Quotient Groups: relating properties of H to properties of G/H

Question

Q: If $G$ is an abelian group, let $H_p$ be the set of all $x\in G$ whose order is a power of p. Prove that $H_p$ is a subgroup of G.

Furthermore, Prove that $G/H_p$ has no elements whose order is a nonzero power of $p$.

A: By Hypothesis we have $H_p = \{x \in G : x^p = e \}$ and it's easy to show that $H_p < G$.

Now, I have a hard time convincing myself of the second part of the statement.

Let's suppose that $xH_p \in G/H_p$ does have an order of nonzero power of $p$, then $(xH)^p = H_p$, it follows that $(xH)^p = x^pH_P = H_p$ then we have $x^p \in H_p$, and since $x^p \in H_p$ then $(x^p)^p = e$ this implies that the only element of $G/H_p$ of order nonzero of $p$ is $H_p$ itself.

Is the above reasoning correct?

Thanks for the Kind Help!

Answer 1

In general the theorem as stated is false. I don't know if you've done Sylow theory, but if $G$ has more than one $p$-Sylow subgroup then $H_p$ is not closed under the operation, since it contains every $p$-Sylow subgroup of $G$ and these are the maximal subgroups of $G$ with every element having order a power of $p$. If you assume $G$ is abelian (and finite) then the theorem holds.

Answer 2

In response to the second part of your question: suppose that $G/H$ contains an element $x$ of order $p$ (which is equivalent to it containing an element of order a nonzero power of $p$). Let $f: G \to G/H$ be the quotient map: then $f(y) = x$, where $y \in G - H$ has order $m*p$ for some positive integer $m$. This is because the order of the image of an element under a homomorphism divides the order of the element. We may assume that $m$ is coprime to $p$ (Why?). Then $y^m \in H$, but that means $x^m = e$ in $G/H$. This is a contradiction because the order of $x$ does not divide $m$.

This theorem is true even if we only assume that $G$ has a unique $p$-Sylow subgroup, which is a weaker condition than being abelian. This assumption is equivalent to $H$ being a subgroup.