 14:11
  • @Arturo : You seem to think that the original wording of the question should stay and that it is better to have the clarification in the answer. Am I correct? – ogerard May 19 '11 at 08:49
  • @ogerard: Well, this was three months ago... It would be fine for the OP to fix it, in which case I would likely remove the final paragraph (if I noticed, granted). My point was to inform Seoral that he did not say what he meant to say, a common problem when using "any". – Arturo Magidin May 19 '11 at 16:10
  • @Arturo: I asked only to better understand etiquette on math.SE. – ogerard May 19 '11 at 16:14
  • @ogerard: It's okay to fix obvious mistakes or grammar/spelling. In this case, this is a subtle logical issue that it's probably best if the OP recognizes it and keeps an eye out for in the future. – Arturo Magidin May 19 '11 at 16:15